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Q. When $1.5 \,mol$ of $CuCl _2 \cdot 2 H _2 O$ is dissolved in enough water to make $1.0 \,L$ of solution Given: $K_f CuCl ^{\oplus}=1.0\left(K_f\right.$ is the formation constant of $\left.CuCl ^{\oplus}\right)$
$\left[ Cu ^{2+}\right]$ in solution is

Equilibrium

Solution:

image
$K_f=\frac{\left[ CuCl ^{\oplus}\right]}{\left[ Cu ^{2+}\right]\left[ Cl ^{\ominus}\right]}=1.0$
$\Rightarrow \frac{x}{(1.5-x)(3.0-x)}=1.0$
$\Rightarrow x=4.5-4.5 x+x^2$
$\Rightarrow x^2-5.5 x+4.5=0$
$x=\frac{5.5 \pm \sqrt{(5.5)^2-4(4.5)}}{2}, x=1.0$ or $4.5$
${\left[ Cu ^{2+}\right]=0.5\, M }$