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 2.64-M HCl to 690. 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 =
pH after mixing=
pH change =

b) 0.152 M C2H3O21-
pH before mixing =
pH after mixing=
pH change =

c) 0.152 M HC2H3O2
pH before mixing =
pH after mixing=
pH change =

d) a buffer solution that is 0.152 M in each C2H3O21- and HC2H3O2
pH before mixing =
pH after mixing=
pH change =

Answer 1

a) pH of water before addition =7

10.0 mL of 2.64-M HCl to 690. mL

total volume = 700mL

new concentration of HCl = initial concentration X initial volume / total volume = 2.64 X 10 / 700 = 0.0377

pH = -log[H+] = -log 0.0377 = 1.423

ph change = 7- 1.423 = 5.577

b) b) 0.152 M C₂H₃O₂^-

initial pH = ?

the acetate ion will hydrolyze as

               CH3COO- + H2O --> CH3COOH + OH-

Initial           0.152                     0               0

Change        -x                       +x               +x

Equilibrium 0.152-x                 x                  x

Kb = Kw /Ka = [CH3COOH] [ OH-] / [CH3COO-] = 10^-14 / 1.8 X 10^-5 = 5.5 X 10^-10 = x² / (0.152-x)

x <<1

Kb = 5.5 X 10^-10 = x² / (0.152

x = 9.14 X 10^-6M = [OH-]

pOH = -log[OH-] = 5.04

pH = 14-5.04 = 8.96

after addition of HCl it will form some acetic acid

moles of acetic acid formed = Moles of HCl added = Molarity X volume in litres = 2.64 X 10 / 1000 = 0.0264

moles of acetate ion present initially = 0.152 X 690/1000 = 0.105 moles

moles of acetate left = Moles of acetate - Moles of acetic acid = 0.105 - 0.0264 = 0.0786

this will form buffer, pka of acetic acid is 4.74

pH = pKa + log [salt] / [acid]

pH = 4.74 + log [0.0786 / 0.0264] = 5.21

pH change = 8.96 - 5.21 = 3.75

c) 0.152 M HC2H3O2

pH before mixing will be due to acetic acid

[H+] = [Ka x concentration of acetic acid]^1/2 = (1.8 X 10-5 x 0.0152)^1/2 = 5.23 X 10^-4 M

Moles of [H+] = Molarity X volume in litres = 5.23 X 10^-4 X 690 / 1000 = 0.000361

pH = -log 5.23 X 10-4 = 3.28

After addition of HCl the mole sof H+ added = Moles of HCl added = Molarity X volume in litres = 2.64 X 10 / 1000 = 0.0264

total moles of H+ = 0.000361 + 0.0264

[H+] = Moles / volume in litres = 0.0268 / 0.7 = 0.038

pH = -log 0.038 = 1.42

pH change = 3.28-1.42 = 1.86

d) a buffer solution that is 0.152 M in each C2H3O21- and HC2H3O2

Initial pH of buffer with acetate and acetic acid in same concentration = pKa of acetic acid = 4.74

after addition of HCl it will react with acetate ion to form acetic acid

moles of acetic acid formed = Moles of HCl added = Molarity X volume in litres = 2.64 X 10 / 1000 = 0.0264

moles of acetate ion present initially = 0.152 X 690/1000 = 0.105 moles

moles of acetic acid present initially = 0.152 X 690/1000 = 0.105 moles

Moles of acetic acid after addition of HCl =0.105 + 0.0264 = 0.1314

Moles of acetate after addition of HCl = 0.105 - 0.0264 =0.0786

pH = pKa + log [salt] / [acid]

pH = 4.74 + log [0.0786] / [0.1314] = 4.52

ph change = 4.74-4.52 = 0.22