Question

What is the change of pH upon addition of 10.0 ml of HCl 3.0 M to 750. ml of the following:

- pure water

- 0.1 M formic acid

- 0.1M acetate of sodium

- a buffer made of 0.1M potassium nitrite and 0.08 M nitrous acid.

Answer 1

Change in pH

a) 10.0 ml of 3 M HCl to 750 ml of pure water

Using M1V1 = M2V2

M1 = 3 M

M2 = ?

V1 = 10 ml

V2 = 10 + 750 = 760 ml

Feed values,

M2 = 3 x 10/760 = 0.040 M

pH = -log[H+] = -log(0.04) = 1.40

b) 10.0 ml of 3 M HCl to 750 ml of 0.1 M sodium acetate

HCl would neutralize sodium aceate

moles of HCl added = 3 M x 0.01 L = 0.03 mols

moles of sodium acetate = 0.1 M x 0.75 L = 0.075 mols

final moles of sodium acetate = 0.075 - 0.03 = 0.045 mols

CH3COO- + H2O <===> CH3COOH + OH-

with x amount reacted,

Kb = [CH3COOH][OH-]/[CH3COO-]

5.75 x 10^-10 = x^2/0.045

x = [OH-] = 5.09 x 10^-6 M

pOH = -log[OH-] = 5.29

pH = 14 - pOH = 8.71

c) 10.0 ml of 3 M HCl to 750 ml of buffer of 0.1 M KNO3 and 0.08 M HNO2

HCl would neutralize KNO3

moles of HCl added = 3 M x 0.01 L = 0.03 mols

initial moles of KNO3 = 0.1 M x 0.75 L = 0.075 mols

initial moles of HNO2 = 0.08 M x 0.75 L = 0.06 mols

final moles of KNO3 = 0.075 - 0.03 = 0.045 mols

final moles of HNO2 = 0.06 + 0.03 = 0.09 mols

concentration of,

[KNO3] = 0.045/0.76 = 0.06 M

[HNO2] = 0.09/0.76 = 0.12 M

pKa for HNO2 = 3.15

Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

      = 3.15 + log(0.06/0.12)

      = 2.85