NEET Test Series from KOTA - 10 Papers In MS WORD
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Ionic Equilibrium
229542
Iron sulphide is heated in air to form $\mathrm{A}$, an oxide of sulphur. $A$ is dissolved in water to give an acid. The basicity of this acid is
229545
Which is the strongest Bronsted base among the following
1 $\mathrm{ClO}_4^{-}$
2 $\mathrm{ClO}_3^{-}$
3 $\mathrm{ClO}_2^{-}$
4 $\mathrm{ClO}^{-}$
Explanation:
Exp: $\begin{aligned} & \mathrm{ClO}_4^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_4 \\ & \mathrm{ClO}_3^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_3 \\ & \mathrm{ClO}_2^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_2 \\ & \mathrm{ClO}^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}^2 \end{aligned}$ The strength of acid is $\mathrm{HClO}_4>\mathrm{HClO}_3>\mathrm{HClO}_2>\mathrm{HClO}$ and strength of conjugate base is $\mathrm{ClO}_4^{-}<\mathrm{ClO}_3^{-}<\mathrm{ClO}_2^{-}<\mathrm{ClO}^{-}$ Hence $\mathrm{ClO}^{-}$is strongest bronsted base.
MPPET-2008
Ionic Equilibrium
229549
The $\mathrm{pk}_{\mathrm{n}}$ of a weak acid is 4.8 . What should be the ratio of $\frac{[\text { acid] }}{\text { [salt] }}$, if a buffer of $\mathrm{pH}=5.8$ is required?
1 0.1
2 10
3 1
4 2
Explanation:
Given the expression for the $\mathrm{pH}$ of the acidic buffer solution is as given below $\begin{aligned} & \mathrm{pH}=\mathrm{Pk}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid }]} \\ & \mathrm{Pk}_{\mathrm{a}}=4.8 \end{aligned}$ Substitute value in the above equation $5.8=4.8+\log \frac{[\text { salt }]}{[\text { acid }]}$ Hence, $\frac{[\text { salt }]}{[\text { acid }]}=\frac{10}{1}=10: 1$ i.e. $\frac{[\text { acid }]}{[\text { salt }]}=\frac{1}{10}=0.1$
1 $(\mathrm{a} \rightarrow$ iii), $(\mathrm{b} \rightarrow \mathrm{i}),(\mathrm{c} \rightarrow$ ii $),(\mathrm{d} \rightarrow$ iv $)$
2 $(\mathrm{a} \rightarrow \mathrm{i}),(\mathrm{b} \rightarrow$ iii) $,(\mathrm{c} \rightarrow$ ii $),(\mathrm{d} \rightarrow$ iv $)$
3 $(\mathrm{a} \rightarrow$ iii), $(\mathrm{b} \rightarrow$ ii $),(\mathrm{c} \rightarrow \mathrm{i}),(\mathrm{d} \rightarrow$ iv $)$
4 $(\mathrm{a} \rightarrow$ iii $),(\mathrm{b} \rightarrow$ ii $),(\mathrm{c} \rightarrow$ iv $),(\mathrm{d} \rightarrow \mathrm{i})$
Explanation:
A conjugate acid is formed when a proton is added to a base. $\begin{aligned} & \mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+} \\ & \mathrm{HCO}_3^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{CO}_3 \\ & \mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{O}^{+} \\ & \mathrm{HSO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{SO}_4 \end{aligned}$
229542
Iron sulphide is heated in air to form $\mathrm{A}$, an oxide of sulphur. $A$ is dissolved in water to give an acid. The basicity of this acid is
229545
Which is the strongest Bronsted base among the following
1 $\mathrm{ClO}_4^{-}$
2 $\mathrm{ClO}_3^{-}$
3 $\mathrm{ClO}_2^{-}$
4 $\mathrm{ClO}^{-}$
Explanation:
Exp: $\begin{aligned} & \mathrm{ClO}_4^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_4 \\ & \mathrm{ClO}_3^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_3 \\ & \mathrm{ClO}_2^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_2 \\ & \mathrm{ClO}^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}^2 \end{aligned}$ The strength of acid is $\mathrm{HClO}_4>\mathrm{HClO}_3>\mathrm{HClO}_2>\mathrm{HClO}$ and strength of conjugate base is $\mathrm{ClO}_4^{-}<\mathrm{ClO}_3^{-}<\mathrm{ClO}_2^{-}<\mathrm{ClO}^{-}$ Hence $\mathrm{ClO}^{-}$is strongest bronsted base.
