The solubility-product constant for barium permanganate, \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\), is \(2.5 \times 10^{-10}\). Assume that solid \(\mathrm{Ba}\left(\mathrm{MnO}_{4}\right)_{2}\) is in equilibrium with a solution of \(\mathrm{KMnO}_{4}\). What concentration of \(\mathrm{KMnO}_{4}\) is required to establish a concentration of \(2.0 \times 10^{-8} \mathrm{M}\) for the \(\mathrm{Ba}^{2+}\) ion in solution?

Write the balanced dissolution equation for barium permanganate: \(Ba(MnO4)2 (s) \rightleftharpoons Ba^{2+} (aq) + 2 MnO4^− (aq)\)

According to the dissociation equation, the expression for the solubility-product constant will be: \(K_{sp} = [Ba^{2+}] [MnO4^-]^2 \)

Given that the concentration of \([Ba^{2+}]\) ions is \(2.0 \times 10^{-8} M\), and there are two \(MnO4^-\) ions formed for each \(Ba^{2+}\), the concentration of \(MnO4^-\) ions will be: \([MnO4^-] = 2 \times [Ba^{2+}] = 2 \times (2.0 \times 10^{-8}) \)

Now that we know the concentration of \(MnO4^-\) ions and \(Ba^{2+}\) ions, we can use the \(K_{sp}\) formula to find the concentration of KMnO4 required: \((2.5 \times 10^{-10}) = (2.0 \times 10^{-8})[MnO4^−]^2\) Divide both sides by \((2.0 \times 10^{-8})\): \(\frac{2.5 \times 10^{-10}}{2.0 \times 10^{-8}} = [MnO4^−]^2\) Calculate the square root to find the concentration of \(MnO4^-\) ions: \([MnO4^-] = \sqrt{\frac{2.5 \times 10^{-10}}{2.0 \times 10^{-8}}} = 1.12 \times 10^{-2} M\) In this problem, the concentration of KMnO4 is equal to the concentration of \(MnO4^-\) ions, thus: The concentration of KMnO4 required to establish the concentration of \(Ba^{2+}\) ions in the solution as \(2.0 \times 10^{-8} M\) is approximately \(1.12 \times 10^{-2} M\).