K_{sp} for Cr(OH)_3 is 2.7 times 10^{-31}. Wh | vedclass

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(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 $x \ mol/

Vedclass · Accepted Answer

$1 	imes 10^{-8}$

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 $x \ mol/L$.Then,$[Cr^{3+}] = x$ and $[OH^-] = 3x$.The solubility product expression is: $K_{sp} = [Cr^{3+}][OH^-]^3$.Substituting the values: $K_{sp} = (x)(3x)^3 = 27x^4$.Given $K_{sp} = 2.7 	imes 10^{-31}$,we have:$27x^4 = 2.7 	imes 10^{-31}$.$x^4 = \frac{2.7 	imes 10^{-31}}{27} = 0.1 	imes 10^{-31} = 1 	imes 10^{-32}$.Taking the fourth root on both sides:$x = (1 	imes 10^{-32})^{1/4} = 1 	imes 10^{-8} \ mol/L$.

$K_{sp}$ for $Cr(OH)_3$ is $2.7 \times 10^{-31}$. What is its solubility in moles / litre?

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