Question

Calculate the pH at the equivalence point in titrating 0.110 M solutions of each of the following with 9.0×10⁻²MNaOH.

A) hydrobromic acid (HBr)

B) chlorous acid (HClO2)

C) benzoic acid (C6H5COOH)

Answer 1

A) Hydrobromic acid

HBr + NaOH -------> H2O + NaBr

this reaction is between a strong acid and a strong base

the salt formed NaBr will not hydrolysed by water

so , at equivalence point

pH = 7

B) Chlorous Acid ( HClO2 )

HClO2 + NaOH ------> NaClO2 + H2O

This reaction is 1:1 molar reaaction

concentration ratio of HClO2 and NaOH = 0.110M/0.09M = 1.2222

So, at equivalence point [ ClO2-] is diluted 2.2222 time

so the final [ ClO2-]=0.110M/2.2222 = 0.0495M

This reaction is between a strong base and a weak acid , so the salt formed is hydrolysed by water

ClO2- + H2O <--------> HClO2 + OH-

Kb = [ HClO2 ] [ OH- ] / [ ClO2-]=9.1*10^-13

X²/0.0495 = 9.1*10^-13

X² = 4.5*10^-14

X = 2.12 *10^-7

[OH-] = 2.12*10^-7M

pOH = 6.67

pH = 14-6.67 = 7.33

C) C6H5 COOH + NaOH ------> C6H5COONa + H2O

This is 1:1 reaction

at equivalence point C6H5COO- is diluted for 2.2222 time

at equivalence point [ C6H5COO- ] = 0.0495M

this reaction is between weak acid and strong base , so the salt C6H5COO- is hydrolysed by water

C6H5COO- + H2O -----> C6H5COOH + OH-

Kb = [ C6H5COOH][OH-]/[C6H5COO-] = 1.59*10^-10

X² / 0.0495 = 1.59*10^-10

X² = 7.9*10^-12

X = 2.81*10^-6

[OH-] = 2.81 * 10^-6

pOH = 5.55

pH = 14 - 5.55

= 8.45