229494
The solubility of $\mathrm{AgCl}_{(\mathrm{s})}$ with solubility product $1.6 \times 10^{-10}$ in $0.1 \mathrm{M} \mathrm{NaCl}$ would be
1 $1.26 \times 10^{-5} \mathrm{M}$
2 $1.6 \times 10^{-9} \mathrm{M}$
3 $1.6 \times 10^{-11} \mathrm{M}$
4 zero
Explanation:
$\begin{aligned} & \mathrm{AgCl} \text { } \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \begin{array}{lll} \mathrm{a} & 0 & 0 \end{array} \\ & \mathrm{a}-\mathrm{S} \quad \mathrm{S} \quad \mathrm{S}+0.1 \\ & \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right] \cdot\left[\mathrm{Cl}^{-}\right] \\ & =[\mathrm{S}] .[0.1+\mathrm{S}] \\ & \end{aligned}$ $\mathrm{K}_{\mathrm{sp}}$ is value seems to very small so $\mathrm{S}$ value can be neglected, with to $0.1 \mathrm{~m}$ $\begin{aligned} \mathrm{K}_{5 \mathrm{pp}} & =\mathrm{S}[0.1] \\ \mathrm{S} & =\frac{\mathrm{K}_{\mathrm{pp}}}{0.1} \\ \mathrm{~S} & =\frac{1.6 \times 10^{-10}}{0.1} \\ \mathrm{~S} & =1.6 \times 10^{-9} \mathrm{M} \end{aligned}$
NEET-2016
Ionic Equilibrium
229495
MY and $\mathrm{NY}_3$ two nearly insoluble salts, have the same $K_{5 p}$ values of $6.2 \times 10^{-13}$ at room temperature. Which statement would be true in regard to $\mathrm{MY}$ and $\mathrm{NY}_3$ ?
1 The salts MY and $\mathrm{NY}_3$ are more soluble in $0.5 \mathrm{M} \mathrm{KY}$ than pure water
2 The addition of the salt of KY to solution of $\mathrm{MY}$ and $\mathrm{NY}_3$ will have no effect on their solubilities
3 The molar solubilities of $\mathrm{MY}$ and $\mathrm{NY}_3$ in water are identical
4 The molar solubility of MY in water is less than that of $\mathrm{NY}_3$
Explanation:
Molar solubility of MY in water is less than that of $\mathrm{NY}_3$ $\begin{aligned} & \mathrm{MY} \longrightarrow \mathrm{M}^{+}+\mathrm{Y}^{-} \\ & \mathrm{NY}_3 \longrightarrow \mathrm{N}^{3+}+3 \mathrm{Y}^{-} \end{aligned}$ For $\mathrm{MY}$ the molar solubility $=\sqrt{\mathrm{K}_{\mathrm{sp}}}$ $\begin{aligned} & =\sqrt{6.2 \times 10^{-13}} \\ & =7.88 \times 10^{-7} \mathrm{~m} . \end{aligned}$ For $\mathrm{NY}_3$ the molar solubility- $\begin{aligned} \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{ep}}}{27}\right)^{1 / 4} & =\left(\frac{1.2 \times 10^{-13}}{27}\right)^{1 / 4} \\ \mathrm{~S} & =3.9 \times 10^{-4} \mathrm{M} \end{aligned}$ Hence molar solubility of $\mathrm{MY}$ water is less than that of $\mathrm{NY}_3$
NEET-I 2016
Ionic Equilibrium
229497
Find out the solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.1 \mathrm{M}$ $\mathrm{NaOH}$. Given that the ionic product if $\mathrm{Ni}(\mathrm{OH})_2$ is $2 \times 10^{-15}$
229494
The solubility of $\mathrm{AgCl}_{(\mathrm{s})}$ with solubility product $1.6 \times 10^{-10}$ in $0.1 \mathrm{M} \mathrm{NaCl}$ would be
1 $1.26 \times 10^{-5} \mathrm{M}$
2 $1.6 \times 10^{-9} \mathrm{M}$
3 $1.6 \times 10^{-11} \mathrm{M}$
4 zero
Explanation:
$\begin{aligned} & \mathrm{AgCl} \text { } \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \begin{array}{lll} \mathrm{a} & 0 & 0 \end{array} \\ & \mathrm{a}-\mathrm{S} \quad \mathrm{S} \quad \mathrm{S}+0.1 \\ & \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right] \cdot\left[\mathrm{Cl}^{-}\right] \\ & =[\mathrm{S}] .[0.1+\mathrm{S}] \\ & \end{aligned}$ $\mathrm{K}_{\mathrm{sp}}$ is value seems to very small so $\mathrm{S}$ value can be neglected, with to $0.1 \mathrm{~m}$ $\begin{aligned} \mathrm{K}_{5 \mathrm{pp}} & =\mathrm{S}[0.1] \\ \mathrm{S} & =\frac{\mathrm{K}_{\mathrm{pp}}}{0.1} \\ \mathrm{~S} & =\frac{1.6 \times 10^{-10}}{0.1} \\ \mathrm{~S} & =1.6 \times 10^{-9} \mathrm{M} \end{aligned}$
NEET-2016
Ionic Equilibrium
229495
MY and $\mathrm{NY}_3$ two nearly insoluble salts, have the same $K_{5 p}$ values of $6.2 \times 10^{-13}$ at room temperature. Which statement would be true in regard to $\mathrm{MY}$ and $\mathrm{NY}_3$ ?
