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 3.78-M HCl to 440. 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) 0.154 M C₂H₃O₂^1-

pH before mixing =
pH after mixing=
pH change =

c) 0.154 M HC₂H₃O₂

pH before mixing =
pH after mixing=
pH change =

d) a buffer solution that is 0.154 M in each C₂H₃O₂^1- and HC₂H₃O₂

pH before mixing =
pH after mixing=
pH change =

Answer 1

a)

pH before mixing = 7.00

resultant HCl molarity = 3.78 x 10 / (440 +10) = 0.084 M

pH = -log [H+]

pH = 1.08

pH after mixing = 1.08

change in pH = -5.92

b)

it is the salt of weak acid and strong base . so pH >7

pH = 7 + 1/2 [pKa + log C]

pH = 7 + 1/2 [4.74 + log 0.154]

pH = 8.96

pH before mxing = pH = 8.96

millimoles of CH3COO- = 440 x 0.154 = 67.76

millimoles of HCl = 3.78 x 10 = 37.8

CH3COO- + HCl -----------------> CH3COOH

67.76           37.8                             0

29.96           0                                  37.8

pH = pKa + log [CH3COO- / CH3COOH]

pH = 4.74 + log (37.8 / 29.96)

pH = 4.64

pH after mixing = 4.64

change in pH = - 4.32