229802
The $\mathrm{pK}_{\mathrm{a}}$ of a weak acid, benzoic acid and $\mathrm{pK}_{\mathrm{b}}$ of a weak base, ammonium hydroxide are 4.25 and 4.75 respectively. Then the $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of ammonium benzoate will be
1 7.10
2 7.50
3 6.75
4 6.50
Explanation:
Ammonium benzoate \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4\right)\) is the salt weak benzoic acid \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH} \mathrm{pK}_{\mathrm{a}}=4.25\right)\) and weak ammonium hydroxide \(\left(\mathrm{NH}_4 \mathrm{OH}, \mathrm{pK}_{\mathrm{b}}=4.75\right)\). So, \(\mathrm{pH}\) of the solution of \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\) will be, \(\begin{aligned} & \mathrm{pH}=7+\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}}-\mathrm{pK}_{\mathrm{b}}\right) \\ & =7+\frac{1}{2}(4.25-4.75)=7-0.25=6.75 \end{aligned}\)
AP EAMCET (Engg.) 18.9.2020 Shift-I
Ionic Equilibrium
229803
If hydrogen electrode dipped in two solutions of $\mathrm{pH}=3$ and $\mathrm{pH}=6$ are connected by a salt bridge, the emf of the resulting cell is
229807
The $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of acetic acid will be [degree of dissociation of acetic acid is $\overline{0.0132}]$
1 4.32
2 3.14
3 1.14
4 2.88
Explanation:
Given that Degree of dissociation of acetic acid $=0.0132$ Concentration $=0.1 \mathrm{M}$ $\therefore$ Dissociation of acetic acid $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}$ So, amount of $\mathrm{H}_3 \mathrm{O}^{+}$ions formed $=0.0132 \times 0.1$ $=0.00132$ So, $\mathrm{pH}$ of acetic acid will be, $\begin{aligned} \mathrm{pH} & =-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\ & =-\log [0.00132] \\ \mathrm{pH} & =2.88 \end{aligned}$
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Ionic Equilibrium
229802
The $\mathrm{pK}_{\mathrm{a}}$ of a weak acid, benzoic acid and $\mathrm{pK}_{\mathrm{b}}$ of a weak base, ammonium hydroxide are 4.25 and 4.75 respectively. Then the $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of ammonium benzoate will be
1 7.10
2 7.50
3 6.75
4 6.50
Explanation:
Ammonium benzoate \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4\right)\) is the salt weak benzoic acid \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH} \mathrm{pK}_{\mathrm{a}}=4.25\right)\) and weak ammonium hydroxide \(\left(\mathrm{NH}_4 \mathrm{OH}, \mathrm{pK}_{\mathrm{b}}=4.75\right)\). So, \(\mathrm{pH}\) of the solution of \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\) will be, \(\begin{aligned} & \mathrm{pH}=7+\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}}-\mathrm{pK}_{\mathrm{b}}\right) \\ & =7+\frac{1}{2}(4.25-4.75)=7-0.25=6.75 \end{aligned}\)
AP EAMCET (Engg.) 18.9.2020 Shift-I
Ionic Equilibrium
229803
If hydrogen electrode dipped in two solutions of $\mathrm{pH}=3$ and $\mathrm{pH}=6$ are connected by a salt bridge, the emf of the resulting cell is
229807
The $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of acetic acid will be [degree of dissociation of acetic acid is $\overline{0.0132}]$
1 4.32
2 3.14
3 1.14
4 2.88
Explanation:
Given that Degree of dissociation of acetic acid $=0.0132$ Concentration $=0.1 \mathrm{M}$ $\therefore$ Dissociation of acetic acid $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}$ So, amount of $\mathrm{H}_3 \mathrm{O}^{+}$ions formed $=0.0132 \times 0.1$ $=0.00132$ So, $\mathrm{pH}$ of acetic acid will be, $\begin{aligned} \mathrm{pH} & =-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\ & =-\log [0.00132] \\ \mathrm{pH} & =2.88 \end{aligned}$
