NEET Test Series from KOTA - 10 Papers In MS WORD
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Ionic Equilibrium
229753
Buffer solutions have constant acidity and alkalinity because
1 These give unionised acid or base on reaction with added acid on alkali
2 Acids and alkalies in these solution are shielded from attack by other ions
3 They have large excess of $\mathrm{H}^{+}$or $\mathrm{OH}^{-}$ions
4 They have fixed value of $\mathrm{pH}$
Explanation:
If small amount of an acid or alkali is added to a buffer solution, it converts them into unionized acid or base. Thus, its $\mathrm{pH}$ remains unaffected or in other words its acidity/alkalinity remains constant. e.g. $\begin{aligned} & \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-} \text {?? } \mathrm{H}_2 \mathrm{O}+\mathrm{HA} \\ & { }^{-} \mathrm{OH}+\mathrm{HA} \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{A}^{-} \end{aligned}$ If acid is added, it reacts with $\mathrm{A}^{-}$to form undissociated HA similarly, if base/alkali is added, $\mathrm{OH}^{-}$combines with $\mathrm{HA}$ to give $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{A}^{-}$and thus, maintains the acidity/alkalinity of buffer solution.
NEET-2012
Ionic Equilibrium
229755
In a buffer solution containing concentration of $\mathrm{B}^{-}$and $\mathrm{HB}$, the $K_b$ for $\mathrm{B}^{-}$is $10^{-10}$. The $\mathrm{pH}$ of buffer solution is
1 10
2 7
3 6
4 4
Explanation:
Given, $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{B}^{-}$is $10^{-10}$ $\therefore \mathrm{pK}_{\mathrm{b}}=-\log 10 \mathrm{~K}_{\mathrm{b}}=-\log \left(10^{-10}\right)=10$ According to the Henderson's equation $\text { We know that } \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { salt }]}{[\text { Acid }]}$ Here, the concentration of the salt and acid is same $\begin{aligned} & \therefore \log _{10} 1=0 \\ & \therefore \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}=10 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-10=4 \end{aligned}$
NEET-2010
Ionic Equilibrium
229758
Solution of $0.1 \mathrm{NH}_4 \mathrm{OH}$ and $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{Cl}$ has pH 9.25. Then find out $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_4 \mathrm{OH}$.
1 9.25
2 4.75
3 3.75
4 8.25
Explanation:
Given that - Normality of $\mathrm{NH}_4 \mathrm{OH}=0.1 \mathrm{~N}$ Molarity of $\mathrm{NH}_4 \mathrm{Cl}=0.1 \mathrm{M}$ $\mathrm{pH}=9.25$ According to the Henderson's equation- $\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { Salt }]}{[\text { Base }]} \\ & \mathrm{pH}+\mathrm{pOH}=14\left(\text { at } 25^{\circ} \mathrm{C}\right) \\ & \mathrm{pOH}=14-9.5=4.75 \\ & 4.75=\mathrm{pK}_{\mathrm{b}}+\log \frac{[1]}{[1]} \\ & \mathrm{pK}_{\mathrm{b}}=4.75 \end{aligned}$
BCECE-2012]
Ionic Equilibrium
229759
A physician wishes to prepare a buffer solution at $\mathbf{p H}=3.85$ that efficiently resists changes in pH yet contains only small concentration of the buffering agents. Which of the following weak acids together with its sodium salt would be best to use?
By the use of Henderson's equation $\mathrm{pH}=\mathrm{pK}_2+\log _{10} \frac{[\text { salt }]}{[\text { acid }]}$ When, $[$ salt $]=[$ acid $]$ $\therefore \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}$ $\because \mathrm{pK}_{\mathrm{n}}=3.58$, thus at ths state $\mathrm{pH}=3.58$ So, acetoacetic acid $\left(\mathrm{pK}_{\mathrm{a}}=3.58\right)$ is best to use.
