Learning Outcomes

Assess the family member staminas of acids and bases according to their ionization constantsRationalize trends in acid–base toughness in relation to molecular structureCarry out equilibrium calculations for weak acid–base systems

The relative toughness of an acid or base is the level to which it ionizes when liquified in water. If the ionization reaction is essentially complete, the acid or base is termed strong; if relatively little bit ionization occurs, the acid or base is weak. As will certainly be obvious throughout the remainder of this chapter, tbelow are many kind of even more weak acids and bases than strong ones. The the majority of widespread solid acids and bases are provided in Table 1.

You are watching: Indicate, by letter, the strongest acid and the weakest acid of those below.

Table 1. Some Common Strong acids and also Strong BasesStrong AcidsStrong Bases
HClO4 perchloric acidLiOH lithium hydroxide
HCl hydrochloric acidNaOH sodium hydroxide
HBr hydrobromic acidKOH potassium hydroxide
HI hydroiodic acidCa(OH)2 calcium hydroxide
HNO3 nitric acidSr(OH)2 strontium hydroxide
H2SO4 sulfuric acidBa(OH)2 barium hydroxide

The family member toughness of acids might be quantified by measuring their equilibrium constants in aqueous remedies. In services of the same concentration, stronger acids ionize to a greater degree, and so yield greater concentrations of hydronium ions than execute weaker acids. The equilibrium continuous for an acid is referred to as the acid-ionization constant, Ka. For the reaction of an acid HA:
extHAleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extA^-left(aq ight)

the acid ionization continuous is written

K_ exta=dfracleft< extH_3 extO^+ ight>left< extA^- ight> ext

wbelow the concentrations are those at equilibrium. Although water is a reactant in the reaction, it is the solvent as well, so we perform not include in the equation. The bigger the Ka of an acid, the bigger the concentration of extH_3 extO^+and also A− relative to the concentration of the nonionized acid, HA, in an equilibrium mixture, and the stronger the acid. An acid is classified as “strong” once it undergoes finish ionization, in which situation the concentration of HA is zero and the acid ionization consistent is immeasurably huge (Ka ≈ ∞). Acids that are partially ionized are referred to as “weak,” and their acid ionization constants might be experimentally measured. A table of ionization constants for weak acids is offered in Ionization Constants of Weak Acids.

To illustrate this idea, three acid ionization equations and Ka values are shown below. The ionization constants rise from initially to last of the noted equations, indicating the relative acid strength increases in the order extCH_3 extCO_2 extH −> = ). Unlike the Ka value, the percent ionization of a weak acid varies through the initial concentration of acid, frequently decreasing as concentration increases. Equilibrium calculations of the sort described later in this chapter have the right to be offered to confirm this actions.


Example 1: Calculation of Percent Ionization from pH

Calculate the percent ionization of a 0.125-M solution of nitrous acid (a weak acid), with a pH of 2.09.


The percent ionization for an acid is:

dfracleft< extH_3 extO^+ ight>_ exteqleft< extHNO_2 ight>_0 imes 100

The chemical equation for the dissociation of the nitrous acid is: extHNO_2left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extNO_2^-left(aq ight)+ extH_3 extO^+left(aq ight). Since 10−pH = left< extH_3 extO^+ ight>, we discover that 10−2.09 = 8.1 × 10−3M, so that percent ionization is:

dfrac8.1 imes 10^-30.125 imes 100=6.5\%

Remember, the logarithm 2.09 indicates a hydronium ion concentration through just two considerable numbers.


Check Your Learning

Calculate the percent ionization of a 0.10-M solution of acetic acid through a pH of 2.89.


As we did with acids, we deserve to meacertain the relative staminas of bases by measuring their base-ionization continuous, (Kb) in aqueous services. In options of the exact same concentration, more powerful bases ionize to a greater extent, and so yield greater hydroxide ion concentrations than perform weaker bases. A stronger base has actually a larger ionization consistent than does a weaker base. For the reactivity of a base, B:

extBleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extHB^+left(aq ight)+ extOH^-left(aq ight),

we write the equation for the ionization constant as:

K_ extb=dfracleft< extHB^+ ight>left< extOH^- ight>left< extB ight>

where the concentrations are those at equilibrium. Aacquire, we execute not incorporate in the equation bereason water is the solvent. The chemical reactions and also ionization constants of the 3 bases displayed are:

extNO_2^-left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extHNO_2left(aq ight)+ extOH^-left(aq ight)qquadK_ extb=2.22 imes 10^-11

extCH_3 extCO_2^-left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extCH_3 extCO_2 extHleft(aq ight)+ extOH^-left(aq ight)qquadK_ extb=5.6 imes 10^-10

extNH_3left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extNH_4^+left(aq ight)+ extOH^-left(aq ight)qquadK_ extb=1.8 imes 10^-5

A table of ionization constants of weak bases appears in Ionization Constants of Weak Bases. As for acids, the family member stamina of a base is likewise reflected in its percent ionization, computed as

\% ext ionization=< extOH^->_eq/< extB>_0 imes100\%

however will differ depending upon the base ionization continuous and the initial concentration of the solution.

