229360
At $90^{\circ} \mathrm{C}$, the concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the value of $K_w$ at this temperature?
1 $10^{-6}$
2 $10^{-8}$
3 $10^{-14}$
4 $10^{-12}$
Explanation:
Given concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6}$ mole $\mathrm{L}^{-1}$ We know that, $\mid \begin{array}{} \because & \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ \therefore & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]} \\ \therefore & \mathrm{K}_{\mathrm{w}}=10^{-6} \times 10^{-6} \\ \text { or } & =10^{-12} \end{array}$
Manipal-2005
Ionic Equilibrium
229356
The $K_{\mathrm{sp}}$ for bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$ is $1.08 \times$ $10^{-73}$. The solubility of $\mathrm{Bi}_2 \mathrm{~S}_3$ in $\mathrm{mol} \mathrm{L}^{-1}$ at $298 \mathrm{~K}$ is
229363
Magnesium fluoride $\mathrm{MgF}_2$ is a slightly soluble salt whose solubility product $\mathrm{K}_{\mathrm{SP}}=3.7 \times 10^{-8}$ What is the approximate solubility of $\mathrm{MgF}_2$ ?
1 $9.2 \times 10^{-5} \mathrm{M}$
2 $1.2 \times 10^{-8} \mathrm{M}$
3 $1.4 \times 10^{-4} \mathrm{M}$
4 $2.1 \times 10^{-3} \mathrm{M}$
Explanation:
Given that, $\mathrm{K}_{\mathrm{sp}}=3.7 \times 10^{-8}$ The dissociation of $\mathrm{MgF}_2$ is $\mathrm{MgF}_2 \rightleftharpoons \mathrm{Mg}^{2+}+2 \mathrm{~F}^{-}$ Let, the solubility product is $\mathrm{s}$. $\begin{array}{ll} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{F}^{-}\right]^2 \\ \therefore & \mathrm{K}_{\mathrm{sp}}=\mathrm{s} \times(2 \mathrm{~s})^2 \\ & 3.7 \times 10^{-8}=4 \mathrm{~s}^3 \\ \text { or } & \mathrm{s}^3=\frac{3.7 \times 10^{-8}}{4}=0.925 \times 10^{-8} \\ & \mathrm{~s}^3=9.25 \times 10^{-9} \\ \text { or } & \mathrm{s}=2.1 \times 10^{-3} \mathrm{M} \end{array}$
229360
At $90^{\circ} \mathrm{C}$, the concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the value of $K_w$ at this temperature?
1 $10^{-6}$
2 $10^{-8}$
3 $10^{-14}$
4 $10^{-12}$
Explanation:
Given concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6}$ mole $\mathrm{L}^{-1}$ We know that, $\mid \begin{array}{} \because & \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ \therefore & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]} \\ \therefore & \mathrm{K}_{\mathrm{w}}=10^{-6} \times 10^{-6} \\ \text { or } & =10^{-12} \end{array}$
Manipal-2005
Ionic Equilibrium
229356
The $K_{\mathrm{sp}}$ for bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$ is $1.08 \times$ $10^{-73}$. The solubility of $\mathrm{Bi}_2 \mathrm{~S}_3$ in $\mathrm{mol} \mathrm{L}^{-1}$ at $298 \mathrm{~K}$ is
229363
Magnesium fluoride $\mathrm{MgF}_2$ is a slightly soluble salt whose solubility product $\mathrm{K}_{\mathrm{SP}}=3.7 \times 10^{-8}$ What is the approximate solubility of $\mathrm{MgF}_2$ ?
