229765
The $\mathrm{pH}$ of the solution containing $50 \mathrm{~mL}$ each of $0.10 \mathrm{M}$ sodium acetate and $0.01 \mathrm{M}$ acetic acid is [Given $\mathrm{pK}_{\mathrm{n}}$ of $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right]$
229766
Which among the following has highest $\mathrm{pH}$ ?
1 $1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
2 $0.1 \mathrm{MNaOH}$
3 $1 \mathrm{M} \mathrm{HCl}$
4 $1 \mathrm{M} \mathrm{NaOH}$
Explanation:
For high value of $\mathrm{pH}$ we required max conc. of $\mathrm{OH}^{-}$ion. So, $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HCl}$ are acid. $0.1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=10^{-1} \mathrm{~m}$ $1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=1 \mathrm{~m}$ So, $1 \mathrm{~m} \mathrm{NaOH}$ has highest $\mathrm{pH}$.
Shift-II
Ionic Equilibrium
229768
The $\mathrm{pH}$ of $0.05 \mathrm{M}$ Solution of a strong dibasic acid is
229770
$0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted by 100 times. The pH of the solution so formed is
1 4
2 1
3 2
4 3
Explanation:
When $0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted 100 times, the molarity of solution will be $\frac{0.1}{100}=0.001 \mathrm{M}$ $\mathrm{HCl}$ is strong electrolyte. It completely dissociates. Its hydrogen ion concentration will be $0.001 \mathrm{M}$. $\begin{aligned} & \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-\log _{10}[0.001] \\ & \mathrm{pH}=3 \log _{10} 10 \\ & \mathrm{pH}=3 \end{aligned}$
Ionic Equilibrium
229771
The $\mathrm{pK}_{\mathrm{a}}$ of a certain weak acid is 4.0 . What should be the [salt] to [acid] ratio, if we have to prepare a buffer with a $\mathbf{p H}=5$ using the acid and one of the salts?
1 $4: 5$
2 $5: 4$
3 $10: 1$
4 $1: 10$
Explanation:
Given that, $\mathrm{pH}=5, \mathrm{pK}_{\mathrm{a}}=4$ According to Henderson equation for $\mathrm{pH}$ of an buffer is- $\begin{gathered} \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid] }} \\ 5=4+\log \frac{\text { [salt] }}{\text { [acid] }} \\ 1=\log \frac{\text { [salt ] }}{\text { [acid] }} \\ \therefore \quad \frac{[\text { salt }]}{[\text { acid }]}=10: 1 \end{gathered}$
229765
The $\mathrm{pH}$ of the solution containing $50 \mathrm{~mL}$ each of $0.10 \mathrm{M}$ sodium acetate and $0.01 \mathrm{M}$ acetic acid is [Given $\mathrm{pK}_{\mathrm{n}}$ of $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right]$
229766
Which among the following has highest $\mathrm{pH}$ ?
1 $1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
2 $0.1 \mathrm{MNaOH}$
3 $1 \mathrm{M} \mathrm{HCl}$
4 $1 \mathrm{M} \mathrm{NaOH}$
Explanation:
For high value of $\mathrm{pH}$ we required max conc. of $\mathrm{OH}^{-}$ion. So, $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HCl}$ are acid. $0.1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=10^{-1} \mathrm{~m}$ $1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=1 \mathrm{~m}$ So, $1 \mathrm{~m} \mathrm{NaOH}$ has highest $\mathrm{pH}$.
Shift-II
Ionic Equilibrium
229768
The $\mathrm{pH}$ of $0.05 \mathrm{M}$ Solution of a strong dibasic acid is
229770
$0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted by 100 times. The pH of the solution so formed is
1 4
2 1
3 2
4 3
Explanation:
When $0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted 100 times, the molarity of solution will be $\frac{0.1}{100}=0.001 \mathrm{M}$ $\mathrm{HCl}$ is strong electrolyte. It completely dissociates. Its hydrogen ion concentration will be $0.001 \mathrm{M}$. $\begin{aligned} & \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-\log _{10}[0.001] \\ & \mathrm{pH}=3 \log _{10} 10 \\ & \mathrm{pH}=3 \end{aligned}$
Ionic Equilibrium
229771
The $\mathrm{pK}_{\mathrm{a}}$ of a certain weak acid is 4.0 . What should be the [salt] to [acid] ratio, if we have to prepare a buffer with a $\mathbf{p H}=5$ using the acid and one of the salts?