Relative Strengths of Conjugate Acid-Base Pairs

Brønsted-Lowry acid-base chemistry is the deliver of protons; thus, logic suggests a relation between the family member toughness of conjugate acid-base pairs. The toughness of an acid or base is quantified in its ionization continuous, Ka or Kb, which represents the level of the acid or base ionization reactivity. For the conjugate acid-base pair HA / A−, ionization equilibrium equations and ionization continuous expressions are

extHAleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extA^-left(aq ight)qquadK_ exta=dfracleft< extH_3 extO^+ ight>left< extA^- ight>left< extHA ight>

extA^-left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extOH^-left(aq ight)+ extHAleft(aq ight)qquadK_ extb=dfracleft< extHA ight>left< extOH ight>left< extA^- ight>

Adding these 2 chemical equations returns the equation for the autoionization for water:

cancel extHAleft(aq ight)+ extH_2 extOleft(l ight)+cancel extA^-left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+cancel extA^-left(aq ight)+ extOH^-left(aq ight)+cancel extHAleft(aq ight)

ext2H_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extOH^-left(aq ight)

As debated in an additional chapter on equilibrium, the equilibrium continuous for a summed reaction is equal to the mathematical product of the equilibrium constants for the added reactions, and also so

K_ exta imes K_ extb=dfracleft< extH_3 extO^+ ight>left< extA^- ight> ext imes dfrac extleft< extOH^- ight>left< extA^- ight>=left< extH_3 extO^+ ight>left< extOH^- ight>=K_ extw

This equation states the relation between ionization constants for any kind of conjugate acid-base pair, namely, their mathematical product is equal to the ion product of water, Kw. By rearvarying this equation, a reciprocal relation between the strengths of a conjugate acid-base pair becomes evident:

K_ exta=K_ extw/K_ extbqquad ext or qquadK_ extb=K_ extw/K_ exta

The inverse proportional relation between Ka and Kb means the stronger the acid or base, the weaker its conjugate partner. Figure 1 illustprices this relation for numerous conjugate acid-base pairs.


Figure 1. This diagram shows the loved one toughness of conjugate acid-base pairs, as suggested by their ionization constants in aqueous solution.


The listing of conjugate acid–base pairs shown in Figure 2 is arranged to present the loved one stamina of each species as compared with water, whose entries are highlighted in each of the table’s columns. In the acid column, those species detailed listed below water are weaker acids than water. These species do not undergo acid ionization in water; they are not Bronsted-Lowry acids. All the species provided over water are stronger acids, delivering proloads to water to some degree when dissolved in an aqueous solution to generate hydronium ions. Species above water however listed below hydronium ion are weak acids, undergoing partial acid ionization, wheres those over hydronium ion are strong acids that are totally ionized in aqueous solution.


Figure 2. The chart shows the family member toughness of conjugate acid-base pairs.


If all these strong acids are completely ionized in water, why does the column indicate they vary in stamina, via nitric acid being the weakest and also perchloric acid the strongest? Notice that the sole acid species current in an aqueous solution of any solid acid is H3O+(aq), meaning that hydronium ion is the strongest acid that may exist in water; any kind of stronger acid will react completely via water to generate hydronium ions. This limit on the acid toughness of solutes in a solution is called a leveling effect. To meacertain the differences in acid strength for “strong” acids, the acids need to be dissolved in a solvent that is much less basic than water. In such solvents, the acids will certainly be “weak,” and also so any differences in the level of their ionization can be figured out. For instance, the binary hydrogen halides HCl, HBr, and HI are solid acids in water but weak acids in ethanol (stamina enhancing HCl a, and so its conjugate base will exhilittle bit a Kb that is basically zero:

strong acid: K_a infty

conjugate base: K_b = K_w/K_a = K_w/infty ≈ 0

A comparable strategy can be used to support the monitoring that conjugate acids of solid bases (Kb ≈ ∞) are of negligible stamina (Ka ≈ 0).


Example 2: Calculating Ionization Constants for Conjugate Acid-Base Pairs

Use the Kb for the nitrite ion, extNO_2^-, to calculate the Ka for its conjugate acid.


Kb for extNO_2^- is given in this section as 2.22 × 10−11. The conjugate acid of extNO_2^- is HNO2; Ka for HNO2 deserve to be calculated making use of the relationship:

K_ exta imes K_ extb=1.0 imes 10^-14=K_ extw

Solving for Ka, we get:

K_ exta=dfracK_ extwK_ extb=dfrac1.0 imes 10^-142.22 imes 10^-11=4.5 imes 10^-4

This answer have the right to be showed by finding the Ka for HNO2 in Ionization Constants of Weak Acids.


Check Your Learning

We deserve to determine the family member acid staminas of extNH_4^+ and HCN by comparing their ionization constants. The ionization constant of HCN is provided in Ionization Constants of Weak Acids as 4 × 10−10. The ionization constant of extNH_4^+ is not noted, however the ionization consistent of its conjugate base, NH3, is provided as 1.8 × 10−5. Determine the ionization consistent of extNH_4^+, and decide which is the stronger acid, HCN or extNH_4^+.


extNH_4^+ is the slightly more powerful acid (Ka for extNH_4^+ = 5.6 × 10−10).

Try It

Explain why the neutralization reactivity of a solid acid and a weak base gives a weakly acidic solution.Explain why the neutralization reactivity of a weak acid and also a strong base provides a weakly fundamental solution.Use this list of vital commercial compounds (and also Figure 2) to answer the following concerns regarding: CaO, Ca(OH)2, CH3CO2H, CO2, HCl, H2CO3, HF, HNO2, HNO3, H3PO4, H2SO4, NH3, NaOH, Na2CO3.Identify the solid Brønsted-Lowry acids and also solid Brønsted-Lowry bases.List those compounds in (a) that deserve to behave as Brønsted-Lowry acids via toughness lying in between those of extH_3 extO^+ and H2O.List those compounds in (a) that can behave as Brønsted-Lowry bases through strengths lying between those of H2O and OH−.Exsimple why the ionization consistent, Ka, for H2SO4 is larger than the ionization constant for H2SO3.Exordinary why the ionization consistent, Ka, for HI is larger than the ionization continuous for HF.What is the ionization constant at 25 °C for the weak acid extCH_3 extNH_3^+, the conjugate acid of the weak base CH3NH2, Kb = 4.4 × 10−4.What is the ionization continuous at 25 °C for the weak acid left( extCH_3 ight)_2 extNH_2^+, the conjugate acid of the weak base (CH3)2NH, Kb = 7.4 × 10−4?