1 The salts MY and $\mathrm{NY}_3$ are more soluble in $0.5 \mathrm{M} \mathrm{KY}$ than pure water
2 The addition of the salt of KY to solution of $\mathrm{MY}$ and $\mathrm{NY}_3$ will have no effect on their solubilities
3 The molar solubilities of $\mathrm{MY}$ and $\mathrm{NY}_3$ in water are identical
4 The molar solubility of MY in water is less than that of $\mathrm{NY}_3$
Explanation:
Molar solubility of MY in water is less than that of $\mathrm{NY}_3$ $\begin{aligned} & \mathrm{MY} \longrightarrow \mathrm{M}^{+}+\mathrm{Y}^{-} \\ & \mathrm{NY}_3 \longrightarrow \mathrm{N}^{3+}+3 \mathrm{Y}^{-} \end{aligned}$ For $\mathrm{MY}$ the molar solubility $=\sqrt{\mathrm{K}_{\mathrm{sp}}}$ $\begin{aligned} & =\sqrt{6.2 \times 10^{-13}} \\ & =7.88 \times 10^{-7} \mathrm{~m} . \end{aligned}$ For $\mathrm{NY}_3$ the molar solubility- $\begin{aligned} \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{ep}}}{27}\right)^{1 / 4} & =\left(\frac{1.2 \times 10^{-13}}{27}\right)^{1 / 4} \\ \mathrm{~S} & =3.9 \times 10^{-4} \mathrm{M} \end{aligned}$ Hence molar solubility of $\mathrm{MY}$ water is less than that of $\mathrm{NY}_3$
NEET-I 2016
Ionic Equilibrium
229497
Find out the solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.1 \mathrm{M}$ $\mathrm{NaOH}$. Given that the ionic product if $\mathrm{Ni}(\mathrm{OH})_2$ is $2 \times 10^{-15}$
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ionic Equilibrium
229494
The solubility of $\mathrm{AgCl}_{(\mathrm{s})}$ with solubility product $1.6 \times 10^{-10}$ in $0.1 \mathrm{M} \mathrm{NaCl}$ would be
1 $1.26 \times 10^{-5} \mathrm{M}$
2 $1.6 \times 10^{-9} \mathrm{M}$
3 $1.6 \times 10^{-11} \mathrm{M}$
4 zero
Explanation:
$\begin{aligned} & \mathrm{AgCl} \text { } \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \begin{array}{lll} \mathrm{a} & 0 & 0 \end{array} \\ & \mathrm{a}-\mathrm{S} \quad \mathrm{S} \quad \mathrm{S}+0.1 \\ & \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right] \cdot\left[\mathrm{Cl}^{-}\right] \\ & =[\mathrm{S}] .[0.1+\mathrm{S}] \\ & \end{aligned}$ $\mathrm{K}_{\mathrm{sp}}$ is value seems to very small so $\mathrm{S}$ value can be neglected, with to $0.1 \mathrm{~m}$ $\begin{aligned} \mathrm{K}_{5 \mathrm{pp}} & =\mathrm{S}[0.1] \\ \mathrm{S} & =\frac{\mathrm{K}_{\mathrm{pp}}}{0.1} \\ \mathrm{~S} & =\frac{1.6 \times 10^{-10}}{0.1} \\ \mathrm{~S} & =1.6 \times 10^{-9} \mathrm{M} \end{aligned}$
NEET-2016
Ionic Equilibrium
229495
MY and $\mathrm{NY}_3$ two nearly insoluble salts, have the same $K_{5 p}$ values of $6.2 \times 10^{-13}$ at room temperature. Which statement would be true in regard to $\mathrm{MY}$ and $\mathrm{NY}_3$ ?
