Question

50 ml of 0.05 M HNO₃ is added to 50 ml of 0.025 M KOH. Calculate the pH of the resultant solution.

Answer 1

Number of moles of HNO₃ = 0.05 × 50 × 10⁻³

= 2.5 × 10⁻³

Number of moles of KOH = 0.025 × 50 × 10⁻³

= 1.25 × 10⁻³

Number of moles of HNO₃ after mixing

= 2.5 × 10⁻³ – 1.25 × 10⁻³

= 1.25 × 10⁻³

∴ Concentration of HNO₃ = `("Number of moles of HNO"_3)/("Volume is litre")`

After mixing, total volume = 100 ml = 100 × 10⁻³ L

∴ [H⁺] = `(1.25 xx 10^-3  "moles")/(100 xx 10^-3  "L")`

= 1.25 × 10⁻² moles L⁻¹

pH = –log [H⁺]

pH = –log (1.25 × 10⁻²)

= 2 – 0.0969

= 1.9031