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Solubility of $B(OH)_2$ in water at $25\,^oC$ is $10^{-7}\ M$ . The value nearest to $K_{sp}$ is
$4 × 10^{-21}\ M^3$
$9 × 10^{-21}\ M^3$
$2 × 10^{-21}\ M^3$
$6 × 10^{-21}\ M^3$
Solution
contribution due to water can not be neglected
$H_2O$ $\to $ $H^+$ $+$ $OH^-$
$x$ $2\times 10^{-7}+x$
$10^{-14}=x\left[2 \times 10^{-7}+x\right]$
$x^{2}+2 \times 10^{-7} x+10^{-14}=0$
$\mathrm{x}=\frac{-2 \times 10^{-7}+\sqrt{4 \times 10^{-14}+4 \times 10^{-14}}}{2}$
$=(\sqrt{2}-1) \times 10^{-7}$
$=0.414 \times 10^{-7}$
$\left[\mathrm{OH}^{-}\right]=2 \times 10^{-7}+0.414 \times 10^{-7}$
$=2.414 \times 10^{-7}$
$\mathrm{Ksp}=\left[\mathrm{B}^{+}\right]\left[\mathrm{OH}^{-}\right]^{2}$
$=\left[10^{-7}\right]\left(2.414 \times 10^{-7}\right)^{2}$
$=5.82 \times 10^{-21}$