2. The salt ionizes in solution, however the anion slightly reacts with water to create the weak acid. This reaction also creates OH−, which causes the solution to be basic. An instance is NaCN. The CN− reacts with water as follows:

extCN^-left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extHCNleft(aq ight)+ extOH^-left(aq ight)

4. The oxidation state of the sulhair in H2SO4 is better than the oxidation state of the sulfur in H2SO3.

6. Kw = Ka × Kb; hence,

eginarrayrclK_ exta&=&fracK_ extwK_ extb\K_ exta&=&frac1.0 imes 10^-144.4 imes 10^-4=2.3 imes 10^-11endarray


Acid- Base Equilibrium Calculations

The chapter on chemical equilibria introduced numerous forms of equilibrium calculations and the miscellaneous mathematical techniques that are beneficial in perdeveloping them. These strategies are generally valuable for equilibrium systems regardmuch less of chemical reactivity class, and also so they may be effectively applied to acid-base equilibrium problems. This area presents a number of example exercises including equilibrium calculations for acid-base devices.


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Figure 3. pH paper indicates that a 0.l-M solution of HCl (beaker on left) has actually a pH of 1. The acid is fully ionized and = 0.1 M. A 0.1-M solution of CH3CO2H (beaker on right) is has a pH of 3 ( = 0.001 M) because the weak acid CH3CO2H is just partly ionized. In this solution, 3CO2H>. (credit: modification of work by Sahar Atwa)


Table 2. Ionization Constants of Some Weak AcidsIonization ReactionKa at 25 °C
extHSO_4^-+ extH_2 extO ightleftharpoons extH_3 extO^++ extSO_4^2-1.2 × 10−2
extHF+ extH_2 extO ightleftharpoons extH_3 extO^++ extF^-7.2 × 10−4
extHNO_2+ extH_2 extO ightleftharpoons extH_3 extO^++ extNO_2^-4.5 × 10−4
extHNCO+ extH_2 extO ightleftharpoons extH_3 extO^++ extNCO^-3.46 × 10−4
extHCO_2 extH+ extH_2 extO ightleftharpoons extH_3 extO^++ extHCO_2^-1.8 × 10−4
extCH_3 extCO_2 extH+ extH_2 extO ightleftharpoons extH_3 extO^++ extCH_3 extCO_2^-1.8 × 10−5
extHCIO+ extH_2 extO ightleftharpoons extH_3 extO^++ extCIO^-3.5 × 10−8
extHBrO+ extH_2 extO ightleftharpoons extH_3 extO^++ extBrO^-2 × 10−9
extHCN+ extH_2 extO ightleftharpoons extH_3 extO^++ extCN^-4 × 10−10

Table 2 gives the ionization constants for numerous weak acids; additional ionization constants can be discovered in Ionization Constants of Weak Acids.


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Figure 4. pH paper indicates that a 0.1-M solution of NH3 (left) is weakly fundamental. The solution has a pOH of 3 ( = 0.001 M) bereason the weak base NH3 only partly reacts via water. A 0.1-M solution of NaOH (right) has actually a pOH of 1 bereason NaOH is a strong base. (credit: change of work by Sahar Atwa)


The ionization constants of several weak bases are offered in Table 3 and in Ionization Constants of Weak Bases.

Table 3. Ionization Constants of Some Weak BasesIonization ReactionKb at 25 °C
left( extCH_3 ight)_2 extNH+ extH_2 extO ightleftharpoons left( extCH_3 ight)_2 extNH_2^++ extOH^-7.4 × 10−4
extCH_3 extNH_2+ extH_2 extO ightleftharpoons extCH_3 extNH_3^++ extOH^-4.4 × 10−4
left( extCH_3 ight)_3 extN+ extH_2 extO ightleftharpoons left( extCH_3 ight)_3 extNH^++ extOH^-6.3 × 10−5
extNH_3+ extH_2 extO ightleftharpoons extNH_4^++ extOH^-1.8 × 10−5
extC_6 extH_5 extNH_2+ extH_2 extO ightleftharpoons extC_6 extN_5 extNH_3^++ extOH^-4.6 × 10−10

Think about It

Both HF and also HCN ionize in water to a limited level. Which of the conjugate bases, F− or CN−, is the more powerful base? See Table 3.


