Ncert Exercises & Solutions

pH of a solution of a strong acid is 5.0. What will be the pH of the solution obtained after diluting the given solution a 100 times?

Given,

pH = 5

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

On diluting the solution 100 times

[H^{+}] = \frac{10^{- 5}}{100} = 10^{- 7} mol L^{- 1}

On calculating the pH using the equation

pH = - \log [H^{+}]

, value of pH comes out to be 7. It is not possible. This indicates that solution is very dilute.

Hence, Total H⁺ ion concentration=H⁺ ions from acid+H⁺ ion from water

[H^{+}] = 10^{- 7} + 10^{- 7} = 2 \times 10^{- 7} M

pH = - \log [2 \times 10^{- 7}]

pH = 7 - 0.3010 = 6.699