229802
The $\mathrm{pK}_{\mathrm{a}}$ of a weak acid, benzoic acid and $\mathrm{pK}_{\mathrm{b}}$ of a weak base, ammonium hydroxide are 4.25 and 4.75 respectively. Then the $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of ammonium benzoate will be
1 7.10
2 7.50
3 6.75
4 6.50
Explanation:
Ammonium benzoate \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4\right)\) is the salt weak benzoic acid \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH} \mathrm{pK}_{\mathrm{a}}=4.25\right)\) and weak ammonium hydroxide \(\left(\mathrm{NH}_4 \mathrm{OH}, \mathrm{pK}_{\mathrm{b}}=4.75\right)\). So, \(\mathrm{pH}\) of the solution of \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\) will be, \(\begin{aligned} & \mathrm{pH}=7+\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}}-\mathrm{pK}_{\mathrm{b}}\right) \\ & =7+\frac{1}{2}(4.25-4.75)=7-0.25=6.75 \end{aligned}\)
AP EAMCET (Engg.) 18.9.2020 Shift-I
Ionic Equilibrium
229803
If hydrogen electrode dipped in two solutions of $\mathrm{pH}=3$ and $\mathrm{pH}=6$ are connected by a salt bridge, the emf of the resulting cell is
229807
The $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of acetic acid will be [degree of dissociation of acetic acid is $\overline{0.0132}]$
1 4.32
2 3.14
3 1.14
4 2.88
Explanation:
Given that Degree of dissociation of acetic acid $=0.0132$ Concentration $=0.1 \mathrm{M}$ $\therefore$ Dissociation of acetic acid $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}$ So, amount of $\mathrm{H}_3 \mathrm{O}^{+}$ions formed $=0.0132 \times 0.1$ $=0.00132$ So, $\mathrm{pH}$ of acetic acid will be, $\begin{aligned} \mathrm{pH} & =-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\ & =-\log [0.00132] \\ \mathrm{pH} & =2.88 \end{aligned}$
229802
The $\mathrm{pK}_{\mathrm{a}}$ of a weak acid, benzoic acid and $\mathrm{pK}_{\mathrm{b}}$ of a weak base, ammonium hydroxide are 4.25 and 4.75 respectively. Then the $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of ammonium benzoate will be
1 7.10
2 7.50
3 6.75
4 6.50
Explanation:
Ammonium benzoate \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COONH}_4\right)\) is the salt weak benzoic acid \(\left(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH} \mathrm{pK}_{\mathrm{a}}=4.25\right)\) and weak ammonium hydroxide \(\left(\mathrm{NH}_4 \mathrm{OH}, \mathrm{pK}_{\mathrm{b}}=4.75\right)\). So, \(\mathrm{pH}\) of the solution of \(\mathrm{C}_6 \mathrm{H}_5 \mathrm{COOH}\) will be, \(\begin{aligned} & \mathrm{pH}=7+\frac{1}{2}\left(\mathrm{pK}_{\mathrm{a}}-\mathrm{pK}_{\mathrm{b}}\right) \\ & =7+\frac{1}{2}(4.25-4.75)=7-0.25=6.75 \end{aligned}\)
AP EAMCET (Engg.) 18.9.2020 Shift-I
Ionic Equilibrium
229803
If hydrogen electrode dipped in two solutions of $\mathrm{pH}=3$ and $\mathrm{pH}=6$ are connected by a salt bridge, the emf of the resulting cell is
229807
The $\mathrm{pH}$ of $0.1 \mathrm{M}$ solution of acetic acid will be [degree of dissociation of acetic acid is $\overline{0.0132}]$
1 4.32
2 3.14
3 1.14
4 2.88
Explanation:
Given that Degree of dissociation of acetic acid $=0.0132$ Concentration $=0.1 \mathrm{M}$ $\therefore$ Dissociation of acetic acid $\mathrm{CH}_3 \mathrm{COOH}+\mathrm{H}_2 \mathrm{O} \rightarrow \mathrm{CH}_3 \mathrm{COO}^{-}+\mathrm{H}_3 \mathrm{O}^{+}$ So, amount of $\mathrm{H}_3 \mathrm{O}^{+}$ions formed $=0.0132 \times 0.1$ $=0.00132$ So, $\mathrm{pH}$ of acetic acid will be, $\begin{aligned} \mathrm{pH} & =-\log \left[\mathrm{H}_3 \mathrm{O}^{+}\right] \\ & =-\log [0.00132] \\ \mathrm{pH} & =2.88 \end{aligned}$