NEET Test Series from KOTA - 10 Papers In MS WORD
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Ionic Equilibrium
229458
The expression for the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$ will be
1 $\mathrm{K}_{\mathrm{sp}}=\mathrm{s}^2$
2 $\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^3$
3 $\mathrm{K}_{\mathrm{sp}}=27 \mathrm{~s}^4$
4 $\mathrm{K}_{\mathrm{sp}}=\mathrm{s}$
Explanation:
Let, $\mathrm{K}_{\mathrm{sp}}$ be the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$. the ionisation of $\mathrm{Ag}_2 \mathrm{CO}_3$ is - $\begin{aligned} & \mathrm{Ag}_2 \mathrm{CO}_3 ? \mathrm{G} \mathrm{Ag}^{+}+\mathrm{Ag}^{2 \mathrm{~s}}+\mathrm{CO}_3{ }^{2-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2 \cdot\left[\mathrm{CO}_3{ }^{2-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^2(\mathrm{~s}) \\ & \mathrm{K}_{\mathrm{sp}}=\left(4 \mathrm{~s}^2\right) .(\mathrm{s}) \\ & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^3 \end{aligned}$
JIPMER-2011
Ionic Equilibrium
229459
The solubility of saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $\mathrm{s} \mathrm{mol} \mathrm{L}^{-1}$. What is its solubility product?
229460
A 0.1 aqueous solution of a weak acid is $2 \%$ ionised. If the ionic product of water is $1 \times 10^{-4}$, the $[\mathrm{OH}]$ is
1 $5 \times 10^{-12} \mathrm{M}$
2 $2 \times 10^{-3} \mathrm{M}$
3 $1 \times 10^{-14} \mathrm{M}$
4 None of these
Explanation:
Degree of dissociation of weak acid - $\alpha=\frac{2}{100}=0.02$ Concentration product of water $=1 \times 10^{-14}$ $\begin{aligned} & {\left[\mathrm{H}^{+}\right] .\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}} \\ & {\left[\mathrm{OH}^{-}\right]=\frac{1 \times 10^{-14}}{0.002}} \\ & {\left[\mathrm{OH}^{-}\right]=5 \times 10^{-12}} \end{aligned}$
JIPMER-2009
Ionic Equilibrium
229461
Solubility product of $\mathrm{PbCl}_2$ at $298 \mathrm{~K}$ is $1 \times 10^{-6}$, at this temperature solubility of $\mathrm{PbCl}_2$ in $\mathrm{mol} / \mathrm{L}$ is
229458
The expression for the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$ will be
1 $\mathrm{K}_{\mathrm{sp}}=\mathrm{s}^2$
2 $\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^3$
3 $\mathrm{K}_{\mathrm{sp}}=27 \mathrm{~s}^4$
4 $\mathrm{K}_{\mathrm{sp}}=\mathrm{s}$
Explanation:
Let, $\mathrm{K}_{\mathrm{sp}}$ be the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$. the ionisation of $\mathrm{Ag}_2 \mathrm{CO}_3$ is - $\begin{aligned} & \mathrm{Ag}_2 \mathrm{CO}_3 ? \mathrm{G} \mathrm{Ag}^{+}+\mathrm{Ag}^{2 \mathrm{~s}}+\mathrm{CO}_3{ }^{2-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2 \cdot\left[\mathrm{CO}_3{ }^{2-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^2(\mathrm{~s}) \\ & \mathrm{K}_{\mathrm{sp}}=\left(4 \mathrm{~s}^2\right) .(\mathrm{s}) \\ & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^3 \end{aligned}$
JIPMER-2011
Ionic Equilibrium
229459
The solubility of saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $\mathrm{s} \mathrm{mol} \mathrm{L}^{-1}$. What is its solubility product?
229460
A 0.1 aqueous solution of a weak acid is $2 \%$ ionised. If the ionic product of water is $1 \times 10^{-4}$, the $[\mathrm{OH}]$ is
1 $5 \times 10^{-12} \mathrm{M}$
2 $2 \times 10^{-3} \mathrm{M}$
3 $1 \times 10^{-14} \mathrm{M}$
4 None of these
Explanation:
Degree of dissociation of weak acid - $\alpha=\frac{2}{100}=0.02$ Concentration product of water $=1 \times 10^{-14}$ $\begin{aligned} & {\left[\mathrm{H}^{+}\right] .\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}} \\ & {\left[\mathrm{OH}^{-}\right]=\frac{1 \times 10^{-14}}{0.002}} \\ & {\left[\mathrm{OH}^{-}\right]=5 \times 10^{-12}} \end{aligned}$
JIPMER-2009
Ionic Equilibrium
229461
Solubility product of $\mathrm{PbCl}_2$ at $298 \mathrm{~K}$ is $1 \times 10^{-6}$, at this temperature solubility of $\mathrm{PbCl}_2$ in $\mathrm{mol} / \mathrm{L}$ is
229458
The expression for the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$ will be
1 $\mathrm{K}_{\mathrm{sp}}=\mathrm{s}^2$
2 $\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^3$
3 $\mathrm{K}_{\mathrm{sp}}=27 \mathrm{~s}^4$
4 $\mathrm{K}_{\mathrm{sp}}=\mathrm{s}$
Explanation:
Let, $\mathrm{K}_{\mathrm{sp}}$ be the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$. the ionisation of $\mathrm{Ag}_2 \mathrm{CO}_3$ is - $\begin{aligned} & \mathrm{Ag}_2 \mathrm{CO}_3 ? \mathrm{G} \mathrm{Ag}^{+}+\mathrm{Ag}^{2 \mathrm{~s}}+\mathrm{CO}_3{ }^{2-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2 \cdot\left[\mathrm{CO}_3{ }^{2-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^2(\mathrm{~s}) \\ & \mathrm{K}_{\mathrm{sp}}=\left(4 \mathrm{~s}^2\right) .(\mathrm{s}) \\ & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^3 \end{aligned}$
JIPMER-2011
Ionic Equilibrium
229459
The solubility of saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $\mathrm{s} \mathrm{mol} \mathrm{L}^{-1}$. What is its solubility product?
