Question

The pH of a 0.10 M solution of Ni(NO₃)₂ is 5.0. Calculate the acid ionization constant of Ni(H₂O)₆²⁺ (aq).

Short answer

The acid ionisation constant of Ni(H₂O)₆²⁺ is$K_a=1.0\times {10}^{-9}$

Step-by-step solution

The reaction of Ni(NO3)2 with water

The following chemical reaction takes place when nickel nitrate reacts with water:

$\mathit{N}\mathit{i}{{\mathbf{\left(}{\mathbf{H}}_{\mathbf{2}}\mathbf{O}\mathbf{\right)}}_{\mathbf{6}}}^{\mathbf{2}\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{2}}\mathit{O}\mathbf{\left(}\mathbf{I}\mathbf{\right)}\mathbf{\to }{\mathbf{\left[}\mathbf{N}\mathbf{i}{\left(H_{2}O\right)}_{\mathbf{5}}\mathbf{O}\mathbf{H}\mathbf{\right]}}^{\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}\mathbf{+}{\mathit{H}}_{\mathbf{3}}{\mathit{O}}^{\mathbf{+}}\mathbf{\left(}\mathbf{a}\mathbf{q}\mathbf{\right)}$

Construction of ICE table

$$\begin{gathered} Ni{{\left({\mathrm{H}}_2\mathrm{O}\right)}_6}^{2+}\left(\mathrm{aq}\right)+H_{2}O\left(\mathrm{I}\right)\to {\left[\mathrm{Ni}{\left({\mathrm{H}}_2\mathrm{O}\right)}_5\mathrm{OH}\right]}^{+}\left(\mathrm{aq}\right)+H_3{O}^{+}\left(\mathrm{aq}\right) \\ I:0.10M-00 \\ C:-x-+x+x \\ E:0.10-x-xx=1.0\times {10}^{-5} \end{gathered}$$

The value of acid ionisation constant

$\begin{array}{rcl}{K}_a& =& \frac{{\left(1.0\times {10}^{-5}\right)}^2}{0.10-0.00001}\\ & =& 1.0\times {10}^{-9}\end{array}$