1 $4: 5$
2 $5: 4$
3 $10: 1$
4 $1: 10$
Explanation:
Given that, $\mathrm{pH}=5, \mathrm{pK}_{\mathrm{a}}=4$ According to Henderson equation for $\mathrm{pH}$ of an buffer is- $\begin{gathered} \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid] }} \\ 5=4+\log \frac{\text { [salt] }}{\text { [acid] }} \\ 1=\log \frac{\text { [salt ] }}{\text { [acid] }} \\ \therefore \quad \frac{[\text { salt }]}{[\text { acid }]}=10: 1 \end{gathered}$
229765
The $\mathrm{pH}$ of the solution containing $50 \mathrm{~mL}$ each of $0.10 \mathrm{M}$ sodium acetate and $0.01 \mathrm{M}$ acetic acid is [Given $\mathrm{pK}_{\mathrm{n}}$ of $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right]$
229766
Which among the following has highest $\mathrm{pH}$ ?
1 $1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
2 $0.1 \mathrm{MNaOH}$
3 $1 \mathrm{M} \mathrm{HCl}$
4 $1 \mathrm{M} \mathrm{NaOH}$
Explanation:
For high value of $\mathrm{pH}$ we required max conc. of $\mathrm{OH}^{-}$ion. So, $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HCl}$ are acid. $0.1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=10^{-1} \mathrm{~m}$ $1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=1 \mathrm{~m}$ So, $1 \mathrm{~m} \mathrm{NaOH}$ has highest $\mathrm{pH}$.
Shift-II
Ionic Equilibrium
229768
The $\mathrm{pH}$ of $0.05 \mathrm{M}$ Solution of a strong dibasic acid is
229770
$0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted by 100 times. The pH of the solution so formed is
1 4
2 1
3 2
4 3
Explanation:
When $0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted 100 times, the molarity of solution will be $\frac{0.1}{100}=0.001 \mathrm{M}$ $\mathrm{HCl}$ is strong electrolyte. It completely dissociates. Its hydrogen ion concentration will be $0.001 \mathrm{M}$. $\begin{aligned} & \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-\log _{10}[0.001] \\ & \mathrm{pH}=3 \log _{10} 10 \\ & \mathrm{pH}=3 \end{aligned}$
Ionic Equilibrium
229771
The $\mathrm{pK}_{\mathrm{a}}$ of a certain weak acid is 4.0 . What should be the [salt] to [acid] ratio, if we have to prepare a buffer with a $\mathbf{p H}=5$ using the acid and one of the salts?
1 $4: 5$
2 $5: 4$
3 $10: 1$
4 $1: 10$
Explanation:
Given that, $\mathrm{pH}=5, \mathrm{pK}_{\mathrm{a}}=4$ According to Henderson equation for $\mathrm{pH}$ of an buffer is- $\begin{gathered} \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid] }} \\ 5=4+\log \frac{\text { [salt] }}{\text { [acid] }} \\ 1=\log \frac{\text { [salt ] }}{\text { [acid] }} \\ \therefore \quad \frac{[\text { salt }]}{[\text { acid }]}=10: 1 \end{gathered}$
229765
The $\mathrm{pH}$ of the solution containing $50 \mathrm{~mL}$ each of $0.10 \mathrm{M}$ sodium acetate and $0.01 \mathrm{M}$ acetic acid is [Given $\mathrm{pK}_{\mathrm{n}}$ of $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right]$
229766
Which among the following has highest $\mathrm{pH}$ ?
1 $1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
2 $0.1 \mathrm{MNaOH}$
3 $1 \mathrm{M} \mathrm{HCl}$
4 $1 \mathrm{M} \mathrm{NaOH}$
Explanation:
For high value of $\mathrm{pH}$ we required max conc. of $\mathrm{OH}^{-}$ion. So, $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HCl}$ are acid. $0.1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=10^{-1} \mathrm{~m}$ $1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=1 \mathrm{~m}$ So, $1 \mathrm{~m} \mathrm{NaOH}$ has highest $\mathrm{pH}$.