Try It

Calculate the equilibrium concentration of the nonionized acids and also all ions in a solution that is 0.25 M in HCO2H and 0.10 M in HClO.Calculate the equilibrium concentration of the nonionized acids and also all ions in a solution that is 0.134 M in HNO2 and also 0.120 M in HBrO.Calculate the equilibrium concentration of the nonionized bases and also all ions in a solution that is 0.25 M in CH3NH2 and 0.10 M in C5H5N (Kb = 1.7 × 10−9).Calculate the equilibrium concentration of the nonionized bases and all ions in a solution that is 0.115 M in NH3 and also 0.100 M in C6H5NH2.
Concern 2

The reactions and also equilibrium constants are:

eginarrayl extHNO_2left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extNO_2^-left(aq ight)K_ exta=4.5 imes 10^-4\ extHBrOleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extBrO^-left(aq ight)K_ exta=2 imes 10^-9endarray

As Ka is much bigger for HNO2 than for HBrO, the initially equilibrium will overcome. The equilibrium expression is K_ exta=fracleft< extNO_2^- ight>left< extH_3 extO^+ ight>left< extHNO_2 ight>=4.5 imes 10^-5

The initial and equilibrium concentrations for this mechanism can be written as follows:

Initial concentration (M)Change (M)Equilibrium (M)
0.13400
xxx
0.134 − xxx

Substituting the equilibrium concentrations right into the equilibrium expression and making the assumption that (0.134 − x) ≈ 0.134 gives:

fracleft< extNO_2^- ight>left< extH_3 extO^+ ight>left< extHNO_2 ight>=fracleft(x ight)left(x ight)left(0.134-x ight)approx fracleft(x ight)left(x ight)0.134=4.5 imes 10^-4

Solving for x offers 7.77 × 10−3M. Due to the fact that this value is 5.8% of 0.134, our assumption is incorrect. As such, we need to use the quadratic formula. Using the data gives the complying with equation: x2 + 4.5 × 10−7x − 6.03 × 10−5 = 0

Using the quadratic formula provides (a = 1, b = 4.5 × 10−4, and also c = −6.03 × 10−5)

eginarrayrcl x&=&frac-bpm sqrtb^ ext2+-4ac2a=frac-left(4.5 imes 10^-4 ight)pm sqrtleft(4.5 imes 10^-4 ight)^ ext2+-4left(1 ight)left(-6.03 imes 10^-5 ight)2left(1 ight)\&=&frac-left(4.5 imes 10^-4 ight)pm left(1.55 imes 10^-2 ight)2=7.54 imes 10^-3Mleft( extpositive root ight)endarray

The equilibrium concentrations are therefore:

left< extH_3 extO^+ ight>=left< extNO_2^- ight>=7.54 imes 10^-3=7.5 imes 10^-3M = 0.134 − 7.54 × 10−3 = 0.1264 = 0.126

have the right to be calculated using Kw:

left< extOH^- ight>=fracK_ extwleft< extH_3 extO^+ ight>=frac1.0 imes 10^-147.54 imes 10^-3=1.33 imes 10^-12=1.3 imes 10^-12M

Finally, usage the various other equilibrium to discover the other concentrations. Assume for that (0.120 − x) ≈ 0.124 M.

K_ exta=fracleft< extBrO^- ight>left< extH_3 extO^+ ight>left< extHBrO ight>=2 imes 10^-9approx fracleft< extBrO^- ight>left(7.54 imes 10^-3 ight)0.120

Solving for gives:

= 3.2 × 10−8 = 3.2 × 10−8M = 0.120 − 3.2 × 10−8 = 0.120 M
Inquiry 4

The reactions and also equilibrium constants are:

eginarrayl extNH_3left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extOH^-left(aq ight)+ extNH_4^+left(aq ight)K_ extb=1.8 imes 10^-5\ extC_6 extH_5 extNH_2left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extOH^-left(aq ight)+ extC_6 extH_5 extNH_3^+left(aq ight)K_ extb=4.6 imes 10^-10endarray

As Kb is a lot larger for NH3 than for C6H5NH2, the initially equilibrium will overcome. The equilibrium expression is K_ extb=fracleft< extNH_4^+ ight>left< extOH^- ight>left< extNH_3 ight>=1.8 imes 10^-5

The initial and also equilibrium concentrations for this mechanism have the right to be composed as follows:

Initial concentration (M)Change (M)Equilibrium (M)
0.11500
xxx
0.115 − xxx

Substituting the equilibrium concentrations into the equilibrium expression and also making the assumption that (0.115 − x) ≈ 0.115 gives:

fracleft< extNH_4^+ ight>left< extOH^- ight>left< extNH_3 ight>=fracleft(x ight)left(x ight)left(0.115-x ight)approx fracleft(x ight)left(x ight)0.115=1.8 imes 10^-5

Solving for x provides 1.44 × 10−3M. Since this worth is less than 5% of 0.115 M, our assumption is correct. The equilibrium concentrations are therefore:

= left< extNO_4^+ ight> = 1.44 × 10−3 = 0.0014 M = 0.115 − 0.00144 = 0.1136 = 0.144 M

left< extH_3 extO^+ ight> have the right to be calculated utilizing Kw:

left< extH_3 extO^+ ight>=fracK_ extwleft< extOH^- ight>=frac1.0 imes 10^-140.00144=6.94 imes 10^-12=6.9 imes 10^-12M

Finally, usage the other equilibrium to find the other concentrations. Assume for that (0.100 − x) ≈ 0.100 M:

left< extH_3 extO^+ ight>=fracK_ extwleft< extOH^- ight>=frac1.0 imes 10^-140.00144=6.94 imes 10^-12=6.9 imes 10^-12M

Solving for left< extC_6 extH_5 extNH_3^+ ight> gives:

left< extC_6 extH_5 extNH_3^+ ight> = 3.9 × 10−8M = 0.100 − 3.19 × 10−8 = 0.100 M

Figure 5. Vinegar is a solution of acetic acid, a weak acid. (credit: change of occupational by “HomeSpot HQ”/Flickr)


