Solubility product Questions in English

(B) For the salt $AB_4$,the dissociation is:
$AB_{4(s)} \rightleftharpoons A^{4+}_{(aq)} + 4B^{-}_{(aq)}$
If $S$ is the molar solubility,then:
$[A^{4+}] = S$
$[B^{-}] = 4S$
The solubility product $K_{sp}$ is given by:
$K_{sp} = [A^{4+}][B^{-}]^4$
$K_{sp} = (S)(4S)^4$
$K_{sp} = S \cdot 256S^4$
$K_{sp} = 256S^5$
$S^5 = \frac{K_{sp}}{256}$
$S = \left(\frac{K_{sp}}{256}\right)^{1/5}$

(C) The dissociation of $Ni(OH)_2$ is given by: $Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^-$
Let the solubility be $S \ mol \ L^{-1}$. Then $[Ni^{2+}] = S$ and $[OH^-] = 2S$.
The solubility product expression is: $K_{sp} = [Ni^{2+}][OH^-]^2 = S(2S)^2 = 4S^3$.
Given $K_{sp} = 2 \times 10^{-15}$,we have $4S^3 = 2 \times 10^{-15}$,so $S^3 = 0.5 \times 10^{-15} = 5 \times 10^{-16}$.
$S = (5 \times 10^{-16})^{1/3} \approx 7.937 \times 10^{-6} \ mol \ L^{-1}$.
$[OH^-] = 2S = 2 \times 7.937 \times 10^{-6} = 1.587 \times 10^{-5} \ mol \ L^{-1}$.
$pOH = -\log[OH^-] = -\log(1.587 \times 10^{-5}) \approx 5 - 0.2 = 4.8$.
Since $pH + pOH = 14$,$pH = 14 - 4.8 = 9.2 \approx 9$.

(D) For a sparingly soluble salt of molecular formula $A_2 X_3$,if solubility in pure water is $y \ M$ at a given temperature:
$A_2 X_3 \rightleftharpoons 2 A^{3+} + 3 X^{2-}$
Initial concentrations: $0, 0$
Equilibrium concentrations: $2y \ M, 3y \ M$
Solubility product,$K_{sp} = [A^{3+}]^2 [X^{2-}]^3$
$K_{sp} = (2y)^2 \times (3y)^3$
$K_{sp} = 4y^2 \times 27y^3 = 108y^5 \ M^5$

(A) When equal volumes are mixed,the concentration of each ion is halved. Let the initial concentrations be $[Ca^{2+}]_i$ and $[F^{-}]_i$. The new concentrations are $[Ca^{2+}] = [Ca^{2+}]_i / 2$ and $[F^{-}] = [F^{-}]_i / 2$.
Precipitation occurs if the ionic product $Q_{sp} = [Ca^{2+}][F^{-}]^2 > K_{sp} = 1.6 \times 10^{-10}$.
For option $A$: $[Ca^{2+}] = 0.5 \times 10^{-2}$,$[F^{-}] = 0.5 \times 10^{-5}$. $Q_{sp} = (0.5 \times 10^{-2})(0.5 \times 10^{-5})^2 = 0.5 \times 0.25 \times 10^{-12} = 1.25 \times 10^{-13} < 1.6 \times 10^{-10}$. Precipitation will not occur.
For option $B$: $[Ca^{2+}] = 0.5 \times 10^{-3}$,$[F^{-}] = 0.5 \times 10^{-3}$. $Q_{sp} = (0.5 \times 10^{-3})(0.5 \times 10^{-3})^2 = 0.125 \times 10^{-9} = 1.25 \times 10^{-10} < 1.6 \times 10^{-10}$. Precipitation will not occur.
For option $C$: $[Ca^{2+}] = 0.5 \times 10^{-4}$,$[F^{-}] = 0.5 \times 10^{-2}$. $Q_{sp} = (0.5 \times 10^{-4})(0.5 \times 10^{-2})^2 = 0.5 \times 0.25 \times 10^{-8} = 1.25 \times 10^{-9} > 1.6 \times 10^{-10}$. Precipitation occurs.
For option $D$: $[Ca^{2+}] = 0.5 \times 10^{-2}$,$[F^{-}] = 0.5 \times 10^{-3}$. $Q_{sp} = (0.5 \times 10^{-2})(0.5 \times 10^{-3})^2 = 0.5 \times 0.25 \times 10^{-8} = 1.25 \times 10^{-9} > 1.6 \times 10^{-10}$. Precipitation occurs.
Note: Both $A$ and $B$ do not precipitate. Given the original question structure,$A$ is the most dilute case.

