Question

Calculate the pH of a solution prepared by mixing 500.0 mL of 0.300 M NaOH (Kb = ∞) and 250.0 mL of 0.150 M HNO2 (Ka = 4.50 X 10-4)

Answer 1

NaOH = 500 mL of 0.3 M , therefore no. of moles of OH^- = (0.3*500)\1000 = 0.15 mole

HNO2 = 250 mL of 0.15 M, therfore no. of moles of H⁺ = (0.15*250)\1000 = 0.0375 mole

thus, no.of moles of alkali left = 0.15-0.0375 = 0.1125 mole

total volume = 500+250 = 750 mL

therefore, no. of moles of alkali i.e. 0.1125 mole left in 750 mL solution,

i.e. concentration of OH^- = no.of moles of alkali *1000\ volume of solution (mL)

= 0.1125*1000\750

[OH^-] =0.15 mole\litre

since, [H⁺][OH^-]=10^-14

   [H⁺]0.15=10^-14

  [H⁺]= 6.66*10^-14

=> pH = - log[H⁺] =-log[ 6.66*10^-14] =13.2

=> pH = 13.2