We are asked to calculate an equilibrium continuous from equilibrium concentrations. At equilibrium, the worth of the equilibrium constant is equal to the reactivity quotient for the reaction:

extCH_3 extCO_2 extHleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extCH_3 extCO_2^-left(aq ight)

K_ exta=dfracleft< extH_3 extO^+ ight>left< extCH_3 extCO_2^- ight>left< extCH_3 extCO_2 extH ight>=dfracleft(0.00118 ight)left(0.00118 ight)0.0787=1.77 imes 10^-5


Check Your Learning

What is the equilibrium consistent for the ionization of the extHSO_4^- ion, the weak acid supplied in some family cleansers:

extHSO_4^-left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extSO_4^2-left(aq ight)

In one mixture of NaHSO4 and also Na2SO4 at equilibrium, left< extH_3 extO^+ ight> = 0.027 M; left< extHSO_4^- ight>=0.29M; and also left< extSO_4^2- ight>=0.13M.


At equilibrium, the value of the equilibrium continuous is equal to the reactivity quotient for the reaction:

extC_8 extH_10 extN_4 extO_2left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extC_8 extH_10 extN_4 extO_2 extH^+left(aq ight)+ extOH^-left(aq ight)K_ extb=dfracleft< extC_8 extH_10 extN_4 extO_2 extH^+ ight>left< extOH^- ight>left< extC_8 extH_10 extN_4 extO_2 ight>=dfracleft(5.0 imes 10^-3 ight)left(2.5 imes 10^-3 ight)0.050=2.5 imes 10^-4


Check Your Learning

What is the equilibrium constant for the ionization of the extHPO_4^2- ion, a weak base:

extHPO_4^2-left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_2 extPO_4^-left(aq ight)+ extOH^-left(aq ight)

In a solution containing a mixture of NaH2PO4 and also Na2HPO4 at equilibrium, = 1.3 × 10−6M; left< extH_2 extPO_4^- ight>=0.042M; and left< extHPO_4^2- ight>=0.341M.


The nitrous acid concentration provided is a formal concentration, one that does not account for any type of chemical equilibria that may be establimelted in solution. Such concentrations are treated as “initial” worths for equilibrium calculations making use of the ICE table strategy. Notice the initial value of hydronium ion is provided as approximately zero because a little concentration of H3O+ is existing (1 × 10−7 M) as a result of the autoprotolysis of water. In many kind of cases, such as all the ones presented in this chapter, this concentration is a lot less than that generated by ionization of the acid (or base) in question and might be neglected.

The pH offered is a logarithmic meacertain of the hydronium ion concentration resulting from the acid ionization of the nitrous acid, and also so it represents an “equilibrium” worth for the ICE table:

left< extH_3 extO^+ ight>=10^-2.34=0.0046M

The ICE table for this mechanism is then:


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Finally, we calculate the worth of the equilibrium continuous utilizing the information in the table:

K_ exta=dfracleft< extH_3 extO^+ ight>left< extNO_2^- ight>left< extHNO_2 ight>=dfracleft(0.0046 ight)left(0.0046 ight)left(0.0470 ight)=4.5 imes 10^-4


Check Your Learning

The pH of a solution of family members ammonia, a 0.950-M solution of NH3, is 11.612. What is Kb for NH3.


Figure 6. The pain of an ant’s sting is caused by formic acid. (credit: John Tann)


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Substituting the equilibrium concentration terms into the Ka expression gives

eginarrayrclK_ exta&=&1.8 imes 10^-4=dfracleft< extH_3 extO^+ ight>left< extHCO_2^- ight>left< extHCO_2 extH ight>\&=&dfracleft(x ight)left(x ight)0.534-x=1.8 imes 10^-4endarray

Due to the fact that the initial concentration of acid is sensibly large and Ka is exceptionally small, we assume that x Finally, the pH is calculated to be

extpH= extlog< extH_3 extO^+>=- extlog(0.0098)=2.01


Check Your Learning

Only a tiny fractivity of a weak acid ionizes in aqueous solution. What is the percent ionization of acetic acid in a 0.100-M solution of acetic acid, CH3CO2H?

extCH_3 extCO_2 extHleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extCH_3 extCO_2^-left(aq ight)qquadK_ exta=1.8 imes 10^-5

(Hint: Determine left< extCH_3 extCO_2^- ight> at equilibrium.) Respeak to that the percent ionization is the fractivity of acetic acid that is ionized × 100, or dfracleft< extCH_3 extCO_2^- ight>left< extCH_3 extCO_2 extH ight>_ extinitial imes 100.


Example 7 mirrors that the concentration of commodities developed by the ionization of a weak base deserve to be identified by the same series of steps offered with a weak acid.


Example 7: Equilibrium Concentrations in a Solution of a Weak Base

Find the concentration of hydroxide ion in a 0.25-M solution of trimethylamine, a weak base:

left( extCH_3 ight)_3 extNleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons left( extCH_3 ight)_3 extNH^+left(aq ight)+ extOH^-left(aq ight)qquadK_ extb=6.3 imes 10^-5


At equilibrium:

K_ extb=dfracleftleft< extOH^- ight>left=dfracleft(x ight)left(x ight)0.25-x=6.3 imes 10^-5

If we assume that x is small family member to 0.25, then we can replace (0.25 − x) in the coming before equation with 0.25. Solving the streamlined equation gives x=4.0 imes 10^-3.