Shift-II
Ionic Equilibrium
229768
The $\mathrm{pH}$ of $0.05 \mathrm{M}$ Solution of a strong dibasic acid is
229770
$0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted by 100 times. The pH of the solution so formed is
1 4
2 1
3 2
4 3
Explanation:
When $0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted 100 times, the molarity of solution will be $\frac{0.1}{100}=0.001 \mathrm{M}$ $\mathrm{HCl}$ is strong electrolyte. It completely dissociates. Its hydrogen ion concentration will be $0.001 \mathrm{M}$. $\begin{aligned} & \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-\log _{10}[0.001] \\ & \mathrm{pH}=3 \log _{10} 10 \\ & \mathrm{pH}=3 \end{aligned}$
Ionic Equilibrium
229771
The $\mathrm{pK}_{\mathrm{a}}$ of a certain weak acid is 4.0 . What should be the [salt] to [acid] ratio, if we have to prepare a buffer with a $\mathbf{p H}=5$ using the acid and one of the salts?
1 $4: 5$
2 $5: 4$
3 $10: 1$
4 $1: 10$
Explanation:
Given that, $\mathrm{pH}=5, \mathrm{pK}_{\mathrm{a}}=4$ According to Henderson equation for $\mathrm{pH}$ of an buffer is- $\begin{gathered} \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid] }} \\ 5=4+\log \frac{\text { [salt] }}{\text { [acid] }} \\ 1=\log \frac{\text { [salt ] }}{\text { [acid] }} \\ \therefore \quad \frac{[\text { salt }]}{[\text { acid }]}=10: 1 \end{gathered}$
229765
The $\mathrm{pH}$ of the solution containing $50 \mathrm{~mL}$ each of $0.10 \mathrm{M}$ sodium acetate and $0.01 \mathrm{M}$ acetic acid is [Given $\mathrm{pK}_{\mathrm{n}}$ of $\left.\mathrm{CH}_3 \mathrm{COOH}=4.57\right]$
229766
Which among the following has highest $\mathrm{pH}$ ?
1 $1 \mathrm{M} \mathrm{H}_2 \mathrm{SO}_4$
2 $0.1 \mathrm{MNaOH}$
3 $1 \mathrm{M} \mathrm{HCl}$
4 $1 \mathrm{M} \mathrm{NaOH}$
Explanation:
For high value of $\mathrm{pH}$ we required max conc. of $\mathrm{OH}^{-}$ion. So, $\mathrm{H}_2 \mathrm{SO}_4$ and $\mathrm{HCl}$ are acid. $0.1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=10^{-1} \mathrm{~m}$ $1 \mathrm{~m} \mathrm{NaOH}=[\mathrm{OH}]=1 \mathrm{~m}$ So, $1 \mathrm{~m} \mathrm{NaOH}$ has highest $\mathrm{pH}$.
Shift-II
Ionic Equilibrium
229768
The $\mathrm{pH}$ of $0.05 \mathrm{M}$ Solution of a strong dibasic acid is
229770
$0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted by 100 times. The pH of the solution so formed is
1 4
2 1
3 2
4 3
Explanation:
When $0.1 \mathrm{M} \mathrm{HCl}$ solution is diluted 100 times, the molarity of solution will be $\frac{0.1}{100}=0.001 \mathrm{M}$ $\mathrm{HCl}$ is strong electrolyte. It completely dissociates. Its hydrogen ion concentration will be $0.001 \mathrm{M}$. $\begin{aligned} & \mathrm{pH}=-\log _{10}\left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-\log _{10}[0.001] \\ & \mathrm{pH}=3 \log _{10} 10 \\ & \mathrm{pH}=3 \end{aligned}$
Ionic Equilibrium
229771
The $\mathrm{pK}_{\mathrm{a}}$ of a certain weak acid is 4.0 . What should be the [salt] to [acid] ratio, if we have to prepare a buffer with a $\mathbf{p H}=5$ using the acid and one of the salts?
1 $4: 5$
2 $5: 4$
3 $10: 1$
4 $1: 10$
Explanation:
Given that, $\mathrm{pH}=5, \mathrm{pK}_{\mathrm{a}}=4$ According to Henderson equation for $\mathrm{pH}$ of an buffer is- $\begin{gathered} \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\text { [salt }]}{\text { [acid] }} \\ 5=4+\log \frac{\text { [salt] }}{\text { [acid] }} \\ 1=\log \frac{\text { [salt ] }}{\text { [acid] }} \\ \therefore \quad \frac{[\text { salt }]}{[\text { acid }]}=10: 1 \end{gathered}$
