Question

Calculate the pH at the equivalence point in titrating 0.110 M solutions of each of the following with 8.0×10−2 MNaOH. chlorous acid (HClO2) benzoic acid (C6H5COOH)

Answer 1

Ans- HClO2 -- Ka = 1.1 x 10^-2

At the equivalence point you have an aqueous solution of NaClO2. If you started with 10.0 mL of 0.110 M HClO2:

mmoles HClO2 = M HClO2 x mL HClO2 = (0.110)(10.0) = 1.1 mmoles HClO2

The reaction between HClO2 and NaOH is in a 1:1 mole ratio: HClO2 + NaOH ==> NaClO2 + H2O.
All of the HClO2 will be converted to NaClO2.

1.00 mmoles HClO2 x (1 mmole NaOH / 1 mmole HClO2) = 1.00 mmoles NaOH

1.00 mmoles NaOH x (1 mL NaOH / 8.0×10⁻² mmoles NaOH) = 1.25*10^-3 mL NaOH used in the titration. The final volume is then 10.0 mL + 1.25*10^-3 mL = 10 mL.

M NaClO2 = mmoles NaClO2 / mL solution = 1.00 / 10 = 0.1 M NaClO2 at the equivalence point.
The salt solution will be basic (i.e. it was made using a STRONG BASE and a weak acid). So ClO2- ion will ionize as follows:

Molarity . . . . .ClO2- + H2O <==> HClO2 + OH-
Initial . . . . . . 0.1 0 0
Change . . . . . . .-x . . . . . . . . . . . . . . . .x . . . . .x
Equilibrium . .0.1-x . . . . . . . . . . . . . x . . . . .x

Kb ClO2- = Kw / Ka HClO2 = (1.0 x 10^-14) / (1.1 x 10^-2) = 9.1 x 10^-13

Kb = [HClO2][OH-] / [ClO2-] = (x)(x) / (0.1-x) = 9.1 x 10^-13

Because Kb is so small, we can neglect the -x term to simplify the math.

x^2 / 0.1 = 9.1 x 10^-13
x^2 = 9.1 x 10^-14
x = 3.01 x 10^-7 = [OH-]
pOH = -log [OH-] = -log (3.01 x 10^-7) = 6.52
pH = 14.00 - pOH = 14.00 - 6.52 = 7.4

(b) C6H5COOH -- Ka = 6.5 x 10^-5

Like the first problem, C6H5COOH and NaOH will react in a 1:1 mole ratio and at the equivalence point the concentration of C6H5COONa will equal 0.1 M. The ICE chart will look the same, too.

Kb C6H5COO- = Kw / Ka C6H5COOH = (1.0 x 10^-14) / (6.5 x 10^-5) = 1.5 x 10^-10

Kb = [C6H5COOH][OH-] / [C6H5COO-] = (x)(x) / (0.1-x) = 1.5 x 10^-10

Again, we can neglect the -x term to get

x^2 / 0.1 = 1.5 x 10^-10
x^2 = 1.5 x 10^-11
x = 3.87x 10^-6 = [OH-]
pOH = -log3.87x 10^-6 = 5.41
pH = 14 - pOH = 8.59