Question

The pH of a 0.01 M weak acid $\mathrm{HX}\left(\mathrm{K}_{\mathrm{a}}=4 \times 10^{-10}\right)$ is found to be 5 . Now the acid solution is diluted with excess of water so that the pH of the solution changes to 6 . The new concentration of the diluted weak acid is given as $\mathrm{x} \times 10^{-4} \mathrm{M}$. The value of x is _______ (nearest integer).

Hint:

Check the consistency of the given data and ensure the calculations are justified.

Answer 1

Correct Answer: 10

1. Initial pH calculation: \[ \mathrm{HX}_{(\mathrm{aq})} \rightleftharpoons \mathrm{H}_{(\mathrm{aq})}^{+}+\mathrm{X}_{(\mathrm{aq})}^{-} \quad \mathrm{K}_{\mathrm{a}}=4 \times 10^{-10} \] \[ 0.01(1-\alpha) \quad 0.01 \alpha \quad 0.01 \alpha \quad \text{Not justified} \] \[ \Rightarrow 0.01 \alpha=10^{-5} \Rightarrow \alpha=10^{-3} \]
2. Calculate $\mathrm{K}_{\mathrm{a}}$: \[ \mathrm{K}_{\mathrm{a}}=0.01 \alpha^{2}=10^{-8} \]