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 3.59-M HCl to 450. 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 = 7.00

pH after mixing= 1.10

pH change = -5.90

b) 0.103 M C2H3O2^1-

pH before mixing =

pH after mixing=

pH change =

c) 0.103 M HC2H3O2

pH before mixing = 2.87

pH after mixing=

pH change =

d) a buffer solution that is 0.103 M in each C2H3O2^1- and HC2H3O2

pH before mixing =

pH after mixing=

pH change =

Answer 1

pH of each of the solutions and the change in pH caused by adding 10.0 mL of 3.59 M HCl to 450 mL of each of thesolutions.

b) 0.103 M CH₃COO-

pH before mixing :Acetate is a weak base.

A^- is a weak base, Kb for it = Kw / Ka (CH₃COOH) = 10^-14 / (1.8*10^-5) = 5.56*10^-10

CH₃COO^-(aq) + H₂O (l) $\rightleftharpoons$  CH₃COOH (aq) + OH^- (aq)

Kb = (CH₃COOH ) (OH^- ) / (CH₃COO-)

[OH^–] = (Kb*   ( CH₃COO^-) ) ^1/2 = (5.56*10^-10*  0.103) ^1/2 = 7.57 *10^-6 M

pH = 14 -pOH = 14 - (-log[7.57*10^-6 ] = 8.88

pH after mixing :

CH₃COO^-(aq) + H₃O⁺ (aq) $\rightleftharpoons$  CH₃COOH (aq) + H₂O (l)

Any such solution containing comparable amounts of a weak acid , and its conjugate
weak base, is act as buffer. So, pH can be calculated following equation :

pH = pKa + log ( ( A^-) / (HA ) )

( CH₃COO^-) = (0.103 M*450 ml - 3.59 M*10 ml )/ 460 ml = 0.023 M

(CH₃COOH ) =  3.59 m*10 ml / 460 ml = 0.078 M

pH = 4.74 + log ( (0.023) / 0.078 ) ) = 4.21

pH change = -4.67

c) 0.103 M CH₃COOH

pH before mixing = 2.87

pH after mixing : adding 10.0 mL of 3.59 M HCl will increase H₃O⁺ concentration :

pH before mixing : 2.87 ; so,  H₃O⁺ = 10^-2.87 M

( H₃O⁺ ) after mixing =  10^-2.87 M* 450 ml + 10 ml * 3.59 M / 460 ml = 0.079

pH = -log (H₃O⁺ ) = 1.10

pH change = - 1.77

d) a buffer solution that is 0.103 M in each CH₃COO^- and CH₃COOH

pH before mixing =

pH can be calculated following equation :

pH = pKa + log ( ( A^-) / (HA ) )

pH = 4.74 + log ( (0.103) / 0.103 ) ) = 4.74

pH after mixing : we have following reaction ,

CH₃COO^-(aq) + H₃O⁺ (aq) $\rightleftharpoons$  CH₃COOH (aq) + H₂O (l)

New, concentrations : ( CH₃COO^-) = (0.103 M*450 ml - 3.59 M*10 ml )/ 460 ml = 0.023 M

(CH₃COOH ) = ( 0.103 M*450 ml+ 3.59 m*10 ml) / 460 ml = 0.179 M

pH = 4.74 + log ( (0.0.023) / 0.179 ) ) = 3.85

pH change = - 0.89