(B) The dissolution of $AgBr$ is represented by the equilibrium: $AgBr(s) \rightleftharpoons Ag^{+}(aq) + Br^{-}(aq)$.
The solubility product expression is $K_{sp} = [Ag^{+}][Br^{-}] = 5.0 \times 10^{-13}$.
In the presence of $0.1 \ M$ $NaBr$,the concentration of $Br^{-}$ ions is dominated by the strong electrolyte $NaBr$,so $[Br^{-}] \approx 0.1 \ M$.
Let the solubility of $AgBr$ be $S$. Then $[Ag^{+}] = S$.
Substituting these values into the $K_{sp}$ expression: $S \times 0.1 = 5.0 \times 10^{-13}$.
Solving for $S$: $S = (5.0 \times 10^{-13}) / 0.1 = 5.0 \times 10^{-12} \ M$.
Therefore,the solubility of $AgBr$ in $0.1 \ M$ $NaBr$ is $5.0 \times 10^{-12} \ M$.

(D) Let $S$ be the solubility of $CaSO_4$ in $mol/L$.
$CaSO_4(s) \rightleftharpoons Ca^{2+}(aq) + SO_4^{2-}(aq)$
$K_{sp} = [Ca^{2+}][SO_4^{2-}] = S^2 = 9 \times 10^{-6}$
$S = \sqrt{9 \times 10^{-6}} = 3 \times 10^{-3} \ M$ (or $mol/L$).
The molar mass of $CaSO_4 = 40 + 32 + (4 \times 16) = 136 \ g/mol$.
Solubility in $g/L = S \times \text{Molar Mass} = 3 \times 10^{-3} \ mol/L \times 136 \ g/mol = 0.408 \ g/L$.
This means $0.408 \ g$ of $CaSO_4$ dissolves in $1 \ L$ of water.
Therefore,the volume required to dissolve $1 \ g$ of $CaSO_4 = \frac{1 \ g}{0.408 \ g/L} \approx 2.45 \ L$.
Hence,the correct option is $D$.

(B) For the salt $A_2B$,the dissociation equilibrium is $A_2B \rightleftharpoons 2A^{+} + B^{2-}$.
Let the solubility be $s \ mol \cdot L^{-1}$.
The concentrations of the ions are $[A^{+}] = 2s$ and $[B^{2-}] = s$.
The solubility product expression is $K_{sp} = [A^{+}]^2 [B^{2-}] = (2s)^2(s) = 4s^3$.
Given $K_{sp} = 3.2 \times 10^{-11}$.
Equating the two: $4s^3 = 3.2 \times 10^{-11}$.
$s^3 = \frac{3.2 \times 10^{-11}}{4} = 0.8 \times 10^{-11} = 8 \times 10^{-12}$.
Taking the cube root: $s = \sqrt[3]{8 \times 10^{-12}} = 2 \times 10^{-4} \ mol \cdot L^{-1}$.

(A) For a salt of type $AB$: $K_{sp} = s^2$,so $s = \sqrt{K_{sp}} = \sqrt{4.0 \times 10^{-20}} = 2.0 \times 10^{-10} \ M$.
For a salt of type $A_2B$: $K_{sp} = 4s^3$,so $s = \sqrt[3]{K_{sp}/4} = \sqrt[3]{3.2 \times 10^{-11} / 4} = \sqrt[3]{8.0 \times 10^{-12}} = 2.0 \times 10^{-4} \ M$.
For a salt of type $AB_3$: $K_{sp} = 27s^4$,so $s = \sqrt[4]{K_{sp}/27} = \sqrt[4]{2.7 \times 10^{-31} / 27} = \sqrt[4]{1.0 \times 10^{-32}} = 1.0 \times 10^{-8} \ M$.
Comparing the solubilities: $2.0 \times 10^{-10} < 1.0 \times 10^{-8} < 2.0 \times 10^{-4}$.
Thus,the increasing order is $AB < AB_3 < A_2B$.