MPPET-2008
Ionic Equilibrium
229549
The $\mathrm{pk}_{\mathrm{n}}$ of a weak acid is 4.8 . What should be the ratio of $\frac{[\text { acid] }}{\text { [salt] }}$, if a buffer of $\mathrm{pH}=5.8$ is required?
1 0.1
2 10
3 1
4 2
Explanation:
Given the expression for the $\mathrm{pH}$ of the acidic buffer solution is as given below $\begin{aligned} & \mathrm{pH}=\mathrm{Pk}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid }]} \\ & \mathrm{Pk}_{\mathrm{a}}=4.8 \end{aligned}$ Substitute value in the above equation $5.8=4.8+\log \frac{[\text { salt }]}{[\text { acid }]}$ Hence, $\frac{[\text { salt }]}{[\text { acid }]}=\frac{10}{1}=10: 1$ i.e. $\frac{[\text { acid }]}{[\text { salt }]}=\frac{1}{10}=0.1$
1 $(\mathrm{a} \rightarrow$ iii), $(\mathrm{b} \rightarrow \mathrm{i}),(\mathrm{c} \rightarrow$ ii $),(\mathrm{d} \rightarrow$ iv $)$
2 $(\mathrm{a} \rightarrow \mathrm{i}),(\mathrm{b} \rightarrow$ iii) $,(\mathrm{c} \rightarrow$ ii $),(\mathrm{d} \rightarrow$ iv $)$
3 $(\mathrm{a} \rightarrow$ iii), $(\mathrm{b} \rightarrow$ ii $),(\mathrm{c} \rightarrow \mathrm{i}),(\mathrm{d} \rightarrow$ iv $)$
4 $(\mathrm{a} \rightarrow$ iii $),(\mathrm{b} \rightarrow$ ii $),(\mathrm{c} \rightarrow$ iv $),(\mathrm{d} \rightarrow \mathrm{i})$
Explanation:
A conjugate acid is formed when a proton is added to a base. $\begin{aligned} & \mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+} \\ & \mathrm{HCO}_3^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{CO}_3 \\ & \mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{O}^{+} \\ & \mathrm{HSO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{SO}_4 \end{aligned}$
229542
Iron sulphide is heated in air to form $\mathrm{A}$, an oxide of sulphur. $A$ is dissolved in water to give an acid. The basicity of this acid is
229545
Which is the strongest Bronsted base among the following
1 $\mathrm{ClO}_4^{-}$
2 $\mathrm{ClO}_3^{-}$
3 $\mathrm{ClO}_2^{-}$
4 $\mathrm{ClO}^{-}$
Explanation:
Exp: $\begin{aligned} & \mathrm{ClO}_4^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_4 \\ & \mathrm{ClO}_3^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_3 \\ & \mathrm{ClO}_2^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_2 \\ & \mathrm{ClO}^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}^2 \end{aligned}$ The strength of acid is $\mathrm{HClO}_4>\mathrm{HClO}_3>\mathrm{HClO}_2>\mathrm{HClO}$ and strength of conjugate base is $\mathrm{ClO}_4^{-}<\mathrm{ClO}_3^{-}<\mathrm{ClO}_2^{-}<\mathrm{ClO}^{-}$ Hence $\mathrm{ClO}^{-}$is strongest bronsted base.
MPPET-2008
Ionic Equilibrium
229549
The $\mathrm{pk}_{\mathrm{n}}$ of a weak acid is 4.8 . What should be the ratio of $\frac{[\text { acid] }}{\text { [salt] }}$, if a buffer of $\mathrm{pH}=5.8$ is required?