229753
Buffer solutions have constant acidity and alkalinity because
1 These give unionised acid or base on reaction with added acid on alkali
2 Acids and alkalies in these solution are shielded from attack by other ions
3 They have large excess of $\mathrm{H}^{+}$or $\mathrm{OH}^{-}$ions
4 They have fixed value of $\mathrm{pH}$
Explanation:
If small amount of an acid or alkali is added to a buffer solution, it converts them into unionized acid or base. Thus, its $\mathrm{pH}$ remains unaffected or in other words its acidity/alkalinity remains constant. e.g. $\begin{aligned} & \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-} \text {?? } \mathrm{H}_2 \mathrm{O}+\mathrm{HA} \\ & { }^{-} \mathrm{OH}+\mathrm{HA} \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{A}^{-} \end{aligned}$ If acid is added, it reacts with $\mathrm{A}^{-}$to form undissociated HA similarly, if base/alkali is added, $\mathrm{OH}^{-}$combines with $\mathrm{HA}$ to give $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{A}^{-}$and thus, maintains the acidity/alkalinity of buffer solution.
NEET-2012
Ionic Equilibrium
229755
In a buffer solution containing concentration of $\mathrm{B}^{-}$and $\mathrm{HB}$, the $K_b$ for $\mathrm{B}^{-}$is $10^{-10}$. The $\mathrm{pH}$ of buffer solution is
1 10
2 7
3 6
4 4
Explanation:
Given, $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{B}^{-}$is $10^{-10}$ $\therefore \mathrm{pK}_{\mathrm{b}}=-\log 10 \mathrm{~K}_{\mathrm{b}}=-\log \left(10^{-10}\right)=10$ According to the Henderson's equation $\text { We know that } \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { salt }]}{[\text { Acid }]}$ Here, the concentration of the salt and acid is same $\begin{aligned} & \therefore \log _{10} 1=0 \\ & \therefore \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}=10 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-10=4 \end{aligned}$
NEET-2010
Ionic Equilibrium
229758
Solution of $0.1 \mathrm{NH}_4 \mathrm{OH}$ and $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{Cl}$ has pH 9.25. Then find out $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_4 \mathrm{OH}$.
1 9.25
2 4.75
3 3.75
4 8.25
Explanation:
Given that - Normality of $\mathrm{NH}_4 \mathrm{OH}=0.1 \mathrm{~N}$ Molarity of $\mathrm{NH}_4 \mathrm{Cl}=0.1 \mathrm{M}$ $\mathrm{pH}=9.25$ According to the Henderson's equation- $\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { Salt }]}{[\text { Base }]} \\ & \mathrm{pH}+\mathrm{pOH}=14\left(\text { at } 25^{\circ} \mathrm{C}\right) \\ & \mathrm{pOH}=14-9.5=4.75 \\ & 4.75=\mathrm{pK}_{\mathrm{b}}+\log \frac{[1]}{[1]} \\ & \mathrm{pK}_{\mathrm{b}}=4.75 \end{aligned}$
BCECE-2012]
Ionic Equilibrium
229759
A physician wishes to prepare a buffer solution at $\mathbf{p H}=3.85$ that efficiently resists changes in pH yet contains only small concentration of the buffering agents. Which of the following weak acids together with its sodium salt would be best to use?
By the use of Henderson's equation $\mathrm{pH}=\mathrm{pK}_2+\log _{10} \frac{[\text { salt }]}{[\text { acid }]}$ When, $[$ salt $]=[$ acid $]$ $\therefore \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}$ $\because \mathrm{pK}_{\mathrm{n}}=3.58$, thus at ths state $\mathrm{pH}=3.58$ So, acetoacetic acid $\left(\mathrm{pK}_{\mathrm{a}}=3.58\right)$ is best to use.
