NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ionic Equilibrium
229824
$\mathrm{pH}$ of a salt solution of weak acid $\left(\mathrm{pK}_{\mathrm{n}}=4\right)$ \& weak base $\left(\mathrm{pK}_{\mathrm{b}}=5\right)$ at $25^{\circ} \mathrm{C}$ is:
229825
The aqueous solution of which of the following salt will have the lowest $\mathrm{pH}$ ?
1 $\mathrm{NaClO}_3$
2 $\mathrm{NaClO}$
3 $\mathrm{NaClO}_4$
4 $\mathrm{NaClO}_2$
Explanation:
Salt of SA + SB : $\mathrm{pH}$ will remain neutral at 7 Salt of $\mathrm{SA}+\mathrm{WB}: \mathrm{pH}<7$ (acidic) Salt of WA $+\mathrm{SB}: \mathrm{pH}>7$ (basic) $\mathrm{NaClO}_4 \Rightarrow$ Neutral $(\mathrm{pH}=7)$ $\mathrm{HClO}_4+\mathrm{NaOH} \rightarrow \mathrm{NaClO}_4+\mathrm{H}_2 \mathrm{O}$ $(\mathrm{SA}) \quad(\mathrm{SB})$ Where, $\begin{aligned} \mathrm{SA} & =\text { strong acid } \\ \mathrm{SB} & =\text { strong base } \\ \mathrm{WB} & =\text { weak base } \\ \mathrm{WA} & =\text { weak acid } \end{aligned}$ $\mathrm{NaClO}_4$, on hydrolysis gives strongest acid $\mathrm{HClO}_4$ as compared to other salt. This is strongly ionized and gives highest concentration of $\mathrm{H}^{+}$therefore, aqueous solution of $\mathrm{NaCl}_4$ will have lowest $\mathrm{pH}$ value.
AIIMS-1996
Ionic Equilibrium
229827
The $\mathrm{pH}$ of $0.001(\mathrm{~N})$ acetic acid solution, which is $10 \%$ dissociated, is
1 3
2 1
3 4
4 2
Explanation:
Given that, Concentration of acetic acid $=0.001 \mathrm{~N}$ for $10 \%$ dissociation, the effective concentration of $\mathrm{H}^{+}$is $1 / 10^{\text {th }}$ of total concentration. Hence, effective concentration $\begin{aligned} & =\frac{1}{10} \times 0.001=0.0001 \mathrm{~N}=1 \times 10^{-4} \mathrm{~N} \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-\log \left[1 \times 10^{-4}\right] \\ & =4 \\ & \end{aligned}$
AIIMS-1996
Ionic Equilibrium
229828
Assertion : In a titration of weak acid and $\mathrm{NaOH}$, the $\mathrm{pH}$ at half equivalence point is $\mathrm{pK}_{\mathrm{a}}$. **Reason :** At half equivalence point, it forms an acidic buffer and the buffer capacity is maximum where [acid] $=$ [salt $]$
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both assertion and Reason are correct, but Reason is not the correct explanation of assertion.
3 If Assertion if correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}$ At the half-equivalence point say we have 10 moles of weak acid, and so there will be 5 moles of strong base as the name suggests (half equivalence). $\mathrm{HA}+\mathrm{OH}^{-} \rightarrow \mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O}$ for the half equivalent point, $[H . A]=\left[\mathrm{A}^{-}\right]$ So, $\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}$
229824
$\mathrm{pH}$ of a salt solution of weak acid $\left(\mathrm{pK}_{\mathrm{n}}=4\right)$ \& weak base $\left(\mathrm{pK}_{\mathrm{b}}=5\right)$ at $25^{\circ} \mathrm{C}$ is:
229825
The aqueous solution of which of the following salt will have the lowest $\mathrm{pH}$ ?
1 $\mathrm{NaClO}_3$
2 $\mathrm{NaClO}$
3 $\mathrm{NaClO}_4$
4 $\mathrm{NaClO}_2$
Explanation:
Salt of SA + SB : $\mathrm{pH}$ will remain neutral at 7 Salt of $\mathrm{SA}+\mathrm{WB}: \mathrm{pH}<7$ (acidic) Salt of WA $+\mathrm{SB}: \mathrm{pH}>7$ (basic) $\mathrm{NaClO}_4 \Rightarrow$ Neutral $(\mathrm{pH}=7)$ $\mathrm{HClO}_4+\mathrm{NaOH} \rightarrow \mathrm{NaClO}_4+\mathrm{H}_2 \mathrm{O}$ $(\mathrm{SA}) \quad(\mathrm{SB})$ Where, $\begin{aligned} \mathrm{SA} & =\text { strong acid } \\ \mathrm{SB} & =\text { strong base } \\ \mathrm{WB} & =\text { weak base } \\ \mathrm{WA} & =\text { weak acid } \end{aligned}$ $\mathrm{NaClO}_4$, on hydrolysis gives strongest acid $\mathrm{HClO}_4$ as compared to other salt. This is strongly ionized and gives highest concentration of $\mathrm{H}^{+}$therefore, aqueous solution of $\mathrm{NaCl}_4$ will have lowest $\mathrm{pH}$ value.