This readjust is less than 5% of the initial concentration (0.25), so the assumption is justified. Respeak to that, for this computation, x is equal to the equilibrium concentration of hydroxide ion in the solution (view earlier tabulation):

left< extOH^- ight>= ext~0+x=x=4.0 imes 10^-3M

=4.0 imes 10^-3M

Then calculate pOH as follows:

extpOH=- extlogleft(4.0 imes 10^-3 ight)=2.40

Using the relation presented in the previous section of this chapter, we get

extpH+ extpOH= extpK_ extw=14.00

which permits the computation of pH:

extpH=14.00- extpOH=14.00 - 2.40=11.60

Tip 3: Check the work

A check of our arithmetic shows that Kb = 6.3 × 10−5.


Check Your Learning

Calculate the hydroxide ion concentration and the percent ionization of a 0.0325-M solution of ammonia, a weak base through a Kb of 1.76 × 10−5.


In some instances, the stamina of the weak acid or base and its formal (initial) concentration lead to an appreciable ionization. Though the ICE strategy continues to be effective for these units, the algebra is a little bit even more affiliated because the simplifying assumption that x is negligible have the right to not be made. Calculations of this kind are demonstrated in Example 8 listed below.


Example 8: Calculating Equilibrium Concentrations without simplifying assumptions

Sodium bisulfate, NaHSO4, is offered in some family members cleansers bereason it consists of the extHSO_4^- ion, a weak acid. What is the pH of a 0.50-M solution of extHSO_4^-?

extHSO_4^-left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extSO_4^2-left(aq ight)qquadK_ exta=1.2 imes 10^-2


As we start resolving for x, we will find this is even more complex than in previous examples. As we comment on these complications we have to not shed track of the truth that it is still the purpose of this step to determine the value of x. At equilibrium:

K_ exta=1.2 imes 10^-2=dfracleft< extH_3 extO^+ ight>left< extSO_4^2- ight>left< extHSO_4^- ight>=dfracleft(x ight)left(x ight)0.50-x

If we assume that x is little and also approximate (0.50 − x) as 0.50, we find x=7.7 imes 10^-2M. When we check the presumption, we calculate:

dfracxleft< extHSO_4^- ight>_ exti

dfracx0.50=dfrac7.7 imes 10^-20.50=0.15left(15\% ight)

The worth of x is not much less than 5% of 0.50, so the assumption is not valid. We require the quadratic formula to uncover x.

K_ exta=1.2 imes 10^-2=dfracleft(x ight)left(x ight)0.50-x

Rearvarying this equation yields

6.0 imes 10^-3-1.2 imes 10^-2x=x^2

Writing the equation in quadratic form gives

x^2+1.2 imes 10^-2x - 6.0 imes 10^-3=0

Solving for the 2 roots of this quadratic equation results in a negative value that might be discarded as physically irpertinent and also a positive value equal to x. As characterized in the ICE table, x is equal to the hydronium concentration.

left< extH_3 extO^+ ight>=~0+x=0+7.2 imes 10^-2M=7.2 imes 10^-2M

extpH=- extlogleft< extH_3 extO^+ ight>=- extlog(7.2 imes 10^-2)=1.14


Check Your LearningCalculate the pH in a 0.010-M solution of caffeine, a weak base: extC_8 extH_10 extN_4 extO_2left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extC_8 extH_10 extN_4 extO_2 extH^+left(aq ight)+ extOH^-left(aq ight)K_ extb=2.5 imes 10^-4

(Hint: It will certainly be vital to convert to left< extH_3 extO^+ ight> or pOH to pH toward the finish of the calculation.)


The reactivity is: extHClOleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extClO^-left(aq ight)

The equilibrium expression is K_ exta=fracleft< extH_3 extO^+ ight>left< extClO^- ight>left< extHClO ight>=3.5 imes 10^-8

The initial and also equilibrium concentrations for this system deserve to be composed as follows:

Initial concentration (M)Change (M)Equilibrium (M)
0.009200
xxx
0.0092 − xxx

Substituting the equilibrium concentrations into the equilibrium expression and also making the presumption that (0.0092 − x) ≈ 0.0092 gives:

fracleft< extH_3 extO^+ ight>left< extClO^- ight>left< extHClO ight>=fracleft(x ight)left(x ight)left(0.0092-x ight)approx fracleft(x ight)left(x ight)0.0092=3.5 imes 10^-8

Solving for x offers 1.79 × 10−5M. This worth is less than 5% of 0.0092, so the presumption that it can be neglected is valid. Hence, the concentrations of solute species at equilibrium are:

left< extH_3 extO^+ ight> = = 1.8 × 10−5M = 0.0092 − 1.79 × 10−5 = 0.00918 = 0.00092 Mleft< extOH^- ight>=fracK_ extwleft< extH_3 extO^+ ight>=frac1.0 imes 10^-141.79 imes 10^-5=5.6 imes 10^-10M

The reactivity is  extC_6 extH_5 extNH_2left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extC_6 extH_5 extNH_3^+left(aq ight)+ extOH^-left(aq ight)

The equilibrium expression is K_ exta=fracleft< extC_6 extH_5 extNH_3^+ ight>left< extOH^- ight>left< extC_6 extH_5 extNH_2 ight>=4.6 imes 10^-10

The initial and equilibrium concentrations for this device can be composed as follows:

Initial concentration (M)Change (M)Equilibrium (M)
0.078400
xxx
0.0784 − xxx

Substituting the equilibrium concentrations into the equilibrium expression and also making the presumption that (0.0784 − x) ≈ 0.0784 gives:

fracleft< extC_6 extH_5 extNH_3^+ ight>left< extOH^- ight>left< extC_6 extH_5 extNH_2 ight>=fracleft(x ight)left(x ight)left(0.0784-x ight)approx fracleft(x ight)left(x ight)0.0784=4.6 imes 10^-10