1 $9.2 \times 10^{-5} \mathrm{M}$
2 $1.2 \times 10^{-8} \mathrm{M}$
3 $1.4 \times 10^{-4} \mathrm{M}$
4 $2.1 \times 10^{-3} \mathrm{M}$
Explanation:
Given that, $\mathrm{K}_{\mathrm{sp}}=3.7 \times 10^{-8}$ The dissociation of $\mathrm{MgF}_2$ is $\mathrm{MgF}_2 \rightleftharpoons \mathrm{Mg}^{2+}+2 \mathrm{~F}^{-}$ Let, the solubility product is $\mathrm{s}$. $\begin{array}{ll} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{F}^{-}\right]^2 \\ \therefore & \mathrm{K}_{\mathrm{sp}}=\mathrm{s} \times(2 \mathrm{~s})^2 \\ & 3.7 \times 10^{-8}=4 \mathrm{~s}^3 \\ \text { or } & \mathrm{s}^3=\frac{3.7 \times 10^{-8}}{4}=0.925 \times 10^{-8} \\ & \mathrm{~s}^3=9.25 \times 10^{-9} \\ \text { or } & \mathrm{s}=2.1 \times 10^{-3} \mathrm{M} \end{array}$
229360
At $90^{\circ} \mathrm{C}$, the concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the value of $K_w$ at this temperature?
1 $10^{-6}$
2 $10^{-8}$
3 $10^{-14}$
4 $10^{-12}$
Explanation:
Given concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6}$ mole $\mathrm{L}^{-1}$ We know that, $\mid \begin{array}{} \because & \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ \therefore & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]} \\ \therefore & \mathrm{K}_{\mathrm{w}}=10^{-6} \times 10^{-6} \\ \text { or } & =10^{-12} \end{array}$
Manipal-2005
Ionic Equilibrium
229356
The $K_{\mathrm{sp}}$ for bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$ is $1.08 \times$ $10^{-73}$. The solubility of $\mathrm{Bi}_2 \mathrm{~S}_3$ in $\mathrm{mol} \mathrm{L}^{-1}$ at $298 \mathrm{~K}$ is
229363
Magnesium fluoride $\mathrm{MgF}_2$ is a slightly soluble salt whose solubility product $\mathrm{K}_{\mathrm{SP}}=3.7 \times 10^{-8}$ What is the approximate solubility of $\mathrm{MgF}_2$ ?
1 $9.2 \times 10^{-5} \mathrm{M}$
2 $1.2 \times 10^{-8} \mathrm{M}$
3 $1.4 \times 10^{-4} \mathrm{M}$
4 $2.1 \times 10^{-3} \mathrm{M}$
Explanation:
Given that, $\mathrm{K}_{\mathrm{sp}}=3.7 \times 10^{-8}$ The dissociation of $\mathrm{MgF}_2$ is $\mathrm{MgF}_2 \rightleftharpoons \mathrm{Mg}^{2+}+2 \mathrm{~F}^{-}$ Let, the solubility product is $\mathrm{s}$. $\begin{array}{ll} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{F}^{-}\right]^2 \\ \therefore & \mathrm{K}_{\mathrm{sp}}=\mathrm{s} \times(2 \mathrm{~s})^2 \\ & 3.7 \times 10^{-8}=4 \mathrm{~s}^3 \\ \text { or } & \mathrm{s}^3=\frac{3.7 \times 10^{-8}}{4}=0.925 \times 10^{-8} \\ & \mathrm{~s}^3=9.25 \times 10^{-9} \\ \text { or } & \mathrm{s}=2.1 \times 10^{-3} \mathrm{M} \end{array}$
229360
At $90^{\circ} \mathrm{C}$, the concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the value of $K_w$ at this temperature?
1 $10^{-6}$
2 $10^{-8}$
3 $10^{-14}$
4 $10^{-12}$
Explanation:
Given concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6}$ mole $\mathrm{L}^{-1}$ We know that, $\mid \begin{array}{} \because & \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ \therefore & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]} \\ \therefore & \mathrm{K}_{\mathrm{w}}=10^{-6} \times 10^{-6} \\ \text { or } & =10^{-12} \end{array}$
Manipal-2005
Ionic Equilibrium
229356
The $K_{\mathrm{sp}}$ for bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$ is $1.08 \times$ $10^{-73}$. The solubility of $\mathrm{Bi}_2 \mathrm{~S}_3$ in $\mathrm{mol} \mathrm{L}^{-1}$ at $298 \mathrm{~K}$ is
229363
Magnesium fluoride $\mathrm{MgF}_2$ is a slightly soluble salt whose solubility product $\mathrm{K}_{\mathrm{SP}}=3.7 \times 10^{-8}$ What is the approximate solubility of $\mathrm{MgF}_2$ ?