AIIMS-1996
Ionic Equilibrium
229827
The $\mathrm{pH}$ of $0.001(\mathrm{~N})$ acetic acid solution, which is $10 \%$ dissociated, is
1 3
2 1
3 4
4 2
Explanation:
Given that, Concentration of acetic acid $=0.001 \mathrm{~N}$ for $10 \%$ dissociation, the effective concentration of $\mathrm{H}^{+}$is $1 / 10^{\text {th }}$ of total concentration. Hence, effective concentration $\begin{aligned} & =\frac{1}{10} \times 0.001=0.0001 \mathrm{~N}=1 \times 10^{-4} \mathrm{~N} \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-\log \left[1 \times 10^{-4}\right] \\ & =4 \\ & \end{aligned}$
AIIMS-1996
Ionic Equilibrium
229828
Assertion : In a titration of weak acid and $\mathrm{NaOH}$, the $\mathrm{pH}$ at half equivalence point is $\mathrm{pK}_{\mathrm{a}}$. **Reason :** At half equivalence point, it forms an acidic buffer and the buffer capacity is maximum where [acid] $=$ [salt $]$
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both assertion and Reason are correct, but Reason is not the correct explanation of assertion.
3 If Assertion if correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}$ At the half-equivalence point say we have 10 moles of weak acid, and so there will be 5 moles of strong base as the name suggests (half equivalence). $\mathrm{HA}+\mathrm{OH}^{-} \rightarrow \mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O}$ for the half equivalent point, $[H . A]=\left[\mathrm{A}^{-}\right]$ So, $\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}$
229824
$\mathrm{pH}$ of a salt solution of weak acid $\left(\mathrm{pK}_{\mathrm{n}}=4\right)$ \& weak base $\left(\mathrm{pK}_{\mathrm{b}}=5\right)$ at $25^{\circ} \mathrm{C}$ is:
229825
The aqueous solution of which of the following salt will have the lowest $\mathrm{pH}$ ?
1 $\mathrm{NaClO}_3$
2 $\mathrm{NaClO}$
3 $\mathrm{NaClO}_4$
4 $\mathrm{NaClO}_2$
Explanation:
Salt of SA + SB : $\mathrm{pH}$ will remain neutral at 7 Salt of $\mathrm{SA}+\mathrm{WB}: \mathrm{pH}<7$ (acidic) Salt of WA $+\mathrm{SB}: \mathrm{pH}>7$ (basic) $\mathrm{NaClO}_4 \Rightarrow$ Neutral $(\mathrm{pH}=7)$ $\mathrm{HClO}_4+\mathrm{NaOH} \rightarrow \mathrm{NaClO}_4+\mathrm{H}_2 \mathrm{O}$ $(\mathrm{SA}) \quad(\mathrm{SB})$ Where, $\begin{aligned} \mathrm{SA} & =\text { strong acid } \\ \mathrm{SB} & =\text { strong base } \\ \mathrm{WB} & =\text { weak base } \\ \mathrm{WA} & =\text { weak acid } \end{aligned}$ $\mathrm{NaClO}_4$, on hydrolysis gives strongest acid $\mathrm{HClO}_4$ as compared to other salt. This is strongly ionized and gives highest concentration of $\mathrm{H}^{+}$therefore, aqueous solution of $\mathrm{NaCl}_4$ will have lowest $\mathrm{pH}$ value.
AIIMS-1996
Ionic Equilibrium
229827
The $\mathrm{pH}$ of $0.001(\mathrm{~N})$ acetic acid solution, which is $10 \%$ dissociated, is
1 3
2 1
3 4
4 2
Explanation:
Given that, Concentration of acetic acid $=0.001 \mathrm{~N}$ for $10 \%$ dissociation, the effective concentration of $\mathrm{H}^{+}$is $1 / 10^{\text {th }}$ of total concentration. Hence, effective concentration $\begin{aligned} & =\frac{1}{10} \times 0.001=0.0001 \mathrm{~N}=1 \times 10^{-4} \mathrm{~N} \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-\log \left[1 \times 10^{-4}\right] \\ & =4 \\ & \end{aligned}$
AIIMS-1996
Ionic Equilibrium
229828
Assertion : In a titration of weak acid and $\mathrm{NaOH}$, the $\mathrm{pH}$ at half equivalence point is $\mathrm{pK}_{\mathrm{a}}$. **Reason :** At half equivalence point, it forms an acidic buffer and the buffer capacity is maximum where [acid] $=$ [salt $]$
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both assertion and Reason are correct, but Reason is not the correct explanation of assertion.
