229507
The solubility of $\mathrm{CaF}_2$ in pure water is $2.3 \times 10-4 \mathrm{~mol} \mathrm{dm}$ Its solubility product will be:
1 $4.6 \times 10^{-4}$
2 $4.6 \times 10^{-8}$
3 $6.9 \times 10^{-12}$
4 $4.9 \times 10^{-11}$
Explanation:
Solubility of $\mathrm{CaF}_2$, in pure water is : $2.3 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}$ $\begin{aligned} & \mathrm{CaF}_2 \square \mathrm{Ca}^{2+}+2 \mathrm{~F}^{-} \\ & {[\mathrm{S}] \quad[2 \mathrm{~S}]} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{F}^{-}\right]^2 \\ & {[\mathrm{~S}] \quad[2 \mathrm{~S}]^2} \\ & K_{5 p \mathrm{p}}=4 \mathrm{~s}^3 \\ & =4 \times\left(2.3 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4.9 \times 10^{-11} \\ & \end{aligned}$
UPTU/UPSEE-2006
Ionic Equilibrium
229508
The solubility product of Barium sulphate is $1.5 \times 10^{-9}$ at $18^{\circ} \mathrm{C}$. Its solubility in water at $18^{\circ} \mathrm{C}$ is:
229510
Solubility product of $\mathrm{Mg}(\mathrm{OH})_2$ at ordinary temp $1.96 \times 10^{-11} . \mathrm{pH}$ of a saturated soln. of $\mathbf{M g}$ $(\mathrm{OH})_2$ will be
229514
The molar solubility (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $\mathrm{MX}_4$ is ' $S$ '. The corresponding solubility product is $K_{\mathrm{sp}} . \mathrm{S}$ in terms of ' $K_{\mathrm{sp}}$ ' is given by the relation
229507
The solubility of $\mathrm{CaF}_2$ in pure water is $2.3 \times 10-4 \mathrm{~mol} \mathrm{dm}$ Its solubility product will be:
1 $4.6 \times 10^{-4}$
2 $4.6 \times 10^{-8}$
3 $6.9 \times 10^{-12}$
4 $4.9 \times 10^{-11}$
Explanation:
Solubility of $\mathrm{CaF}_2$, in pure water is : $2.3 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}$ $\begin{aligned} & \mathrm{CaF}_2 \square \mathrm{Ca}^{2+}+2 \mathrm{~F}^{-} \\ & {[\mathrm{S}] \quad[2 \mathrm{~S}]} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{F}^{-}\right]^2 \\ & {[\mathrm{~S}] \quad[2 \mathrm{~S}]^2} \\ & K_{5 p \mathrm{p}}=4 \mathrm{~s}^3 \\ & =4 \times\left(2.3 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4.9 \times 10^{-11} \\ & \end{aligned}$
UPTU/UPSEE-2006
Ionic Equilibrium
229508
The solubility product of Barium sulphate is $1.5 \times 10^{-9}$ at $18^{\circ} \mathrm{C}$. Its solubility in water at $18^{\circ} \mathrm{C}$ is:
229510
Solubility product of $\mathrm{Mg}(\mathrm{OH})_2$ at ordinary temp $1.96 \times 10^{-11} . \mathrm{pH}$ of a saturated soln. of $\mathbf{M g}$ $(\mathrm{OH})_2$ will be
229514
The molar solubility (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $\mathrm{MX}_4$ is ' $S$ '. The corresponding solubility product is $K_{\mathrm{sp}} . \mathrm{S}$ in terms of ' $K_{\mathrm{sp}}$ ' is given by the relation
NEET Test Series from KOTA - 10 Papers In MS WORD
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Ionic Equilibrium
229507
The solubility of $\mathrm{CaF}_2$ in pure water is $2.3 \times 10-4 \mathrm{~mol} \mathrm{dm}$ Its solubility product will be:
1 $4.6 \times 10^{-4}$
2 $4.6 \times 10^{-8}$
3 $6.9 \times 10^{-12}$
4 $4.9 \times 10^{-11}$
Explanation:
Solubility of $\mathrm{CaF}_2$, in pure water is : $2.3 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}$ $\begin{aligned} & \mathrm{CaF}_2 \square \mathrm{Ca}^{2+}+2 \mathrm{~F}^{-} \\ & {[\mathrm{S}] \quad[2 \mathrm{~S}]} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{F}^{-}\right]^2 \\ & {[\mathrm{~S}] \quad[2 \mathrm{~S}]^2} \\ & K_{5 p \mathrm{p}}=4 \mathrm{~s}^3 \\ & =4 \times\left(2.3 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4.9 \times 10^{-11} \\ & \end{aligned}$
UPTU/UPSEE-2006
Ionic Equilibrium
229508
The solubility product of Barium sulphate is $1.5 \times 10^{-9}$ at $18^{\circ} \mathrm{C}$. Its solubility in water at $18^{\circ} \mathrm{C}$ is:
229510
Solubility product of $\mathrm{Mg}(\mathrm{OH})_2$ at ordinary temp $1.96 \times 10^{-11} . \mathrm{pH}$ of a saturated soln. of $\mathbf{M g}$ $(\mathrm{OH})_2$ will be
229514
The molar solubility (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $\mathrm{MX}_4$ is ' $S$ '. The corresponding solubility product is $K_{\mathrm{sp}} . \mathrm{S}$ in terms of ' $K_{\mathrm{sp}}$ ' is given by the relation
229507
The solubility of $\mathrm{CaF}_2$ in pure water is $2.3 \times 10-4 \mathrm{~mol} \mathrm{dm}$ Its solubility product will be:
1 $4.6 \times 10^{-4}$
2 $4.6 \times 10^{-8}$
3 $6.9 \times 10^{-12}$
4 $4.9 \times 10^{-11}$
Explanation:
Solubility of $\mathrm{CaF}_2$, in pure water is : $2.3 \times 10^{-4} \mathrm{~mol} \mathrm{dm}^{-3}$ $\begin{aligned} & \mathrm{CaF}_2 \square \mathrm{Ca}^{2+}+2 \mathrm{~F}^{-} \\ & {[\mathrm{S}] \quad[2 \mathrm{~S}]} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ca}^{2+}\right] \cdot\left[\mathrm{F}^{-}\right]^2 \\ & {[\mathrm{~S}] \quad[2 \mathrm{~S}]^2} \\ & K_{5 p \mathrm{p}}=4 \mathrm{~s}^3 \\ & =4 \times\left(2.3 \times 10^{-4}\right)^3 \\ & \mathrm{~K}_{\mathrm{sp}}=4.9 \times 10^{-11} \\ & \end{aligned}$
UPTU/UPSEE-2006
Ionic Equilibrium
229508
The solubility product of Barium sulphate is $1.5 \times 10^{-9}$ at $18^{\circ} \mathrm{C}$. Its solubility in water at $18^{\circ} \mathrm{C}$ is:
229510
Solubility product of $\mathrm{Mg}(\mathrm{OH})_2$ at ordinary temp $1.96 \times 10^{-11} . \mathrm{pH}$ of a saturated soln. of $\mathbf{M g}$ $(\mathrm{OH})_2$ will be
229514
The molar solubility (in $\mathrm{mol} \mathrm{L}^{-1}$ ) of a sparingly soluble salt $\mathrm{MX}_4$ is ' $S$ '. The corresponding solubility product is $K_{\mathrm{sp}} . \mathrm{S}$ in terms of ' $K_{\mathrm{sp}}$ ' is given by the relation