Up to what pH must a solution containing a pr | vedclass

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(C) The solubility product expression for $Cr(OH)_3$ is: $K_{sp} = [Cr^{3+}][OH^-]^3$.Given $[Cr^{3+}] = 0.1 \ mol/L$ and $K_{sp} = 6 	imes 10^{-31}$

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Up to $4.2$

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(C) The solubility product expression for $Cr(OH)_3$ is: $K_{sp} = [Cr^{3+}][OH^-]^3$.Given $[Cr^{3+}] = 0.1 \ mol/L$ and $K_{sp} = 6 	imes 10^{-31}$,we can find the concentration of $OH^-$ required to keep the precipitate in solution:$[OH^-]^3 = \frac{K_{sp}}{[Cr^{3+}]} = \frac{6 	imes 10^{-31}}{0.1} = 6 	imes 10^{-30}$.Taking the cube root:$[OH^-] = (6 	imes 10^{-30})^{1/3} \approx 1.817 	imes 10^{-10} \ mol/L$.Now,calculate $pOH$:$pOH = -\log[OH^-] = -\log(1.817 	imes 10^{-10}) = 10 - \log(1.817) \approx 10 - 0.26 = 9.74$.Finally,calculate $pH$:$pH = 14 - pOH = 14 - 9.74 = 4.26$.Rounding to the nearest provided option,the value is approximately $4.2$.

Up to what $pH$ must a solution containing a precipitate of $Cr(OH)_3$ be adjusted so that the precipitate just dissolves? (Given: $[Cr^{3+}] = 0.1 \ mol/L$,$K_{sp} = 6 \times 10^{-31}$)

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