Calculate the concentration of‏ $[\ce{SO_3^-^2}]$ present in 0.010M solution of $\ce{H_2SO_3}$ in pure water?

($Ka_1=0.017)$, ($Ka_2=10^{-7.19}$)

I found this approach in reference text for solving the problem :

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But I confused with the derivation of Eq.2, so I do my calculation according to Eq.1 to find the concentration of sulfite ion equal ‏ $2.20×10^{-7}$ M.

SO :

I choose another approach to find the concentration of sulfite ion depending on substitution in this formula : $$Ka_2 = \dfrac{\ce{[H+][SO3^{2-}]}}{\ce{[HSO3^-]}}\ $$ Thus

$$\mathrm{10^{-7.19}} = \frac{(0.00706 + {[SO_3^{-2}]}){[SO_3^{-2}]}}{(0.00706 - {[SO_3^{-2}]})}$$

From which I obtain $$[\ce{SO_3^{-2}}] = 6.46 \times 10^{-8}$$
Which different from the answer obtained from the first approach.

I appreciate any help.


You messed up calc in eq 2 in multiple ways...

$$K_1 = 0.017$$

$$K_2 = 10^{-7.19} = 6.5\times10^{-8}$$

$$K_1K_2 = 1.11\times10^{-9}$$


$$K_1 = \dfrac{\ce{[H^+][HSO3^-]}}{\ce{[H2SO3]}}\tag{1}$$

$$K_2 = \dfrac{\ce{[H^+][SO3^{2-}]}}{\ce{[HSO3^-]}}\tag{2}$$

$$K_1K_2 = \dfrac{\ce{[H^+][HSO3^-]}}{\ce{[H2SO3]}}\times\dfrac{\ce{[H^+][SO3^{2-}]}}{\ce{[HSO3^-]}} =\dfrac{\ce{[H^+]^2[SO3^{2-}]}}{\ce{[H2SO3]}}\tag{3}$$

rearranging (3)

$$ \ce{[SO3^{2-}]}= \dfrac{K_1K_2\ce{[H2SO3]}}{\ce{[H^+]^2}}= \dfrac{K_1K_2(0.010-0.0076)}{0.0076^2} = 4.6\times10^{-8}\tag{4}$$

  • $\begingroup$ MaxC According to the last solution I think the solution in the reference text incorrect .thanks for the support $\endgroup$ – Adnan AL-Amleh Jul 21 '18 at 23:22

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