Sulphurous acid ($$H_2SO_3$$) has $$Ka_1 = 1.7 \times 10^{-2}$$ and $$Ka_2 = 6.4 \times 10^{-8}$$. The pH of 0.588 M $$H_2SO_3$$ is ________. (Round off to the Nearest Integer).

For sulphurous acid ($$H_2SO_3$$) with $$K_{a1} = 1.7 \times 10^{-2}$$ and $$K_{a2} = 6.4 \times 10^{-8}$$, since $$K_{a1} \gg K_{a2}$$, the pH is primarily determined by the first dissociation.

The first dissociation is: $$H_2SO_3 \rightleftharpoons H^+ + HSO_3^-$$. Let the concentration of $$H^+$$ produced be $$x$$. Then: $$K_{a1} = \frac{x^2}{0.588 - x} = 1.7 \times 10^{-2}$$.

Since $$K_{a1}$$ is not negligible compared to the concentration, we cannot ignore $$x$$ in the denominator. Rearranging: $$x^2 + 0.017x - 0.009996 = 0$$.

Using the quadratic formula: $$x = \frac{-0.017 + \sqrt{(0.017)^2 + 4(0.009996)}}{2} = \frac{-0.017 + \sqrt{0.000289 + 0.039984}}{2} = \frac{-0.017 + \sqrt{0.040273}}{2}$$.

$$x = \frac{-0.017 + 0.2007}{2} = \frac{0.1837}{2} = 0.09185$$ M.

$$pH = -\log(0.09185) = -\log(9.185 \times 10^{-2}) = 2 - \log(9.185) = 2 - 0.963 = 1.037$$.

Rounding off to the nearest integer, the pH is $$1$$.