1. The pH of a 9.75×10 -2 M solution of a weak monoprotic acid is 1.40. Calculat

Question

1. The pH of a 9.75×10-2 M solution of a weak monoprotic acid is 1.40. Calculate the percent dissociation of the acid to 3 significant figures.

2. The pH of a 5.19×10-2 M solution of a weak base that will accept only one proton is 8.48. Calculate the percent dissociation of the base to 3 significant figures.

3.Calculate the pH (to two decimal places) of a 7.98×10-3 M solution of sodium dihydrogen citrate. A table of pKa values can be found here. If the 5% approximation is valid, use the assumption to compute pH.

Explanation / Answer

1. The pH of a 9.75×10-2 M solution of a weak monoprotic acid is 1.40. Calculate the percent dissociation of the acid to 3 significant figures.

Ans : The pH = -log[H+]

so [H+] = 0.0398

let acid is HA

HA ---> H + A-

[H+] = Degree of dissociation X [HA]

Degree of dissociation = 0.0398 / 0.0975 = 0.408

2. The pH of a 5.19×10-2 M solution of a weak base that will accept only one proton is 8.48. Calculate the percent dissociation of the base to 3 significant figures.

2) The pH = 8.48

So pOH = 14-8.48 = 5.52

So [OH-] = 3.02 X 10^-6

Degree of dissocation = [OH-] / [Base] = 3.02 X 10^-6 / 5.19 X 10^-2 = 5.81 X 10^-3

3.Calculate the pH (to two decimal places) of a 7.98×10-3 M solution of sodium dihydrogen citrate. A table of pKa values can be found here. If the 5% approximation is valid, use the assumption to compute pH.

Ans: [Salt] = 7.98 X 10^-3

pH = ?

Ka = 7.4 X 10^-4

NaCA + H2O --> CAH + NaOH

Kb = [Citric acid] [OH-] / [Salt]

Kb = Kw / Ka = 10^-14 / 7.4 X 10^-4 = 1.35 X 10^-11

1.35 X 10^-11 = x^2 / 7.98 X 10^-3 - x

x <<1

1.35 X 10^-11 = x^2 / 7.98 X 10^-3
x^2 = 10.77 X 10^-14

x = 3.28 X 10^-7 = [OH-]

pOH = - log3.28 X 10^-7 = 6.48

pH = 14 - 6.48 =7.52

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