# How do you calculate the pH of an aqueous solution of 0.0726 M potassium sulfite?

Jul 11, 2018

You ain't given us any data with respect to $p {K}_{a}$ of $\text{bisulfite}$, or $p {K}_{b}$ of $S {O}_{3}^{2 -}$....I gets $p H \cong 10$

#### Explanation:

This site reports that the $p {K}_{a}$ of $H S {O}_{3}^{-}$ is $6.97$....and therefore, the $p {K}_{b}$ of $S {O}_{3}^{2 -}$, its conjugate base is $14 - 6.97 = 7.03$..(because $p {K}_{a} + p {K}_{b} = 14$ in aqueous solution)….

And so we address the following equilibrium....

$S {O}_{3}^{2 -} + {H}_{2} O \left(l\right) r i g h t \le f t h a r p \infty n s H S {O}_{3}^{-} + H {O}^{-}$

And thus ${K}_{b} = {10}^{- 7.03} = \frac{\left[H S {O}_{3}^{-}\right] \left[H {O}^{-}\right]}{\left[S {O}_{3}^{2 -}\right]}$

Now INITIALLY, $\left[S {O}_{3}^{2 -}\right] = 0.0726 \cdot m o l \cdot {L}^{-} 1$...and if $x \cdot m o l \cdot {L}^{-} 1$ sulfite ion associates then....

${K}_{b} = {10}^{- 7.03} = \frac{{x}^{2}}{0.0726 - x}$...we propose that $0.0726 \text{>>} x$...and so....

${x}_{1} = \sqrt{{10}^{- 7.03} \times 0.0726} = 8.23 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$...and we use this value for a successive approximation...

${x}_{2} = \sqrt{{10}^{- 7.03} \times \left(0.0726 - 8.23 \times {10}^{-} 5\right)} = 8.23 \times {10}^{-} 5 \cdot m o l \cdot {L}^{-} 1$

And since the approximations have converged, I am prepared to accept this as the true value....

But $x = \left[H {O}^{-}\right]$...$p O H = - {\log}_{10} \left(8.23 \times {10}^{-} 5\right) = 4.08$....$p H = 14 - p O H = 14 - 4.08 = 9.92$

(Why? Because $p H + p O H = 14$ in aqueous solution under standard conditions...)