The solubility product of Cr(OH)_3 at 298 K | vedclass

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Question

(A) The dissociation of $Cr(OH)_3$ is given by: $Cr(OH)_{3(s)} \rightleftharpoons Cr^{3+}_{(aq)} + 3OH^-_{(aq)}$.Let the solubility of $Cr(OH)_3$ be $

Vedclass · Accepted Answer

$(18 	imes 10^{-31})^{1/4}$

Vedclass · Answer

(A) The dissociation of $Cr(OH)_3$ is given by: $Cr(OH)_{3(s)} ightleftharpoons Cr^{3+}_{(aq)} + 3OH^-_{(aq)}$.Let the solubility of $Cr(OH)_3$ be $s \ mol/L$.Then,$[Cr^{3+}] = s$ and $[OH^-] = 3s$.The solubility product expression is $K_{sp} = [Cr^{3+}][OH^-]^3$.Substituting the values: $K_{sp} = (s)(3s)^3 = 27s^4$.Given $K_{sp} = 6.0 	imes 10^{-31}$,we have $27s^4 = 6.0 	imes 10^{-31}$.Thus,$s^4 = \frac{6.0 	imes 10^{-31}}{27} = \frac{60 	imes 10^{-32}}{27} \approx 2.22 	imes 10^{-32}$.The concentration of hydroxide ions is $[OH^-] = 3s$.Since $s = (2.22 	imes 10^{-32})^{1/4}$,then $[OH^-] = 3 	imes (2.22 	imes 10^{-32})^{1/4} = (3^4 	imes 2.22 	imes 10^{-32})^{1/4} = (81 	imes 2.22 	imes 10^{-32})^{1/4} = (179.82 	imes 10^{-32})^{1/4} = (17.982 	imes 10^{-31})^{1/4} \approx (18 	imes 10^{-31})^{1/4}$.

The solubility product of $Cr(OH)_3$ at $298 \ K$ is $6.0 \times 10^{-31}$. The concentration of hydroxide ions in a saturated solution of $Cr(OH)_3$ will be

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