Solubility of B(OH)_2 in water at 25,^{circ}C | vedclass

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(D) The dissociation of $B(OH)_2$ is $B(OH)_2 \rightleftharpoons B^{2+} + 2OH^-$.Given solubility $s = 10^{-7} \ M$,so $[B^{2+}] = 10^{-7} \ M$ and $[

Vedclass · Accepted Answer

$6 	imes 10^{-21} \ M^3$

Vedclass · Answer

(D) The dissociation of $B(OH)_2$ is $B(OH)_2 ightleftharpoons B^{2+} + 2OH^-$.Given solubility $s = 10^{-7} \ M$,so $[B^{2+}] = 10^{-7} \ M$ and $[OH^-]_{salt} = 2 	imes 10^{-7} \ M$.Since the concentration is very low,the contribution of $OH^-$ from water must be considered.Let $[H^+] = x$,then $[OH^-]_{total} = 2 	imes 10^{-7} + x$.From the ionic product of water,$K_w = [H^+][OH^-]_{total} = 10^{-14}$.$x(2 	imes 10^{-7} + x) = 10^{-14} \implies x^2 + 2 	imes 10^{-7}x - 10^{-14} = 0$.Solving for $x$ using the quadratic formula: $x = \frac{-2 	imes 10^{-7} + \sqrt{4 	imes 10^{-14} + 4 	imes 10^{-14}}}{2} = (\sqrt{2} - 1) 	imes 10^{-7} \approx 0.414 	imes 10^{-7} \ M$.Total $[OH^-] = 2 	imes 10^{-7} + 0.414 	imes 10^{-7} = 2.414 	imes 10^{-7} \ M$.$K_{sp} = [B^{2+}][OH^-]^2 = (10^{-7})(2.414 	imes 10^{-7})^2 \approx 5.82 	imes 10^{-21}$.The nearest value is $6 	imes 10^{-21} \ M^3$.

Solubility of $B(OH)_2$ in water at $25\,^{\circ}C$ is $10^{-7} \ M$. The value nearest to $K_{sp}$ is

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