229753
Buffer solutions have constant acidity and alkalinity because
1 These give unionised acid or base on reaction with added acid on alkali
2 Acids and alkalies in these solution are shielded from attack by other ions
3 They have large excess of $\mathrm{H}^{+}$or $\mathrm{OH}^{-}$ions
4 They have fixed value of $\mathrm{pH}$
Explanation:
If small amount of an acid or alkali is added to a buffer solution, it converts them into unionized acid or base. Thus, its $\mathrm{pH}$ remains unaffected or in other words its acidity/alkalinity remains constant. e.g. $\begin{aligned} & \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-} \text {?? } \mathrm{H}_2 \mathrm{O}+\mathrm{HA} \\ & { }^{-} \mathrm{OH}+\mathrm{HA} \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{A}^{-} \end{aligned}$ If acid is added, it reacts with $\mathrm{A}^{-}$to form undissociated HA similarly, if base/alkali is added, $\mathrm{OH}^{-}$combines with $\mathrm{HA}$ to give $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{A}^{-}$and thus, maintains the acidity/alkalinity of buffer solution.
NEET-2012
Ionic Equilibrium
229755
In a buffer solution containing concentration of $\mathrm{B}^{-}$and $\mathrm{HB}$, the $K_b$ for $\mathrm{B}^{-}$is $10^{-10}$. The $\mathrm{pH}$ of buffer solution is
1 10
2 7
3 6
4 4
Explanation:
Given, $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{B}^{-}$is $10^{-10}$ $\therefore \mathrm{pK}_{\mathrm{b}}=-\log 10 \mathrm{~K}_{\mathrm{b}}=-\log \left(10^{-10}\right)=10$ According to the Henderson's equation $\text { We know that } \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { salt }]}{[\text { Acid }]}$ Here, the concentration of the salt and acid is same $\begin{aligned} & \therefore \log _{10} 1=0 \\ & \therefore \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}=10 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-10=4 \end{aligned}$
NEET-2010
Ionic Equilibrium
229758
Solution of $0.1 \mathrm{NH}_4 \mathrm{OH}$ and $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{Cl}$ has pH 9.25. Then find out $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_4 \mathrm{OH}$.
1 9.25
2 4.75
3 3.75
4 8.25
Explanation:
Given that - Normality of $\mathrm{NH}_4 \mathrm{OH}=0.1 \mathrm{~N}$ Molarity of $\mathrm{NH}_4 \mathrm{Cl}=0.1 \mathrm{M}$ $\mathrm{pH}=9.25$ According to the Henderson's equation- $\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { Salt }]}{[\text { Base }]} \\ & \mathrm{pH}+\mathrm{pOH}=14\left(\text { at } 25^{\circ} \mathrm{C}\right) \\ & \mathrm{pOH}=14-9.5=4.75 \\ & 4.75=\mathrm{pK}_{\mathrm{b}}+\log \frac{[1]}{[1]} \\ & \mathrm{pK}_{\mathrm{b}}=4.75 \end{aligned}$
BCECE-2012]
Ionic Equilibrium
229759
A physician wishes to prepare a buffer solution at $\mathbf{p H}=3.85$ that efficiently resists changes in pH yet contains only small concentration of the buffering agents. Which of the following weak acids together with its sodium salt would be best to use?
By the use of Henderson's equation $\mathrm{pH}=\mathrm{pK}_2+\log _{10} \frac{[\text { salt }]}{[\text { acid }]}$ When, $[$ salt $]=[$ acid $]$ $\therefore \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}$ $\because \mathrm{pK}_{\mathrm{n}}=3.58$, thus at ths state $\mathrm{pH}=3.58$ So, acetoacetic acid $\left(\mathrm{pK}_{\mathrm{a}}=3.58\right)$ is best to use.
