Question

Calculate the pH of a 0.10 M solution of CoCl₃. The K_a value for Co(H₂O)₆ ³⁺ is 1.0 × 10^-5.

Short answer

pH of the given solution comes out to be 3.

Step-by-step solution

 Calculation of Ka for Hydrolysis

Since the Co(H₂O)₆³⁺ion is a stronger acid than water, the dominate equilibrium will be:

$$\begin{gathered} Co{{{\left(H_{2}O\right)}_6}^{3+}}_{\left(aq\right)}\rightleftarrows Co\left(OH\right){{{\left(H_{2}O\right)}_5}^{2+}}_{\left(aq\right)}+{H^{+}}_{\left(aq\right)} \\ {K}_a=\frac{\left[Co\left(OH\right){{{\left(H_{2}O\right)}_5}^{2+}}_{\left(aq\right)}\right]+\left[{H^{+}}_{\left(aq\right)}\right]}{\left[Co{{{\left(H_{2}O\right)}_6}^{3+}}_{\left(aq\right)}\right]} \\ {K}_a=1\times {10}^{-5} \end{gathered}$$

Calculation of pH for Given Salt

$$\begin{gathered} Co{{{\left(H_{2}O\right)}_6}^{3+}}_{\left(aq\right)}\rightleftarrows Co\left(OH\right){{{\left(H_{2}O\right)}_5}^{2+}}_{\left(aq\right)}+{H^{+}}_{\left(aq\right)} \\ 0.100 \\ 0.1-xxx \\ {K}_a=\frac{{\left[x\right]}^2}{\left[0.1-x\right]} \\ {K}_a=1\times {10}^{-5} \\ x=1\times {10}^{-3} \\ pH=-\log \left[H^{+}\right] \\ pH=-\log \left({10}^{-3}\right) \\ pH=3 \end{gathered}$$