The absolute solubility product for Chromium(III) hydroxide, , is experimentally determined to be . — Ionic Equilibrium Chemistry Question
Question
The absolute solubility product $K_{sp}$ for Chromium(III) hydroxide, $Cr(OH)_3$, is experimentally determined to be $1.6 \times 10^{-30}$. The exact molar solubility of this specific compound in pure water can be mathematically formulated as $\left(\frac{1.6 \times 10^{-30}}{y}\right)^{1/4}$. Find the precise integer value of the denominator $y$.
💡 Solution & Explanation
The exact dissolution equation is $Cr(OH)_3 (s) \rightleftharpoons Cr^{3+} (aq) + 3OH^- (aq)$. Let the molar solubility be defined as $s$. At equilibrium, the concentration of $[Cr^{3+}] = s$, and the concentration of $[OH^-] = 3s$. The absolute solubility product expression is $K_{sp} = [Cr^{3+}][OH^-]^3 = (s)(3s)^3 = s(27s^3) = 27s^4$. To successfully isolate the solubility $s$, we rearrange the equation: $s^4 = \frac{K_{sp}}{27} \Rightarrow s = \left(\frac{K_{sp}}{27}\right)^{1/4}$. Substituting the given $K_{sp}$ seamlessly yields $\left(\frac{1.6 \times 10^{-30}}{27}\right)^{1/4}$. Thus, the denominator $y$ is exactly 27.