3 If Assertion if correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}$ At the half-equivalence point say we have 10 moles of weak acid, and so there will be 5 moles of strong base as the name suggests (half equivalence). $\mathrm{HA}+\mathrm{OH}^{-} \rightarrow \mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O}$ for the half equivalent point, $[H . A]=\left[\mathrm{A}^{-}\right]$ So, $\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}$
229824
$\mathrm{pH}$ of a salt solution of weak acid $\left(\mathrm{pK}_{\mathrm{n}}=4\right)$ \& weak base $\left(\mathrm{pK}_{\mathrm{b}}=5\right)$ at $25^{\circ} \mathrm{C}$ is:
229825
The aqueous solution of which of the following salt will have the lowest $\mathrm{pH}$ ?
1 $\mathrm{NaClO}_3$
2 $\mathrm{NaClO}$
3 $\mathrm{NaClO}_4$
4 $\mathrm{NaClO}_2$
Explanation:
Salt of SA + SB : $\mathrm{pH}$ will remain neutral at 7 Salt of $\mathrm{SA}+\mathrm{WB}: \mathrm{pH}<7$ (acidic) Salt of WA $+\mathrm{SB}: \mathrm{pH}>7$ (basic) $\mathrm{NaClO}_4 \Rightarrow$ Neutral $(\mathrm{pH}=7)$ $\mathrm{HClO}_4+\mathrm{NaOH} \rightarrow \mathrm{NaClO}_4+\mathrm{H}_2 \mathrm{O}$ $(\mathrm{SA}) \quad(\mathrm{SB})$ Where, $\begin{aligned} \mathrm{SA} & =\text { strong acid } \\ \mathrm{SB} & =\text { strong base } \\ \mathrm{WB} & =\text { weak base } \\ \mathrm{WA} & =\text { weak acid } \end{aligned}$ $\mathrm{NaClO}_4$, on hydrolysis gives strongest acid $\mathrm{HClO}_4$ as compared to other salt. This is strongly ionized and gives highest concentration of $\mathrm{H}^{+}$therefore, aqueous solution of $\mathrm{NaCl}_4$ will have lowest $\mathrm{pH}$ value.
AIIMS-1996
Ionic Equilibrium
229827
The $\mathrm{pH}$ of $0.001(\mathrm{~N})$ acetic acid solution, which is $10 \%$ dissociated, is
1 3
2 1
3 4
4 2
Explanation:
Given that, Concentration of acetic acid $=0.001 \mathrm{~N}$ for $10 \%$ dissociation, the effective concentration of $\mathrm{H}^{+}$is $1 / 10^{\text {th }}$ of total concentration. Hence, effective concentration $\begin{aligned} & =\frac{1}{10} \times 0.001=0.0001 \mathrm{~N}=1 \times 10^{-4} \mathrm{~N} \\ & \mathrm{pH}=-\log \left[\mathrm{H}^{+}\right] \\ & \mathrm{pH}=-\log \left[1 \times 10^{-4}\right] \\ & =4 \\ & \end{aligned}$
AIIMS-1996
Ionic Equilibrium
229828
Assertion : In a titration of weak acid and $\mathrm{NaOH}$, the $\mathrm{pH}$ at half equivalence point is $\mathrm{pK}_{\mathrm{a}}$. **Reason :** At half equivalence point, it forms an acidic buffer and the buffer capacity is maximum where [acid] $=$ [salt $]$
1 If both Assertion and Reason are correct and the Reason is the correct explanation of Assertion.
2 If both assertion and Reason are correct, but Reason is not the correct explanation of assertion.
3 If Assertion if correct but Reason is incorrect.
4 If both the Assertion and Reason are incorrect.
Explanation:
$\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}+\log \frac{\left[\mathrm{A}^{-}\right]}{[\mathrm{HA}]}$ At the half-equivalence point say we have 10 moles of weak acid, and so there will be 5 moles of strong base as the name suggests (half equivalence). $\mathrm{HA}+\mathrm{OH}^{-} \rightarrow \mathrm{A}^{-}+\mathrm{H}_2 \mathrm{O}$ for the half equivalent point, $[H . A]=\left[\mathrm{A}^{-}\right]$ So, $\mathrm{pH}=\mathrm{pK}_{\mathrm{a}}$
