a solution that has 1.40% #C_2H_5NH_2# by mass and also 1.18% #C_2H_5NH_3Br# by mass

a solution that is 12.5 g of #HC_2H_3O_2# and also 14.0 g of #NaC_2H_3O_2# in 150.0 mL of solution? 1. The basic buffer

The chemical equilibrium for a simple buffer is

#"B + H"_2"O" ⇌ "BH"^+ + "OH"^"-"#

The Henderson-Hasselbalch equation for this buffer is

#color(blue)(|bar(ul(color(white)(a/a) "pOH" = "p"K_"b" + log((<"BH"^+>)/(<"B">))color(white)(a/a)|)))" "#

For ethylamine, (#"B"#), #"p"K_"b" = 3.19#.

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Since both #"B"# and #"BH"^+# are in the very same solution, the ratio of their molarities is the very same as the proportion of their moles.

Assume that we have actually 1000 g of solution.

Then we have actually 14.0 g #"CH"_3"CH"_2"NH"_2 ("B")# and also 11.8 g #"CH"_3"CH"_2"NH"_3"Br" ("BH"^+)#.

#"Moles of B" = 14.0 color(red)(cancel(color(black)("g B"))) × "1 mol B"/(45.08 color(red)(cancel(color(black)("g B")))) = "0.3106 mol B"#

#"Moles BH"^+ = 11.8 color(red)(cancel(color(black)("g BH"^+))) × ("1 mol BH"^+)/(126.00 color(red)(cancel(color(black)("g BH"^+)))) = "0.093 65 mol BH"^+#

#"pOH" = "p"K_b + log((<"BH"^+>)/(<"B">)) = 3.19 + log(("0.093 65" color(red)(cancel(color(black)("mol"))))/(0.3106 color(red)(cancel(color(black)("mol")))))=3.19 + log(0.3015) = "3.19 – 0.52" = 2.67#

#"pH" = "14.00 – pOH" = "14.00 – 2.67" = 11.33#

2. The acidic buffer

The chemical equilibrium for an acidic buffer is

#"HA" + "H"_2"O" ⇌ "H"_3"O"^+ +"A"^"-"#

The Henderson-Hasselbalch equation for this buffer is

#color(blue)(|bar(ul(color(white)(a/a) "pH" = "p"K_"a" + log((<"A"^"-">)/(<"HA">))color(white)(a/a)|)))" "#

For acetic acid (#"HA"#), #"p"K_"a" = 4.75#.

Due to the fact that both #"HA"# and #"A"^"-"# are in the exact same solution, the proportion of their molarities is the very same as the ratio of their moles.

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We have actually 12.5 g #"HC"_2"H"_3"O"_2 ("HA")# and also 14.0 g #"NaC"_2"H"_3"O"_2 ("A"^"-")#.

#"Moles of HA" = 12.5 color(red)(cancel(color(black)("g HA"))) × "1 mol HA"/(60.05 color(red)(cancel(color(black)("g HA")))) = "0.2082 mol HA"#

#"Moles of A"^"-"=14.0 color(red)(cancel(color(black)("g A"^"-"))) × (1 "mol A"^"-")/(82.03 color(red)(cancel(color(black)("g A"^"-")))) = "0.1707 mol A"^"-"#

#"pH" = "p"K_"a" + log((<"A"^"-">)/(<"HA">)) = 4.75 + log((0.1707 color(red)(cancel(color(black)("mol"))))/(0.2082 color(red)(cancel(color(black)("mol"))))) = 4.75 + log(0.8199) = "4.75 – 0.09" = 4.66#