1 The salts MY and $\mathrm{NY}_3$ are more soluble in $0.5 \mathrm{M} \mathrm{KY}$ than pure water
2 The addition of the salt of KY to solution of $\mathrm{MY}$ and $\mathrm{NY}_3$ will have no effect on their solubilities
3 The molar solubilities of $\mathrm{MY}$ and $\mathrm{NY}_3$ in water are identical
4 The molar solubility of MY in water is less than that of $\mathrm{NY}_3$
Explanation:
Molar solubility of MY in water is less than that of $\mathrm{NY}_3$ $\begin{aligned} & \mathrm{MY} \longrightarrow \mathrm{M}^{+}+\mathrm{Y}^{-} \\ & \mathrm{NY}_3 \longrightarrow \mathrm{N}^{3+}+3 \mathrm{Y}^{-} \end{aligned}$ For $\mathrm{MY}$ the molar solubility $=\sqrt{\mathrm{K}_{\mathrm{sp}}}$ $\begin{aligned} & =\sqrt{6.2 \times 10^{-13}} \\ & =7.88 \times 10^{-7} \mathrm{~m} . \end{aligned}$ For $\mathrm{NY}_3$ the molar solubility- $\begin{aligned} \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{ep}}}{27}\right)^{1 / 4} & =\left(\frac{1.2 \times 10^{-13}}{27}\right)^{1 / 4} \\ \mathrm{~S} & =3.9 \times 10^{-4} \mathrm{M} \end{aligned}$ Hence molar solubility of $\mathrm{MY}$ water is less than that of $\mathrm{NY}_3$
NEET-I 2016
Ionic Equilibrium
229497
Find out the solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.1 \mathrm{M}$ $\mathrm{NaOH}$. Given that the ionic product if $\mathrm{Ni}(\mathrm{OH})_2$ is $2 \times 10^{-15}$
229494
The solubility of $\mathrm{AgCl}_{(\mathrm{s})}$ with solubility product $1.6 \times 10^{-10}$ in $0.1 \mathrm{M} \mathrm{NaCl}$ would be
1 $1.26 \times 10^{-5} \mathrm{M}$
2 $1.6 \times 10^{-9} \mathrm{M}$
3 $1.6 \times 10^{-11} \mathrm{M}$
4 zero
Explanation:
$\begin{aligned} & \mathrm{AgCl} \text { } \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \begin{array}{lll} \mathrm{a} & 0 & 0 \end{array} \\ & \mathrm{a}-\mathrm{S} \quad \mathrm{S} \quad \mathrm{S}+0.1 \\ & \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right] \cdot\left[\mathrm{Cl}^{-}\right] \\ & =[\mathrm{S}] .[0.1+\mathrm{S}] \\ & \end{aligned}$ $\mathrm{K}_{\mathrm{sp}}$ is value seems to very small so $\mathrm{S}$ value can be neglected, with to $0.1 \mathrm{~m}$ $\begin{aligned} \mathrm{K}_{5 \mathrm{pp}} & =\mathrm{S}[0.1] \\ \mathrm{S} & =\frac{\mathrm{K}_{\mathrm{pp}}}{0.1} \\ \mathrm{~S} & =\frac{1.6 \times 10^{-10}}{0.1} \\ \mathrm{~S} & =1.6 \times 10^{-9} \mathrm{M} \end{aligned}$
NEET-2016
Ionic Equilibrium
229495
MY and $\mathrm{NY}_3$ two nearly insoluble salts, have the same $K_{5 p}$ values of $6.2 \times 10^{-13}$ at room temperature. Which statement would be true in regard to $\mathrm{MY}$ and $\mathrm{NY}_3$ ?
