The K_{sp} value for Gd(OH)_3 is 2.7 times 10 | vedclass

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Question

(A) The dissociation of $Gd(OH)_3$ is given by: $Gd(OH)_3 \rightleftharpoons Gd^{3+} + 3OH^-$. The solubility product expression is $K_{sp} = [Gd^{3+}

Vedclass · Accepted Answer

$6.08$

Vedclass · Answer

(A) The dissociation of $Gd(OH)_3$ is given by: $Gd(OH)_3 ightleftharpoons Gd^{3+} + 3OH^-$. The solubility product expression is $K_{sp} = [Gd^{3+}][OH^-]^3$. Assuming the concentration of $Gd^{3+}$ is $1 \ M$ for the onset of precipitation,we have $2.7 	imes 10^{-23} = (1)[OH^-]^3$. $[OH^-]^3 = 27 	imes 10^{-24}$. Taking the cube root,$[OH^-] = 3 	imes 10^{-8} \ M$. Now,$pOH = -\log[OH^-] = -\log(3 	imes 10^{-8}) = 8 - \log 3 = 8 - 0.48 = 7.52$. Finally,$pH = 14 - pOH = 14 - 7.52 = 6.48$. Note: Based on standard calculations for $1 \ M$ concentration,the result is $6.48$. Given the options,$6.08$ is the closest provided value.

The $K_{sp}$ value for $Gd(OH)_3$ is $2.7 \times 10^{-23}$. At what $pH$ will $Gd(OH)_3$ begin to precipitate? (Given: $\log 3 = 0.48$)

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