Question

Use the \(K_{\mathrm{a}}\) values listed in Table 13.3 for polyprotic acids. Chromic acid, \(\mathrm{H}_{2} \mathrm{CrO}_{4}\), is commonly obtained by adding concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) to a dichromate. It is used to clean glass. For chromic acid, \(K_{\mathrm{a}}\) is \(1.8 \times 10^{-1}\) and \(3.2 \times 10^{-7}\) for its first and second dissociation, respectively. Write the overall equation and calculate \(K\) for the complete dissociation of chromic acid.

Short answer

Answer: The overall equation for the complete dissociation of chromic acid is: \(\mathrm{H}_{2}\mathrm{CrO}_{4} \rightleftharpoons 2\mathrm{H}^{+} + \mathrm{CrO}_{4}^{2-}\). The constant for its complete dissociation (K) is: \(5.76 \times 10^{-8}\).

Step-by-step solution

Write the first dissociation equation for chromic acid

The first dissociation of chromic acid will release one proton (\(\mathrm{H}^{+}\)) and produce the negative ion HCrO4-: \(\mathrm{H}_{2}\mathrm{CrO}_{4} \rightleftharpoons \mathrm{H}^{+} + \mathrm{H}\mathrm{CrO}_{4}^{-}\) with \(K_{a1} = 1.8 \times 10^{-1}\).

Write the second dissociation equation for chromic acid

The second dissociation will release another proton from the negative ion HCrO4-: \(\mathrm{H}\mathrm{CrO}_{4}^{-} \rightleftharpoons \mathrm{H}^{+} + \mathrm{CrO}_{4}^{2-}\) with \(K_{a2} = 3.2 \times 10^{-7}\).

Write the overall equation for the complete dissociation of chromic acid

To find the overall equation for the complete dissociation of chromic acid, we add both dissociation steps: \(\mathrm{H}_{2}\mathrm{CrO}_{4} \rightleftharpoons \mathrm{H}^{+} + \mathrm{H}\mathrm{CrO}_{4}^{-} \rightleftharpoons 2\mathrm{H}^{+} + \mathrm{CrO}_{4}^{2-}\)

Calculate the constant K for the complete dissociation

For the complete dissociation, K will be the product of Ka values for each dissociation step: \(K = K_{a1} \times K_{a2}\) \(K = (1.8 \times 10^{-1})(3.2 \times 10^{-7})\) \(K = 5.76 \times 10^{-8}\) So, the overall equation for the complete dissociation of chromic acid is: \(\mathrm{H}_{2}\mathrm{CrO}_{4} \rightleftharpoons 2\mathrm{H}^{+} + \mathrm{CrO}_{4}^{2-}\) and the constant for the complete dissociation K is: \(5.76 \times 10^{-8}\).