(D) The dissociation of $Zr_3(PO_4)_4$ is given by: $Zr_3(PO_4)_4(s) \rightleftharpoons 3Zr^{4+}(aq) + 4PO_4^{3-}(aq)$.
Let the solubility be $S$. Then $[Zr^{4+}] = 3S$ and $[PO_4^{3-}] = 4S$.
The solubility product $K_{sp}$ is expressed as: $K_{sp} = [Zr^{4+}]^3 [PO_4^{3-}]^4$.
Substituting the values: $K_{sp} = (3S)^3 (4S)^4$.
$K_{sp} = (27S^3) (256S^4) = 6912S^7$.
Therefore,$S^7 = \frac{K_{sp}}{6912}$.
$S = (\frac{K_{sp}}{6912})^{\frac{1}{7}}$.

(C) $NaOH$ is a strong electrolyte,so $[OH^-] = 1.0 \ M$.
Let the solubility of $Ni(OH)_2$ be $s \ M$.
$Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^-(aq)$
$K_{sp} = [Ni^{2+}][OH^-]^2$
$1.9 \times 10^{-15} = (s)(1.0 + 2s)^2$
Since $s$ is very small,we can approximate $(1.0 + 2s) \approx 1.0$.
$1.9 \times 10^{-15} = s \times (1.0)^2$
$s = 1.9 \times 10^{-15} \ M$.

(D) For the saturated solution of $MgCO_3$,the solubility product expression is $K_{sp}(MgCO_3) = [Mg^{2+}][CO_3^{2-}]$.
Given $[Mg^{2+}] = 3.2 \times 10^{-5} \ M$ and $K_{sp}(MgCO_3) = 1.6 \times 10^{-6}$.
Substituting these values: $1.6 \times 10^{-6} = (3.2 \times 10^{-5}) \times [CO_3^{2-}]$.
Therefore,$[CO_3^{2-}] = \frac{1.6 \times 10^{-6}}{3.2 \times 10^{-5}} = 0.5 \times 10^{-1} = 0.05 \ M$.
Now,for the saturated solution of $Ag_2CO_3$,the solubility product expression is $K_{sp}(Ag_2CO_3) = [Ag^{+}]^2[CO_3^{2-}]$.
Given $K_{sp}(Ag_2CO_3) = 8.0 \times 10^{-12}$ and $[CO_3^{2-}] = 0.05 \ M$.
Substituting these values: $8.0 \times 10^{-12} = [Ag^{+}]^2 \times (0.05)$.
$[Ag^{+}]^2 = \frac{8.0 \times 10^{-12}}{0.05} = 160 \times 10^{-12} = 1.6 \times 10^{-10}$.
$[Ag^{+}] = \sqrt{1.6 \times 10^{-10}} = \sqrt{1.6} \times 10^{-5} \ M$.

(D) The solubility equilibrium for $PbCl_2$ is: $PbCl_2(s) \rightleftharpoons Pb^{2+}(aq) + 2Cl^-(aq)$.
Let the solubility be $s \ mol/L$. Then $K_{sp} = [Pb^{2+}][Cl^-]^2 = (s)(2s)^2 = 4s^3$.
Given $K_{sp} = 3.2 \times 10^{-8}$,so $4s^3 = 3.2 \times 10^{-8} \implies s^3 = 0.8 \times 10^{-8} = 8 \times 10^{-9}$.
Thus,$s = 2 \times 10^{-3} \ mol/L$.
Molar mass of $PbCl_2 = 207 + 2 \times 35.5 = 278 \ g/mol$.
Mass of $PbCl_2$ in $1 \ L$ solution $= s \times \text{Molar mass} = 2 \times 10^{-3} \times 278 = 0.556 \ g/L$.
To dissolve $0.1 \ g$ of $PbCl_2$,the volume required $V = \frac{\text{mass}}{\text{solubility in } g/L} = \frac{0.1}{0.556} \approx 0.1798 \ L$.
Converting to $mL$,$V \approx 0.1798 \times 1000 \approx 180 \ mL$.