1 $9.2 \times 10^{-5} \mathrm{M}$
2 $1.2 \times 10^{-8} \mathrm{M}$
3 $1.4 \times 10^{-4} \mathrm{M}$
4 $2.1 \times 10^{-3} \mathrm{M}$
Explanation:
Given that, $\mathrm{K}_{\mathrm{sp}}=3.7 \times 10^{-8}$ The dissociation of $\mathrm{MgF}_2$ is $\mathrm{MgF}_2 \rightleftharpoons \mathrm{Mg}^{2+}+2 \mathrm{~F}^{-}$ Let, the solubility product is $\mathrm{s}$. $\begin{array}{ll} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{F}^{-}\right]^2 \\ \therefore & \mathrm{K}_{\mathrm{sp}}=\mathrm{s} \times(2 \mathrm{~s})^2 \\ & 3.7 \times 10^{-8}=4 \mathrm{~s}^3 \\ \text { or } & \mathrm{s}^3=\frac{3.7 \times 10^{-8}}{4}=0.925 \times 10^{-8} \\ & \mathrm{~s}^3=9.25 \times 10^{-9} \\ \text { or } & \mathrm{s}=2.1 \times 10^{-3} \mathrm{M} \end{array}$
229360
At $90^{\circ} \mathrm{C}$, the concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6} \mathrm{~mol} \mathrm{~L}^{-1}$. What is the value of $K_w$ at this temperature?
1 $10^{-6}$
2 $10^{-8}$
3 $10^{-14}$
4 $10^{-12}$
Explanation:
Given concentration of $\mathrm{H}_3 \mathrm{O}^{+}$in pure water is $10^{-6}$ mole $\mathrm{L}^{-1}$ We know that, $\mid \begin{array}{} \because & \mathrm{K}_{\mathrm{W}}=\left[\mathrm{H}^{+}\right]\left[\mathrm{OH}^{-}\right] \\ \therefore & {\left[\mathrm{H}_3 \mathrm{O}^{+}\right]=\left[\mathrm{OH}^{-}\right]} \\ \therefore & \mathrm{K}_{\mathrm{w}}=10^{-6} \times 10^{-6} \\ \text { or } & =10^{-12} \end{array}$
Manipal-2005
Ionic Equilibrium
229356
The $K_{\mathrm{sp}}$ for bismuth sulphide $\left(\mathrm{Bi}_2 \mathrm{~S}_3\right)$ is $1.08 \times$ $10^{-73}$. The solubility of $\mathrm{Bi}_2 \mathrm{~S}_3$ in $\mathrm{mol} \mathrm{L}^{-1}$ at $298 \mathrm{~K}$ is
229363
Magnesium fluoride $\mathrm{MgF}_2$ is a slightly soluble salt whose solubility product $\mathrm{K}_{\mathrm{SP}}=3.7 \times 10^{-8}$ What is the approximate solubility of $\mathrm{MgF}_2$ ?
1 $9.2 \times 10^{-5} \mathrm{M}$
2 $1.2 \times 10^{-8} \mathrm{M}$
3 $1.4 \times 10^{-4} \mathrm{M}$
4 $2.1 \times 10^{-3} \mathrm{M}$
Explanation:
Given that, $\mathrm{K}_{\mathrm{sp}}=3.7 \times 10^{-8}$ The dissociation of $\mathrm{MgF}_2$ is $\mathrm{MgF}_2 \rightleftharpoons \mathrm{Mg}^{2+}+2 \mathrm{~F}^{-}$ Let, the solubility product is $\mathrm{s}$. $\begin{array}{ll} & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Mg}^{+2}\right]\left[\mathrm{F}^{-}\right]^2 \\ \therefore & \mathrm{K}_{\mathrm{sp}}=\mathrm{s} \times(2 \mathrm{~s})^2 \\ & 3.7 \times 10^{-8}=4 \mathrm{~s}^3 \\ \text { or } & \mathrm{s}^3=\frac{3.7 \times 10^{-8}}{4}=0.925 \times 10^{-8} \\ & \mathrm{~s}^3=9.25 \times 10^{-9} \\ \text { or } & \mathrm{s}=2.1 \times 10^{-3} \mathrm{M} \end{array}$
