Ncert Exercises & Solutions

Calculate the pH of a solution formed by mixing equal volumes of two solutions A and B of a strong acid having pH = 6 and pH = 4 respectively.

pH of solutiion A=6. Hence,

[H^{+}] = 10^{- 6} mol L^{- 1}

pH of solution B=4. Hence,

[H^{+}] = 10^{- 4} mol L^{- 1}

On mixing 1 L of each solution, molar concentration of total H⁺ is halved.

Total,

[H^{+}] = \frac{10^{- 6} + 10^{- 4}}{2} mol L^{- 1}

[H^{+}] = \frac{1.01 \times 10^{- 4}}{2} = 5.05 \times 10^{- 5} mol L^{- 1}

[H^{+}] = 5.0 \times 10^{- 5} mol L^{- 1}

pH = - \log [H^{+}] \Rightarrow pH = - \log (5.0 \times 10^{- 5})

pH = - [\log 5 + (- 5 \log 10)] \Rightarrow pH = - \log 5 + 5

pH = 5 - \log 5 = 5 - 0.6990 \Rightarrow pH = 4.3010 \approx 4.3

Thus, the pH of resulting solution is 4.3.