1 0.1
2 10
3 1
4 2
Explanation:
Given the expression for the $\mathrm{pH}$ of the acidic buffer solution is as given below $\begin{aligned} & \mathrm{pH}=\mathrm{Pk}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid }]} \\ & \mathrm{Pk}_{\mathrm{a}}=4.8 \end{aligned}$ Substitute value in the above equation $5.8=4.8+\log \frac{[\text { salt }]}{[\text { acid }]}$ Hence, $\frac{[\text { salt }]}{[\text { acid }]}=\frac{10}{1}=10: 1$ i.e. $\frac{[\text { acid }]}{[\text { salt }]}=\frac{1}{10}=0.1$
1 $(\mathrm{a} \rightarrow$ iii), $(\mathrm{b} \rightarrow \mathrm{i}),(\mathrm{c} \rightarrow$ ii $),(\mathrm{d} \rightarrow$ iv $)$
2 $(\mathrm{a} \rightarrow \mathrm{i}),(\mathrm{b} \rightarrow$ iii) $,(\mathrm{c} \rightarrow$ ii $),(\mathrm{d} \rightarrow$ iv $)$
3 $(\mathrm{a} \rightarrow$ iii), $(\mathrm{b} \rightarrow$ ii $),(\mathrm{c} \rightarrow \mathrm{i}),(\mathrm{d} \rightarrow$ iv $)$
4 $(\mathrm{a} \rightarrow$ iii $),(\mathrm{b} \rightarrow$ ii $),(\mathrm{c} \rightarrow$ iv $),(\mathrm{d} \rightarrow \mathrm{i})$
Explanation:
A conjugate acid is formed when a proton is added to a base. $\begin{aligned} & \mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+} \\ & \mathrm{HCO}_3^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{CO}_3 \\ & \mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{O}^{+} \\ & \mathrm{HSO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{SO}_4 \end{aligned}$
229542
Iron sulphide is heated in air to form $\mathrm{A}$, an oxide of sulphur. $A$ is dissolved in water to give an acid. The basicity of this acid is
229545
Which is the strongest Bronsted base among the following
1 $\mathrm{ClO}_4^{-}$
2 $\mathrm{ClO}_3^{-}$
3 $\mathrm{ClO}_2^{-}$
4 $\mathrm{ClO}^{-}$
Explanation:
Exp: $\begin{aligned} & \mathrm{ClO}_4^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_4 \\ & \mathrm{ClO}_3^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_3 \\ & \mathrm{ClO}_2^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}_2 \\ & \mathrm{ClO}^{-} \stackrel{\mathrm{H}^{+}}{\longrightarrow} \mathrm{HClO}^2 \end{aligned}$ The strength of acid is $\mathrm{HClO}_4>\mathrm{HClO}_3>\mathrm{HClO}_2>\mathrm{HClO}$ and strength of conjugate base is $\mathrm{ClO}_4^{-}<\mathrm{ClO}_3^{-}<\mathrm{ClO}_2^{-}<\mathrm{ClO}^{-}$ Hence $\mathrm{ClO}^{-}$is strongest bronsted base.
MPPET-2008
Ionic Equilibrium
229549
The $\mathrm{pk}_{\mathrm{n}}$ of a weak acid is 4.8 . What should be the ratio of $\frac{[\text { acid] }}{\text { [salt] }}$, if a buffer of $\mathrm{pH}=5.8$ is required?
1 0.1
2 10
3 1
4 2
Explanation:
Given the expression for the $\mathrm{pH}$ of the acidic buffer solution is as given below $\begin{aligned} & \mathrm{pH}=\mathrm{Pk}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid }]} \\ & \mathrm{Pk}_{\mathrm{a}}=4.8 \end{aligned}$ Substitute value in the above equation $5.8=4.8+\log \frac{[\text { salt }]}{[\text { acid }]}$ Hence, $\frac{[\text { salt }]}{[\text { acid }]}=\frac{10}{1}=10: 1$ i.e. $\frac{[\text { acid }]}{[\text { salt }]}=\frac{1}{10}=0.1$
1 $(\mathrm{a} \rightarrow$ iii), $(\mathrm{b} \rightarrow \mathrm{i}),(\mathrm{c} \rightarrow$ ii $),(\mathrm{d} \rightarrow$ iv $)$
2 $(\mathrm{a} \rightarrow \mathrm{i}),(\mathrm{b} \rightarrow$ iii) $,(\mathrm{c} \rightarrow$ ii $),(\mathrm{d} \rightarrow$ iv $)$
3 $(\mathrm{a} \rightarrow$ iii), $(\mathrm{b} \rightarrow$ ii $),(\mathrm{c} \rightarrow \mathrm{i}),(\mathrm{d} \rightarrow$ iv $)$
4 $(\mathrm{a} \rightarrow$ iii $),(\mathrm{b} \rightarrow$ ii $),(\mathrm{c} \rightarrow$ iv $),(\mathrm{d} \rightarrow \mathrm{i})$
Explanation:
A conjugate acid is formed when a proton is added to a base. $\begin{aligned} & \mathrm{NH}_3+\mathrm{H}^{+} \longrightarrow \mathrm{NH}_4^{+} \\ & \mathrm{HCO}_3^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{CO}_3 \\ & \mathrm{H}_2 \mathrm{O}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{O}^{+} \\ & \mathrm{HSO}_4^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{H}_2 \mathrm{SO}_4 \end{aligned}$