1 The salts MY and $\mathrm{NY}_3$ are more soluble in $0.5 \mathrm{M} \mathrm{KY}$ than pure water
2 The addition of the salt of KY to solution of $\mathrm{MY}$ and $\mathrm{NY}_3$ will have no effect on their solubilities
3 The molar solubilities of $\mathrm{MY}$ and $\mathrm{NY}_3$ in water are identical
4 The molar solubility of MY in water is less than that of $\mathrm{NY}_3$
Explanation:
Molar solubility of MY in water is less than that of $\mathrm{NY}_3$ $\begin{aligned} & \mathrm{MY} \longrightarrow \mathrm{M}^{+}+\mathrm{Y}^{-} \\ & \mathrm{NY}_3 \longrightarrow \mathrm{N}^{3+}+3 \mathrm{Y}^{-} \end{aligned}$ For $\mathrm{MY}$ the molar solubility $=\sqrt{\mathrm{K}_{\mathrm{sp}}}$ $\begin{aligned} & =\sqrt{6.2 \times 10^{-13}} \\ & =7.88 \times 10^{-7} \mathrm{~m} . \end{aligned}$ For $\mathrm{NY}_3$ the molar solubility- $\begin{aligned} \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{ep}}}{27}\right)^{1 / 4} & =\left(\frac{1.2 \times 10^{-13}}{27}\right)^{1 / 4} \\ \mathrm{~S} & =3.9 \times 10^{-4} \mathrm{M} \end{aligned}$ Hence molar solubility of $\mathrm{MY}$ water is less than that of $\mathrm{NY}_3$
NEET-I 2016
Ionic Equilibrium
229497
Find out the solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.1 \mathrm{M}$ $\mathrm{NaOH}$. Given that the ionic product if $\mathrm{Ni}(\mathrm{OH})_2$ is $2 \times 10^{-15}$
229494
The solubility of $\mathrm{AgCl}_{(\mathrm{s})}$ with solubility product $1.6 \times 10^{-10}$ in $0.1 \mathrm{M} \mathrm{NaCl}$ would be
1 $1.26 \times 10^{-5} \mathrm{M}$
2 $1.6 \times 10^{-9} \mathrm{M}$
3 $1.6 \times 10^{-11} \mathrm{M}$
4 zero
Explanation:
$\begin{aligned} & \mathrm{AgCl} \text { } \mathrm{Ag}^{+}+\mathrm{Cl}^{-} \\ & \begin{array}{lll} \mathrm{a} & 0 & 0 \end{array} \\ & \mathrm{a}-\mathrm{S} \quad \mathrm{S} \quad \mathrm{S}+0.1 \\ & \mathrm{~K}_{\mathrm{sp}}=\left[\mathrm{Ag}^{+}\right] \cdot\left[\mathrm{Cl}^{-}\right] \\ & =[\mathrm{S}] .[0.1+\mathrm{S}] \\ & \end{aligned}$ $\mathrm{K}_{\mathrm{sp}}$ is value seems to very small so $\mathrm{S}$ value can be neglected, with to $0.1 \mathrm{~m}$ $\begin{aligned} \mathrm{K}_{5 \mathrm{pp}} & =\mathrm{S}[0.1] \\ \mathrm{S} & =\frac{\mathrm{K}_{\mathrm{pp}}}{0.1} \\ \mathrm{~S} & =\frac{1.6 \times 10^{-10}}{0.1} \\ \mathrm{~S} & =1.6 \times 10^{-9} \mathrm{M} \end{aligned}$
NEET-2016
Ionic Equilibrium
229495
MY and $\mathrm{NY}_3$ two nearly insoluble salts, have the same $K_{5 p}$ values of $6.2 \times 10^{-13}$ at room temperature. Which statement would be true in regard to $\mathrm{MY}$ and $\mathrm{NY}_3$ ?
1 The salts MY and $\mathrm{NY}_3$ are more soluble in $0.5 \mathrm{M} \mathrm{KY}$ than pure water
2 The addition of the salt of KY to solution of $\mathrm{MY}$ and $\mathrm{NY}_3$ will have no effect on their solubilities
3 The molar solubilities of $\mathrm{MY}$ and $\mathrm{NY}_3$ in water are identical
4 The molar solubility of MY in water is less than that of $\mathrm{NY}_3$
Explanation:
Molar solubility of MY in water is less than that of $\mathrm{NY}_3$ $\begin{aligned} & \mathrm{MY} \longrightarrow \mathrm{M}^{+}+\mathrm{Y}^{-} \\ & \mathrm{NY}_3 \longrightarrow \mathrm{N}^{3+}+3 \mathrm{Y}^{-} \end{aligned}$ For $\mathrm{MY}$ the molar solubility $=\sqrt{\mathrm{K}_{\mathrm{sp}}}$ $\begin{aligned} & =\sqrt{6.2 \times 10^{-13}} \\ & =7.88 \times 10^{-7} \mathrm{~m} . \end{aligned}$ For $\mathrm{NY}_3$ the molar solubility- $\begin{aligned} \mathrm{S}=\left(\frac{\mathrm{K}_{\mathrm{ep}}}{27}\right)^{1 / 4} & =\left(\frac{1.2 \times 10^{-13}}{27}\right)^{1 / 4} \\ \mathrm{~S} & =3.9 \times 10^{-4} \mathrm{M} \end{aligned}$ Hence molar solubility of $\mathrm{MY}$ water is less than that of $\mathrm{NY}_3$
NEET-I 2016
Ionic Equilibrium
229497
Find out the solubility of $\mathrm{Ni}(\mathrm{OH})_2$ in $0.1 \mathrm{M}$ $\mathrm{NaOH}$. Given that the ionic product if $\mathrm{Ni}(\mathrm{OH})_2$ is $2 \times 10^{-15}$