(B) The solubility product $(K_{sp})$ values for the given metal sulfides are:
$A$. $PbS$: $8.0 \times 10^{-28}$ $(II)$
$B$. $HgS$: $4.0 \times 10^{-53}$ $(I)$
$C$. $MnS$: $2.5 \times 10^{-13}$ $(IV)$
$D$. $ZnS$: $1.6 \times 10^{-24}$ $(III)$
Therefore,the correct matching sequence is $A-II, B-I, C-IV, D-III$.

(C) The dissociation of $Ca(OH)_2$ is given by: $Ca(OH)_2(s) \rightleftharpoons Ca^{2+}(aq) + 2OH^-(aq)$.
$K_{sp} = [Ca^{2+}][OH^-]^2 = 5.5 \times 10^{-6}$.
In $0.10 \ M \ NaOH$ solution,the concentration of $OH^-$ ions is $0.10 \ M$ due to complete dissociation of $NaOH$.
Let the molar solubility of $Ca(OH)_2$ be $S$.
Then $[Ca^{2+}] = S$ and $[OH^-] = (0.10 + 2S) \approx 0.10 \ M$ (since $S$ is very small).
Substituting these values into the $K_{sp}$ expression:
$5.5 \times 10^{-6} = S \times (0.10)^2$.
$5.5 \times 10^{-6} = S \times 0.01$.
$S = \frac{5.5 \times 10^{-6}}{0.01} = 5.5 \times 10^{-4} \ M$.

(A) The dissociation of calcium phosphate is given by: $Ca_3(PO_4)_2(s) \rightleftharpoons 3Ca^{2+}(aq) + 2PO_4^{3-}(aq)$.
Let the molar solubility be $S$.
Then,$[Ca^{2+}] = 3S$ and $[PO_4^{3-}] = 2S$.
The solubility product expression is $K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2$.
Substituting the values: $K_{sp} = (3S)^3 (2S)^2$.
$K_{sp} = (27S^3) (4S^2) = 108S^5$.
Thus,the correct option is $(A)$.

(D) The dissociation of $Ni(OH)_2$ is given by:
$Ni(OH)_2(s) \rightleftharpoons Ni^{2+}(aq) + 2OH^{-}(aq)$
Let the solubility be $s \ mol \ L^{-1}$.
Then,$[Ni^{2+}] = s$ and $[OH^{-}] = 2s$.
The solubility product expression is:
$K_{sp} = [Ni^{2+}][OH^{-}]^2$
$K_{sp} = (s)(2s)^2 = 4s^3$
Given $K_{sp} = 4.0 \times 10^{-15}$.
$4s^3 = 4.0 \times 10^{-15}$
$s^3 = 1.0 \times 10^{-15}$
$s = \sqrt[3]{1.0 \times 10^{-15}} = 1.0 \times 10^{-5} \ mol \ L^{-1}$

(D) The dissociation of $Ca_3(PO_4)_2$ is given by: $Ca_3(PO_4)_2 \rightleftharpoons 3 Ca^{2+} + 2 PO_4^{3-}$
If the solubility is $x \ mol \ L^{-1}$,then the concentration of $Ca^{2+}$ is $3x \ mol \ L^{-1}$ and the concentration of $PO_4^{3-}$ is $2x \ mol \ L^{-1}$.
The solubility product $K_{sp}$ is defined as: $K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2$
Substituting the values: $K_{sp} = (3x)^3 (2x)^2 = (27x^3) (4x^2) = 108x^5$.

(C) The dissociation of $NaOH$ is: $NaOH \rightarrow Na^+ + OH^-$. Since $NaOH$ is a strong electrolyte,$[OH^-] = 1.0 \ M$.
Let the solubility of $Ni(OH)_2$ be $s \ M$. The dissociation is: $Ni(OH)_2 \rightleftharpoons Ni^{2+} + 2OH^-$.
The concentration of $Ni^{2+}$ is $s$ and the total concentration of $OH^-$ is $(2s + 1.0) \ M$.
Given $K_{sp} = [Ni^{2+}][OH^-]^2 = 1.9 \times 10^{-15}$.
Substituting the values: $s(2s + 1.0)^2 = 1.9 \times 10^{-15}$.
Since $s$ is very small,we can neglect $2s$ compared to $1.0$.
Thus,$s(1.0)^2 = 1.9 \times 10^{-15}$.
$s = 1.9 \times 10^{-15} \ M$.