Solving for x provides 6.01 × 10−6M. This value is less than 5% of 0.0784, so the assumption that it deserve to be neglected is valid. Hence, the concentrations of solute species at equilibrium are:

left< extCH_3 extCO_2^- ight> = = 6.0 × 10−6M = 0.0784 − 6.01 × 10−6 = 0.007839 = 0.00784left< extH_3 extO^+ ight>=fracK_ extwleft< extOH^- ight>=frac1.0 imes 10^-146.01 imes 10^-6=1.7 imes 10^-9M

The reaction is extHCNleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extH_3 extO^+left(aq ight)+ extCN^-left(aq ight).

The equilibrium expression is K_ exta=fracleft< extH_3 extO^+ ight>left< extCN^- ight>left< extHCN ight>=4 imes 10^-10

The initial and equilibrium concentrations for this mechanism deserve to be created as follows:

Initial concentration (M)Change (M)Equilibrium (M)
0.081000
xxx
0.0810 − xxx

Substituting the equilibrium concentrations into the equilibrium expression and also making the assumption that (0.0810 − x) ≈ 0.0810 gives:

fracleft< extH_3 extO^+ ight>left< extCN^- ight>left< extHCN ight>=fracleft(x ight)left(x ight)left(0.0810-x ight)approx fracleft(x ight)left(x ight)0.0810=4 imes 10^-10

Solving for x provides 5.69 × 10−6M. This value is less than 5% of 0.0810, so the assumption that it deserve to be neglected is valid. Therefore, the concentrations of solute species at equilibrium are:

left< extH_3 extO^+ ight> = = 5.7 × 10−6M = 0.0810 − 5.69 × 10−6 = 0.08099 = 0.0810 Mleft< extOH^- ight>=fracK_ extwleft< extH_3 extO^+ ight>=frac1.0 imes 10^-145.69 imes 10^-6=1.8 imes 10^-9M

The reaction is left( extCH_3 ight)_3 extNleft(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons left( extCH_3 ight)_3 extNH^+left(aq ight)+ extOH^-left(aq ight)

The equilibrium expression is K_ extb=fracleftleft< extOH^- ight>left=7.4 imes 10^-5

The initial and also equilibrium concentrations for this mechanism can be written as follows:

<(CH3)3N><(CH3)3NH+>Initial concentration (M)Change (M)Equilibrium (M)
0.1100
xxx
0.11 − xxx

Substituting the equilibrium concentrations right into the equilibrium expression and also making the presumption that (0.11 − x) ≈ 0.11 gives:

fracleftleft< extOH^- ight>left=fracleft(x ight)left(x ight)left(0.11-x ight)approx fracleft(x ight)left(x ight)0.11=7.4 imes 10^-5

Solving for x offers 2.85 × 10−3M. This worth is less than 5% of 0.11, so the assumption that it deserve to be neglected is valid. Therefore, the concentrations of solute species at equilibrium are:

left = = 2.9 × 10−3M<(CH3)3N> = 0.11 − 2.85 × 10−3 = 0.107 = 0.11 Mleft< extH_3 extO^+ ight>=fracK_ extwleft< extOH^- ight>=frac1.0 imes 10^-142.85 imes 10^-3=3.5 imes 10^-12M
Equipment 5

0.120 M extFeleft( extH_2 extO ight)_6^2+ a weak acid, Ka = 1.6 × 10−7


The reactivity is  extFeleft( extH_2 extO ight)_6^2+left(aq ight)+ extH_2 extOleft(l ight) ightleftharpoons extFeleft( extH_2 extO ight)_5left( extOH ight)^+left(aq ight)+ extH_3 extO^+left(aq ight)

The equilibrium expression is K_ exta=fracleft< extFeleft( extH_2 extO ight)_5left( extOH ight)^+ ight>left< extH_3 extO^+ ight>left< extFeleft( extH_2 extO ight)_6^2+ ight>=1.6 imes 10^-7

The initial and also equilibrium concentrations for this device deserve to be created as follows:

Initial concentration (M)Change (M)Equilibrium (M)
0.12000
xxx
0.120 − xxx

Substituting the equilibrium concentrations into the equilibrium expression and also making the presumption that (0.120 − x) ≈ 0.120 gives:

fracleft< extFeleft( extH_2 extO ight)_5left( extOH ight)^+ ight>left< extH_3 extO^+ ight>left< extFeleft( extH_2 extO ight)_6^2+ ight>=fracleft(x ight)left(x ight)left(0.120-x ight)approx fracleft(x ight)left(x ight)0.120=1.6 imes 10^-7

Solving for x gives 1.39 × 10−4M. This value is less than 5% of 0.120, so the assumption that it can be neglected is valid. Hence, the concentrations of solute species at equilibrium are:

left< extFeleft( extH_2 extO ight)_5left( extOH ight)^+ ight> = left< extH_3 extO^+ ight> = 1.4 × 10−4Mleft< extFeleft( extH_2 extO ight)_6^2+ ight> = 0.120 − 1.39 × 10−4 = 0.1199 = 0.120 Mleft< extOH^- ight>=fracK_ extwleft< extH_3 extO^+ ight>=frac1.0 imes 10^-141.39 imes 10^-4=7.2 imes 10^-11M

Effect of Molecular Structure on Acid-Base Strength

In the lack of any leveling impact, the acid strength of binary compounds of hydrogen through nonmetals (A) boosts as the H-A bond toughness decreases dvery own a group in the periodic table. For team 7A, the order of raising acidity is HF 2O 2S 2Se 2Te.

