The pH of a 0.01 M weak acid HX ($$K_a = 4 \times 10^{-10}$$) 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 $$x \times 10^{-4}$$ M. The value of x is ________ (nearest integer).

For a weak monoprotic acid HX the dissociation equilibrium is
$$HX \rightleftharpoons H^{+}+X^{-}$$

The acid-dissociation constant is defined as
$$K_a = \frac{[H^{+}][X^{-}]}{[HX]} \qquad -(1)$$

Let the molarity of the acid after dilution be $$C$$ M and let the equilibrium hydrogen-ion concentration be $$h$$ M.
At equilibrium:
$$[H^{+}] = h, \quad [X^{-}] = h, \quad [HX] = C-h$$

Substituting these values in equation $$(1)$$,
$$K_a = \frac{h^{2}}{C-h} \qquad -(2)$$

Re-arranging $$(2)$$ to express $$C$$ in terms of $$h$$:
$$C-h = \frac{h^{2}}{K_a}\quad\Longrightarrow\quad C = h + \frac{h^{2}}{K_a} \qquad -(3)$$

After dilution the pH of the solution is given to be $$6$$, therefore
$$h = [H^{+}] = 10^{-6}\;{\rm M}$$

Putting $$h = 10^{-6}\;{\rm M}$$ and $$K_a = 4\times10^{-10}$$ in equation $$(3)$$:
$$C = 10^{-6} + \frac{(10^{-6})^{2}}{4\times10^{-10}} = 10^{-6} + \frac{10^{-12}}{4\times10^{-10}}$$

Calculate the second term:
$$\frac{10^{-12}}{4\times10^{-10}} = \frac{1}{4}\times10^{-2} = 2.5\times10^{-3}$$

Thus
$$C = 10^{-6} + 2.5\times10^{-3} \approx 2.501\times10^{-3}\,{\rm M}$$

The small term $$10^{-6}$$ may be ignored for quick estimation, giving $$C \approx 2.5\times10^{-3}\,{\rm M}$$.

Express this concentration in the form $$x\times10^{-4}\,{\rm M}$$:
$$2.5\times10^{-3}\,{\rm M} = 25\times10^{-4}\,{\rm M}$$

Hence $$x = 25$$ (nearest integer).

Answer: 25