(A) The solubility of a sparingly soluble salt like $AgCl$ is affected by the common ion effect and the salt effect.
$1$. In $H_2O$,the solubility is $S = \sqrt{K_{sp}}$.
$2$. In $1 \ M \ NaCl$ and $1 \ M \ CaCl_2$,the common ion $Cl^-$ is present. Since $CaCl_2$ provides $2 \ M \ Cl^-$ ions and $NaCl$ provides $1 \ M \ Cl^-$ ions,the solubility is lowest in $CaCl_2$ due to the strongest common ion effect.
$3$. In $1 \ M \ NaNO_3$,the solubility is slightly higher than in pure water due to the salt effect (or diverse ion effect),which increases the ionic strength of the solution.
$4$. Therefore,the order of increasing solubility is: $CaCl_2 < NaCl < H_2O < NaNO_3$.

(D) For $MX$: $K_{sp} = S^2 = 4.0 \times 10^{-8} \Rightarrow S = 2 \times 10^{-4} \text{ mol dm}^{-3}$.
For $MX_2$: $K_{sp} = 4S^3 = 3.2 \times 10^{-14}$ $\Rightarrow S^3 = 8 \times 10^{-15}$ $\Rightarrow S = 2 \times 10^{-5} \text{ mol dm}^{-3}$.
For $M_3X$: $K_{sp} = 27S^4 = 2.7 \times 10^{-15}$ $\Rightarrow S^4 = 10^{-16}$ $\Rightarrow S = 1 \times 10^{-4} \text{ mol dm}^{-3}$.
Comparing the solubilities: $2 \times 10^{-4} > 1 \times 10^{-4} > 2 \times 10^{-5}$.
Thus,the order of solubility is $MX > M_3X > MX_2$.

(D) For $SrCO_3$,the solubility product expression is $K_{sp} = [Sr^{2+}][CO_3^{2-}]$.
Given $[CO_3^{2-}] = 1.2 \times 10^{-3} \ M$ and $K_{sp}(SrCO_3) = 7.0 \times 10^{-10}$,we calculate the concentration of $Sr^{2+}$:
$[Sr^{2+}] = \frac{7.0 \times 10^{-10}}{1.2 \times 10^{-3}} = 5.83 \times 10^{-7} \ M$.
For $SrF_2$,the solubility product expression is $K_{sp} = [Sr^{2+}][F^{-}]^2$.
Substituting the values: $7.9 \times 10^{-10} = (5.83 \times 10^{-7}) \times [F^{-}]^2$.
$[F^{-}]^2 = \frac{7.9 \times 10^{-10}}{5.83 \times 10^{-7}} = 1.355 \times 10^{-3}$.
$[F^{-}] = \sqrt{1.355 \times 10^{-3}} \approx 3.68 \times 10^{-2} \ M \approx 3.7 \times 10^{-2} \ M$.

(B) For the sparingly soluble salt $MX_4$:
$MX_{4(s)} \rightleftharpoons M^{4+}_{(aq)} + 4X^{-}_{(aq)}$
If the molar solubility is $S$,then:
$[M^{4+}] = S$
$[X^{-}] = 4S$
The solubility product $K_{sp}$ is given by:
$K_{sp} = [M^{4+}][X^{-}]^4$
$K_{sp} = (S)(4S)^4$
$K_{sp} = S \cdot 256S^4$
$K_{sp} = 256S^5$
$S^5 = \frac{K_{sp}}{256}$
$S = \left(\frac{K_{sp}}{256}\right)^{1/5}$

(B) For a salt of $MX_{2}$ type,the dissociation is given by: $MX_{2} \rightleftharpoons M^{2+} + 2X^-$.
If $s$ is the solubility in $mol / L$,then $[M^{2+}] = s$ and $[X^-] = 2s$.
The solubility product $K_{sp}$ is given by: $K_{sp} = [M^{2+}][X^-]^2 = (s)(2s)^2 = 4s^3$.
Given $K_{sp} = 3.2 \times 10^{-8}$.
So,$4s^3 = 3.2 \times 10^{-8}$.
$s^3 = \frac{3.2 \times 10^{-8}}{4} = 0.8 \times 10^{-8} = 8 \times 10^{-9}$.
Taking the cube root,$s = \sqrt[3]{8 \times 10^{-9}} = 2 \times 10^{-3} \ mol / L$.