Across a row in the routine table, the acid strength of binary hydrogen compounds rises through boosting electronegativity of the nonsteel atom because the polarity of the H-A bond boosts. Hence, the order of raising acidity (for removal of one proton) across the second row is CH4 3 2O 4 3 2S

Figure 7. As you move from left to best and dvery own the routine table, the acid stamina boosts. As you move from best to left and up, the base stamina boosts.


Ternary Acids and Bases

Compounds containing oxygen and also one or even more hydroxyl (OH) groups deserve to be acidic, standard, or amphoteric, relying on the position in the regular table of the main atom E, the atom bonded to the hydroxyl team. Such compounds have the general formula OnE(OH)m, and also incorporate sulfuric acid, O2S(OH)2, sulfurous acid, OS(OH)2, nitric acid, O2NOH, perchloric acid, O3ClOH, aluminum hydroxide, Al(OH)3, calcium hydroxide, Ca(OH)2, and potassium hydroxide, KOH:

If the central atom, E, has actually a low electronegativity, its attraction for electrons is low. Little tendency exists for the main atom to create a solid covalent bond via the oxygen atom, and also bond a in between the element and oxygen is even more easily broken than bond b between oxygen and hydrogen. Hence bond a is ionic, hydroxide ions are released to the solution, and the material behaves as a base—this is the case via Ca(OH)2 and also KOH. Lower electronegativity is characteristic of the more metallic elements; hence, the metallic facets create ionic hydroxides that are by interpretation standard compounds.

If, on the various other hand, the atom E has actually a relatively high electronegativity, it strongly attracts the electrons it shares through the oxygen atom, making bond a fairly strongly covalent. The oxygen-hydrogen bond, bond b, is thereby weakened because electrons are displaced toward E. Bond b is polar and readily releases hydrogen ions to the solution, so the product behaves as an acid. High electronegativities are characteristic of the more nonmetallic elements. Therefore, nonmetallic facets develop covalent compounds containing acidic −OH groups that are referred to as oxyacids.

Increasing the oxidation variety of the central atom E additionally rises the acidity of an oxyacid bereason this rises the attractivity of E for the electrons it shares through oxygen and also thereby weakens the O-H bond. Sulfuric acid, H2SO4, or O2S(OH)2 (with a sulhair oxidation number of +6), is even more acidic than sulfurous acid, H2SO3, or OS(OH)2 (via a sulfur oxidation number of +4). Likewise nitric acid, HNO3, or O2NOH (N oxidation number = +5), is even more acidic than nitrous acid, HNO2, or ONOH (N oxidation number = +3). In each of these pairs, the oxidation number of the central atom is larger for the stronger acid (Figure 8).


Figure 8. As the oxidation variety of the main atom E increases, the acidity additionally rises.


Try It

Propionic acid, C2H5CO2H (Ka = 1.34 × 10−5), is offered in the manufacture of calcium propionate, a food preservative. What is the hydronium ion concentration in a 0.698-M solution of C2H5CO2H?White vinegar is a 5.0% by mass solution of acetic acid in water. If the thickness of white vinegar is 1.007 g/cm3, what is the pH?The ionization constant of lactic acid, CH3CH(OH)CO2H, an acid discovered in the blood after strenuous exercise, is 1.36 × 10−4. If 20.0 g of lactic acid is offered to make a solution with a volume of 1.00 L, what is the concentration of hydronium ion in the solution?Nicotine, C10H14N2, is a base that will accept 2 prolots (K1 = 7 × 10−7, K2 = 1.4 × 10−11). What is the concentration of each species present in a 0.050-M solution of nicotine?The pH of a 0.20-M solution of HF is 1.92. Determine Ka for HF from these information.The pH of a 0.15-M solution of extHSO_4^- is 1.43. Determine Ka for extHSO_4^- from these data.The pH of a 0.10-M solution of caffeine is 11.16. Determine Kb for caffeine from these data:The pH of a solution of family members ammonia, a 0.950 M solution of NH3, is 11.612. Determine Kb for NH3 from these data.

See more: How To Smoke A Black And Mild : For Dummies, The Smoke Is Small, But Not The Risk


Sexactly how Solution to Inquiry 2




Key Concepts and also Summary

The toughness of Brønsted-Lowry acids and also bases in aqueous services can be determined by their acid or base ionization constants. Stronger acids form weaker conjugate bases, and weaker acids create more powerful conjugate bases. Hence strong acids are totally ionized in aqueous solution bereason their conjugate bases are weaker bases than water. Weak acids are just partially ionized bereason their conjugate bases are strong enough to complete effectively through water for possession of protons. Strong bases react via water to quantitatively create hydroxide ions. Weak bases offer only small amounts of hydroxide ion. The staminas of the binary acids boost from left to right across a period of the regular table (CH4 3 2O 2SO3 2SO4). The toughness of oxyacids also rise as the electronegativity of the central element rises .

Key EquationsK_ exta=fracleft< extH_3 extO^+ ight>left< extA^- ight>left< extHA ight>K_ extb=fracleft< extHB^+ ight>left< extOH^- ight>left< extB ight>Ka × Kb = 1.0 × 10−14 = Kw extPercent ionization=fracleft< extH_3 extO^+ ight>_ exteqleft< extHA ight>_0 imes 100

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