Question

Calculate the pH of each of the solutions and the change in pH to 0.01 pH units caused by adding 10.0 mL of 2.81-M HCl to 340. mL of each of the following solutions. Change is defined as final minus initial, so if the pH drops upon mixing the change is negative.

a) water pH before mixing = Correct: Your answer is correct.

pH after mixing= Incorrect: Your answer is incorrect.

pH change = Incorrect: Your answer is incorrect.

b) 0.171 M C2H3O21-

pH before mixing =

pH after mixing=

pH change =

c) 0.171 M HC2H3O2

pH before mixing =

pH after mixing=

pH change =

d) a buffer solution that is 0.171 M in each C2H3O21- and HC2H3O2

pH before mixing =

pH after mixing=

pH change =

Answer 1

pH calculations,

a) water

pH before mixing = 7

pH after mixing, pH = -log(2.81 M x 10 ml/350 ml) = 1.09

pH change = 1.09 - 7 = -5.91

b) 0.171 M C2H3O2-

hydrolysis of salt,

C2H3O2- + H2O <==> HC2H3O2 + OH-

let x amount hydrolyzed

Kb = 1 x 10^-14/1.8 x 10^-5 = [HC2H3O2][OH-]/[C2H3O2-]

5.55 x 10^-10 = x^2/0.171

x = [OH-] = 9.75 x 10^-6 M

pH before mixing,

pH = 14 - log(9.75 x 10^-6) = 8.99

after addition of HCl = 2.81 M x 10 ml = 28.1 mmol

initial moles of C2H3O2- = 0.171 M x 340 ml = 58.14 mmol

neutralization gives HC2H3O2 = 28.1 mmol

C2H3O2- remained = 58.14 - 28.1 = 30.04 mmol

pH after mixing,

pH = pKa + log(base/acid)

     = 4.75 + log(30.04/28.1) = 4.78

pH change = 4.78 - 8.99 = -4.21

c) 0.171 M HC2H3O2

pH before mixing,

HC2H3O2 + H2O <==> C2H3O2- + H3O+

Ka = [C2H3O2-][H3O+]/[HC2H3O2]

1.8 x 10^-5 = x^2/0.171

x = [H3O+] = 1.75 x 10^-3 M

pH = -log[H3O+] = 2.75

pH after mixing

HCl is a strong acid, thus

pH = -log(2.81 M x 10 ml/350 ml) = 1.09

pH change = 1.09 - 2.75 = -1.66

d) buffer

pH before mixing

pH = pKa + log(base/acid)

     = 4.75 + log(0.171/0.171) = 4.75

pH after mixing,

pH = 4.75 + log[(0.171 M x 340 ml - 2.81 M x 10 ml)/(0.171 M x 340 ml + 2.81 M x 10 ml)] = 4.29

pH change = 4.29 - 4.75 = -0.46