Question

Calculate the pH of the solution which results from mixing:

25.0mL of 0.100M HNO2 with 15.0mL of 0.100M NaOH

Answer 1

no. of mole = molarity volume of solution in liter

no. of mole of HNO₂  = 0.100 0.025 = 0.0025 mole

no. of mole of NaOH = 0.100 0.015 = 0.0015 mole

Acid base neutrilization reaction take place beween NaOH and HNO₂ is

HNO₂ + NaOH NaNO₂  + H₂O

0.0015 mole of HNO₂ react with 0.0015 mole of NaOH

no. of mole of HNO₂ remain in solution = 0.0025 - 0.0015 = 0.0010 mole

total volume of solution = 15 + 25 = 40 ml = 0.040 L

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

molarty of HNO₂ = 0.0010 / 0.040 = 0.025 M

HNO₂ is weak acid dissociate as

HNO₂ + H₂O H₃O⁺  + NO₂^-

Ka of HNO₂ = 4.5 10^-4

K_a = [NO₂^-][H₃O⁺] / [HNO₂]

but  [NO₂^-] = [H₃O⁺] = x

K_a = [x][x] / [HNO₂]

Substitute the value in equation

4.510^-6 = [x]²/ 0.025

[x]² = 4.510^-40.025 = 1.12510^-5

[x] = 0.003354 M

Concentration of H₃O⁺ = 0.003354 M

pH = - log[H₃O⁺]

pH = - log (0.003354)

pH = 2.47

The pH of resulting solution = 2.47