(D) The dissociation of $Ca_3(PO_4)_2$ is given by:
$Ca_3(PO_4)_2{_{\text{(s)}}} \rightleftharpoons 3Ca^{2+}{_{\text{(aq)}}} + 2PO_4^{3-}{_{\text{(aq)}}}$
If the solubility is $y \text{ mol/L}$,then the concentration of ions at equilibrium is:
$[Ca^{2+}] = 3y$
$[PO_4^{3-}] = 2y$
The solubility product expression is:
$K_{sp} = [Ca^{2+}]^3 [PO_4^{3-}]^2$
Substituting the values:
$K_{sp} = (3y)^3 (2y)^2$
$K_{sp} = (27y^3) (4y^2)$
$K_{sp} = 108y^5$

(A) The solubility of metal chlorides in water depends on their lattice energy and hydration energy.
$NaCl$ is highly soluble due to high hydration energy.
$HgCl_{2}$ and $CuCl_{2}$ are soluble in water.
$Hg_{2}Cl_{2}$ (mercurous chloride),$CuCl$ (cuprous chloride),and $AgCl$ (silver chloride) are known to be insoluble in water.

(D) For the precipitation of $YS_{(s)}$,the condition is $[Y^{2+}][S^{2-}] \geq K_{sp}(YS)$.
Given $[Y^{2+}] = 0.01 \ M$ and $K_{sp}(YS) = 4 \times 10^{-16}$,we have $[S^{2-}] \geq \frac{4 \times 10^{-16}}{0.01} = 4 \times 10^{-14} \ M$.
For the dissociation of $H_2S$,the equilibrium is $H_2S_{(aq)} \rightleftharpoons 2H^{+}_{(aq)} + S^{2-}_{(aq)}$.
The equilibrium constant expression is $\frac{[S^{2-}][H^{+}]^2}{[H_2S]} = K_{a1} \times K_{a2} = 1.0 \times 10^{-21}$.
Substituting the values: $\frac{(4 \times 10^{-14})[H^{+}]^2}{0.1} = 1.0 \times 10^{-21}$.
$[H^{+}]^2 = \frac{1.0 \times 10^{-22}}{4 \times 10^{-14}} = 0.25 \times 10^{-8} = 25 \times 10^{-10}$.
$[H^{+}] = 5 \times 10^{-5} \ M$.
$pH = -\log[H^{+}] = -\log(5 \times 10^{-5}) = 5 - \log 5 = 5 - 0.70 = 4.3$.
The nearest integer value for the pH is $4$.

(D) For $Ag_2CrO_4$ (sparingly soluble salt of type $A_2B$): $K_{sp} = (2s_1)^2(s_1) = 4s_1^3 = 32x$.
Thus,$s_1^3 = 8x$,which gives $s_1 = 2\sqrt[3]{x}$.
For $AgBr$ (sparingly soluble salt of type $AB$): $K_{sp} = s_2^2 = 4y$.
Thus,$s_2 = \sqrt{4y} = 2\sqrt{y}$.
The ratio of molarities is $\frac{s_1}{s_2} = \frac{2\sqrt[3]{x}}{2\sqrt{y}} = \frac{\sqrt[3]{x}}{\sqrt{y}}$.

(A) The molar solubility $S$ is given by $S = \frac{x}{y} \text{ mol L}^{-1}$.
For the dissociation of the salt: $M_{3}A_{2} \rightleftharpoons 3M^{2+} + 2A^{3-}$.
The solubility product $K_{sp}$ is calculated as $K_{sp} = [3S]^{3} \cdot [2S]^{2} = 27S^{3} \cdot 4S^{2} = 108S^{5}$.
The molar concentration of the anion is $[A^{3-}] = 2S = 2 \left( \frac{x}{y} \right)$.
The required ratio is $\frac{[A^{3-}]}{K_{sp}} = \frac{2S}{108S^{5}} = \frac{1}{54S^{4}}$.
Substituting $S = \frac{x}{y}$,we get $\frac{1}{54(x/y)^{4}} = \frac{y^{4}}{54x^{4}}$.

(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,$pH = 14 - pOH = 14 - 4.699 = 9.301$.