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Ionic Equilibrium
229753
Buffer solutions have constant acidity and alkalinity because
1 These give unionised acid or base on reaction with added acid on alkali
2 Acids and alkalies in these solution are shielded from attack by other ions
3 They have large excess of $\mathrm{H}^{+}$or $\mathrm{OH}^{-}$ions
4 They have fixed value of $\mathrm{pH}$
Explanation:
If small amount of an acid or alkali is added to a buffer solution, it converts them into unionized acid or base. Thus, its $\mathrm{pH}$ remains unaffected or in other words its acidity/alkalinity remains constant. e.g. $\begin{aligned} & \mathrm{H}_3 \mathrm{O}^{+}+\mathrm{A}^{-} \text {?? } \mathrm{H}_2 \mathrm{O}+\mathrm{HA} \\ & { }^{-} \mathrm{OH}+\mathrm{HA} \longrightarrow \mathrm{H}_2 \mathrm{O}+\mathrm{A}^{-} \end{aligned}$ If acid is added, it reacts with $\mathrm{A}^{-}$to form undissociated HA similarly, if base/alkali is added, $\mathrm{OH}^{-}$combines with $\mathrm{HA}$ to give $\mathrm{H}_2 \mathrm{O}$ and $\mathrm{A}^{-}$and thus, maintains the acidity/alkalinity of buffer solution.
NEET-2012
Ionic Equilibrium
229755
In a buffer solution containing concentration of $\mathrm{B}^{-}$and $\mathrm{HB}$, the $K_b$ for $\mathrm{B}^{-}$is $10^{-10}$. The $\mathrm{pH}$ of buffer solution is
1 10
2 7
3 6
4 4
Explanation:
Given, $\mathrm{K}_{\mathrm{b}}$ for $\mathrm{B}^{-}$is $10^{-10}$ $\therefore \mathrm{pK}_{\mathrm{b}}=-\log 10 \mathrm{~K}_{\mathrm{b}}=-\log \left(10^{-10}\right)=10$ According to the Henderson's equation $\text { We know that } \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { salt }]}{[\text { Acid }]}$ Here, the concentration of the salt and acid is same $\begin{aligned} & \therefore \log _{10} 1=0 \\ & \therefore \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}=10 \\ & \mathrm{pH}=14-\mathrm{pOH}=14-10=4 \end{aligned}$
NEET-2010
Ionic Equilibrium
229758
Solution of $0.1 \mathrm{NH}_4 \mathrm{OH}$ and $0.1 \mathrm{M} \mathrm{NH}_4 \mathrm{Cl}$ has pH 9.25. Then find out $\mathrm{pK}_{\mathrm{b}}$ of $\mathrm{NH}_4 \mathrm{OH}$.
1 9.25
2 4.75
3 3.75
4 8.25
Explanation:
Given that - Normality of $\mathrm{NH}_4 \mathrm{OH}=0.1 \mathrm{~N}$ Molarity of $\mathrm{NH}_4 \mathrm{Cl}=0.1 \mathrm{M}$ $\mathrm{pH}=9.25$ According to the Henderson's equation- $\begin{aligned} & \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log \frac{[\text { Salt }]}{[\text { Base }]} \\ & \mathrm{pH}+\mathrm{pOH}=14\left(\text { at } 25^{\circ} \mathrm{C}\right) \\ & \mathrm{pOH}=14-9.5=4.75 \\ & 4.75=\mathrm{pK}_{\mathrm{b}}+\log \frac{[1]}{[1]} \\ & \mathrm{pK}_{\mathrm{b}}=4.75 \end{aligned}$
BCECE-2012]
Ionic Equilibrium
229759
A physician wishes to prepare a buffer solution at $\mathbf{p H}=3.85$ that efficiently resists changes in pH yet contains only small concentration of the buffering agents. Which of the following weak acids together with its sodium salt would be best to use?
By the use of Henderson's equation $\mathrm{pH}=\mathrm{pK}_2+\log _{10} \frac{[\text { salt }]}{[\text { acid }]}$ When, $[$ salt $]=[$ acid $]$ $\therefore \mathrm{pH}=\mathrm{pK}_{\mathrm{a}}$ $\because \mathrm{pK}_{\mathrm{n}}=3.58$, thus at ths state $\mathrm{pH}=3.58$ So, acetoacetic acid $\left(\mathrm{pK}_{\mathrm{a}}=3.58\right)$ is best to use.