229460
A 0.1 aqueous solution of a weak acid is $2 \%$ ionised. If the ionic product of water is $1 \times 10^{-4}$, the $[\mathrm{OH}]$ is
1 $5 \times 10^{-12} \mathrm{M}$
2 $2 \times 10^{-3} \mathrm{M}$
3 $1 \times 10^{-14} \mathrm{M}$
4 None of these
Explanation:
Degree of dissociation of weak acid - $\alpha=\frac{2}{100}=0.02$ Concentration product of water $=1 \times 10^{-14}$ $\begin{aligned} & {\left[\mathrm{H}^{+}\right] .\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}} \\ & {\left[\mathrm{OH}^{-}\right]=\frac{1 \times 10^{-14}}{0.002}} \\ & {\left[\mathrm{OH}^{-}\right]=5 \times 10^{-12}} \end{aligned}$
JIPMER-2009
Ionic Equilibrium
229461
Solubility product of $\mathrm{PbCl}_2$ at $298 \mathrm{~K}$ is $1 \times 10^{-6}$, at this temperature solubility of $\mathrm{PbCl}_2$ in $\mathrm{mol} / \mathrm{L}$ is
229458
The expression for the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$ will be
1 $\mathrm{K}_{\mathrm{sp}}=\mathrm{s}^2$
2 $\mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^3$
3 $\mathrm{K}_{\mathrm{sp}}=27 \mathrm{~s}^4$
4 $\mathrm{K}_{\mathrm{sp}}=\mathrm{s}$
Explanation:
Let, $\mathrm{K}_{\mathrm{sp}}$ be the solubility product of $\mathrm{Ag}_2 \mathrm{CO}_3$. the ionisation of $\mathrm{Ag}_2 \mathrm{CO}_3$ is - $\begin{aligned} & \mathrm{Ag}_2 \mathrm{CO}_3 ? \mathrm{G} \mathrm{Ag}^{+}+\mathrm{Ag}^{2 \mathrm{~s}}+\mathrm{CO}_3{ }^{2-} \\ & \mathrm{K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right]^2 \cdot\left[\mathrm{CO}_3{ }^{2-}\right] \\ & \mathrm{K}_{\mathrm{sp}}=(2 \mathrm{~s})^2(\mathrm{~s}) \\ & \mathrm{K}_{\mathrm{sp}}=\left(4 \mathrm{~s}^2\right) .(\mathrm{s}) \\ & \mathrm{K}_{\mathrm{sp}}=4 \mathrm{~s}^3 \end{aligned}$
JIPMER-2011
Ionic Equilibrium
229459
The solubility of saturated solution of $\mathrm{Ag}_2 \mathrm{CrO}_4$ is $\mathrm{s} \mathrm{mol} \mathrm{L}^{-1}$. What is its solubility product?
229460
A 0.1 aqueous solution of a weak acid is $2 \%$ ionised. If the ionic product of water is $1 \times 10^{-4}$, the $[\mathrm{OH}]$ is
1 $5 \times 10^{-12} \mathrm{M}$
2 $2 \times 10^{-3} \mathrm{M}$
3 $1 \times 10^{-14} \mathrm{M}$
4 None of these
Explanation:
Degree of dissociation of weak acid - $\alpha=\frac{2}{100}=0.02$ Concentration product of water $=1 \times 10^{-14}$ $\begin{aligned} & {\left[\mathrm{H}^{+}\right] .\left[\mathrm{OH}^{-}\right]=1 \times 10^{-14}} \\ & {\left[\mathrm{OH}^{-}\right]=\frac{1 \times 10^{-14}}{0.002}} \\ & {\left[\mathrm{OH}^{-}\right]=5 \times 10^{-12}} \end{aligned}$
JIPMER-2009
Ionic Equilibrium
229461
Solubility product of $\mathrm{PbCl}_2$ at $298 \mathrm{~K}$ is $1 \times 10^{-6}$, at this temperature solubility of $\mathrm{PbCl}_2$ in $\mathrm{mol} / \mathrm{L}$ is
