Question

Selenious acid, \(\mathrm{H}_{2} \mathrm{SeO}_{3},\) is primarily used to chemically darken copper, brass, and bronze. It is a diprotic acid with the following \(K_{\mathrm{a}}\) values: \(K_{\mathrm{a} 1}=2.7 \times 10^{-3}\) and \(\mathrm{K}_{\mathrm{a} 2}=5.0 \times 10^{-8} .\) What is the \(\mathrm{pH}\) of a \(2.89 \mathrm{M}\) solution of selenious acid? Estimate \(\left[\mathrm{HSeO}_{3}^{-}\right]\) and \(\left[\mathrm{SeO}_{3}^{2-}\right]\)

Short answer

Based on the given 2.89 M solution of selenious acid with Ka1 = 2.7 x 10^(-3) and Ka2 = 5.0 x 10^(-8), calculate the pH of the solution and estimate the concentrations of HSeO3- and SeO32-.

Step-by-step solution

Determine the Concentration of \(\mathrm{H}^{+}\)

Since selenious acid is a diprotic acid, it loses two protons in two stages. However, we will consider only the first ionization, as the second ionization will have a negligible effect on the pH due to the large difference between the two provided Ka values. We will use the following equation for the first ionization: \(\mathrm{H}_{2} \mathrm{SeO}_{3} \rightleftarrows \mathrm{H}^{+} + \mathrm{HSeO}_{3}^{-}\) And the given value of \(K_{\mathrm{a} 1}=2.7 \times 10^{-3}\) Next, we will set up an ICE (Initial, Change, Equilibrium) table for the first ionization: | | H2SeO3 | H+ | HSeO3- | |--- | --- | --- |--- | | Initial | 2.89 M | 0 | 0 | | Change | -x | +x | +x | | Equilibrium | 2.89-x | x | x | Now, we have: \(K_{\mathrm{a} 1} = \frac{[\mathrm{H}^{+}][\mathrm{HSeO}_{3}^{-}]}{[\mathrm{H}_{2} \mathrm{SeO}_{3}]} \implies 2.7 \times 10^{-3} = \frac{x^2}{2.89-x}\)

Solve for \(\mathrm{H}^{+}\) concentration

Approximating, since K is small, we can assume that the change in x will be negligible, so \(2.89 - x \approx 2.89\). Now we have: \(2.7 \times 10^{-3} = \frac{x^2}{2.89}\) Solve for x: \(x^2 = 2.7 \times 10^{-3} \times 2.89\) \(x = \sqrt{2.7 \times 10^{-3} \times 2.89}\) \(x \approx 0.0778\) Thus, the concentration of H+ ions is approximately 0.0778 M.

Calculate the \(\mathrm{pH}\)

Now that we have the concentration of H+ ions, we can calculate the pH using the following formula: \(\mathrm{pH} = -\log_{10} [\mathrm{H}^{+}]\) \(pH = -\log_{10} (0.0778) \approx 1.11\) So, the pH of the 2.89 M solution of selenious acid is approximately 1.11.

Estimate \(\left[\mathrm{HSeO}_{3}^{-}\right]\) concentration

As we assumed in step 1, the concentration of H+ and HSeO3- at equilibrium are almost the same, so: \(\left[\mathrm{HSeO}_{3}^{-}\right] \approx x = 0.0778 \text{ M}\)

Estimate \(\left[\mathrm{SeO}_{3}^{2-}\right]\) concentration

We will estimate the concentration of SeO32- ions by considering the second ionization. From the second ionization equation: \(\mathrm{HSeO}_{3}^{-} \rightleftarrows \mathrm{H}^{+} + \mathrm{SeO}_{3}^{2-}\) And the given value of \(K_{\mathrm{a} 2}=5.0 \times 10^{-8}\) Using the ICE table for the second ionization: | | HSeO3- | H+ | SeO32- | |--- | --- | --- |--- | | Initial | 0.0778 M | 0.0778 | 0 | | Change | -y | +y | +y | | Equilibrium | 0.0778-y | 0.0778+y| y | Now, we have: \(K_{\mathrm{a} 2} = \frac{[\mathrm{H}^{+}][\mathrm{SeO}_{3}^{2-}]}{[\mathrm{HSeO}_{3}^{-}]} \implies 5.0 \times 10^{-8} = \frac{y(0.0778+y)}{0.0778-y}\) Approximating, as K is small, we can assume the change in y is negligible. Now we have: \(5.0 \times 10^{-8} = \frac{y(0.0778)}{0.0778}\) Solve for y: \(y = 5.0 \times 10^{-8} \text{ M}\) Thus, the concentration of SeO32- ions is approximately \(5.0 \times 10^{-8}\) M. In summary, the pH of the 2.89 M solution of selenious acid is approximately 1.11, and the concentrations of HSeO3- and SeO32- are approximately 0.0778 M and \(5.0 \times 10^{-8}\) M, respectively.