(D) The equilibrium is $BiO(OH)(s) \rightleftharpoons BiO^{+}(aq) + OH^{-}(aq)$.The solubility product constant is $K = [BiO^{+}][OH^{-}] = 4 \times 1

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(D) The equilibrium is $BiO(OH)(s) \rightleftharpoons BiO^{+}(aq) + OH^{-}(aq)$.The solubility product constant is $K = [BiO^{+}][OH^{-}] = 4 \times 10^{-10}$.Assuming $[BiO^{+}] = [OH^{-}] = s$, then $s^2 = 4 \times 10^{-10}$, which gives $s = 2 \times 10^{-5} \text{ M}$.Thus, $[OH^{-}] = 2 \times 10^{-5} \text{ M}$.Calculating $pOH$: $pOH = -\log(2 \times 10^{-5}) = 5 - \log 2 = 5 - 0.3010 = 4.699$.Finally, $pH = 14 - pOH = 14 - 4.699 = 9.301$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

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In a qualitative analysis, $Bi^{3+}$ is detected by the appearance of a precipitate of $BiO(OH)(s)$. Calculate the $pH$ when the following equilibrium exists at $298 \text{ K}$:
$BiO(OH)(s) \rightleftharpoons BiO^{+}(aq) + OH^{-}(aq)$,
$K = 4 \times 10^{-10}$
(Given: $\log 2 = 0.3010$)

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A
$4.699$
B
$5.286$
C
$8.714$
D
$9.301$

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6-2.Equilibrium-II (Ionic Equilibrium)

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In a qualitative analysis, $Bi^{3+}$ is detected by the appearance of a precipitate of $BiO(OH)(s)$. Calculate the $pH$ when the following equilibrium exists at $298 \text{ K}$:$BiO(OH)(s) \rightleftharpoons BiO^{+}(aq) + OH^{-}(aq)$,$K = 4 \times 10^{-10}$(Given: $\log 2 = 0.3010$)

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Solubility product of $CaC_2O_4$ at a given temperature in pure water is $4 \times 10^{-9} \ (mol \ L^{-1})^2$. Solubility of $CaC_2O_4$ at the same temperature is

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In a qualitative analysis, $Bi^{3+}$ is detected by the appearance of a precipitate of $BiO(OH)(s)$. Calculate the $pH$ when the following equilibrium exists at $298 \text{ K}$:
$BiO(OH)(s) \rightleftharpoons BiO^{+}(aq) + OH^{-}(aq)$,
$K = 4 \times 10^{-10}$
(Given: $\log 2 = 0.3010$)