Question

A 20.00 mL solution of 0.0840M HOCl is titrated with 0.0560M KOH solution. Calculate the pH after each of the following additions of the KOH solution:

a. 15.00mL

b. 30.0mL

c. 32.0mL

Answer 1

no. of mole = molarity volume of solution in liter

mole of HOCl = (0.0840)(0.020L) = 0.00168 mole

reaction of HOCl with KOH take place as follow

KOH + HOCl KOCl + H₂O

According to reaction 1 mole of KOH react with 1 mole of HOCl

a.) mole of KOH in 15.00 ml = 0.015L 0.0560 = 0.00084 mole

then 0.00084 mole of KOH react with 0.00084 mole of HOCl

mole of HOCl remained unreacted = 0.00168 - 0.00084 = 0.00084 mole

HOCl  dissociated as

HOCl   H⁺ +OCl^-

mole of H⁺ in solution = 0.00084

total volume of solution = 20ml + 15ml = 35 ml = 0.035L

malarity = no. of mole / volume of solution in liter

[H⁺] = 0.00084/0.035 = 0.024

pH = -log[H⁺] = -log(0.024) = 1.619

pH of solution = 1.619

b.) mole of KOH in 30.00 ml = 0.030L 0.0560 = 0.00168 mole

then 0.00168 mole of KOH react with 0.00168 mole of HOCl

mole of HOCl remained unreacted = 0.00168 - 0.00168 = 0.00 mol

no any mole of H⁺ and OH^- in solution it is nutrilization point pH of solution is 7

c.) mole of KOH in 32.00 ml = 0.032L 0.0560 = 0.001792 mole

then 0.00168 mole of HOCl react with 0.00168 mole of KOH

mole of KOH remained unreacted = 0.001792 - 0.00168 = 0.000112 mole

KOH dissociated as

KOH K⁺ + OH^-

mole of OH^- in solution = 0.000112

total volume of solution = 20ml + 32ml = 52 ml = 0.052L

malarity = no. of mole / volume of solution in liter

[OH^-] = 0.000112/0.052 = 0.002153

pOH = -log[OH^-] = -log(0.002153) = 2.67

pOH of solution = 2.67

pH = 14 - pOH = 14 - 2.67 = 11.33

pH of solution = 11.33