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 3.35-M HCl to 550. 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) 0.134 M C₂H₃O₂^1-

pH before mixing =
pH after mixing=
pH change =

c) 0.134 M HC₂H₃O₂

pH before mixing =
pH after mixing=
pH change =

d) a buffer solution that is 0.134 M in each C₂H₃O₂^1- and HC₂H₃O₂

pH before mixing =
pH after mixing=
pH change =

Answer 1

adding 10.0 mL of 3.35-M HCl to 550. mL of each of the following solutions.

0.010 litres moles in 3.35 M = 0.0335 moles of H+ ion added
=============

a) water
0.0335 moles of H+/ 0.550 litres = 0.0609 Molar H+
pH = 1.215

pH before mixing = 7.00
pH after mixing= 1.215
pH change = -5.78
================

b) 0.134 M C2H3O21-
before Kb of acetate = Kw / Ka acetic acid = (1.0x10^-14 / 1.8x10^-5)

Kb = 5.56 x10^-10 = [acetic acid ] [OH-] /[acetate]

5.56 x10^-10 = [X ] [X] / [0.134]

X = [OH-] = 8.63x10^-6

pOH = 5.063

pH = 8.936
--------------------------------

adding 0.0335 Molar H+ turns 0.0335 Molar of the acetate into acetic acid
after mixing
H C2H3O2 --> H+ & C2H3O2-
[0.0335] ----> [X] & [0.10]

Ka = 1.8x10^-5 = [H+] [C2H3O2] / [acid]

1.8x10^-5 = [H+] [0.10] / [0.0335]

H+ = 6.0x10^-6

pH = 5.22

pH before mixing = 8.936
pH after mixing= 5.22
pH change = -3.71
====================

c) 0.134 M HC2H3O2
before
Ka = 1.8x10^-5 = [H+] [C2H3O2] / [acid]

1.8x10^-5 = [x] [x] / [0.134]

H+ = 1.55x10^-3

pH = 2.80

pH before mixing = 2.80
pH after mixing= 1.49 (same as (a)
pH change = -1.31
=============

d) a buffer solution that is 0.134 M in each C2H3O21- and HC2H3O2

afterwards, th equilibrium shifts to the left
HC2H3O2 <--- H+ & C2H3O21-

[0.134 + 0.0335] <-- H+ & [ 0.134 - 0.0335]

Ka = 1.8x10^-5 = [H+] [0.10] / [0.1675]

H+ = 3.0x10^-5

pH = 4.522

pH before mixing = pKa = 4.74
pH after mixing= 4.522
pH change = - 0.222