Buffer solution Questions in English

(A) buffer solution is formed by a mixture of a weak acid and its conjugate base (salt of the weak acid with a strong base).
$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$.
Therefore,the mixture of $CH_3COOH$ and $CH_3COONa$ acts as an acidic buffer solution in water.

(C) Using Henderson-Hasselbalch equation:
$pH = pK_{a} + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$
Given: $[\text{Salt}] = 0.2 \ M$,$[\text{Acid}] = 0.1 \ M$,$pK_{a} = 4.7$
$pH = 4.7 + \log_{10} \left( \frac{0.2}{0.1} \right)$
$pH = 4.7 + \log_{10} (2)$
Since $\log_{10} (2) \approx 0.3010$
$pH = 4.7 + 0.3010 = 5.001 \approx 5$

(C) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation:
$pH = pK_a + \log_{10} \frac{[\text{salt}]}{[\text{acid}]}$
Given: $pK_a = 4.74$,$[\text{salt}] = 0.05 \ M$,$[\text{acid}] = 0.01 \ M$
Substituting the values:
$pH = 4.74 + \log_{10} \left( \frac{0.05}{0.01} \right)$
$pH = 4.74 + \log_{10} (5)$
Since $\log_{10} (5) \approx 0.70$
$pH = 4.74 + 0.70 = 5.44$

(A) For an acidic buffer,the Henderson-Hasselbalch equation is given by:
$pH = pK_{a} + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$
Since the solution is prepared by mixing equimolar amounts,$[\text{Salt}] = [\text{Acid}]$,therefore $\frac{[\text{Salt}]}{[\text{Acid}]} = 1$.
$pH = pK_{a} + \log_{10}(1) = pK_{a} + 0 = pK_{a}$
Given $K_{a} = 1.78 \times 10^{-5}$,we calculate $pK_{a}$:
$pK_{a} = -\log_{10}(1.78 \times 10^{-5})$
$pK_{a} = -(\log_{10} 1.78 + \log_{10} 10^{-5})$
$pK_{a} = -(0.25 - 5) = 4.75$
Thus,the $pH$ of the buffer solution is $4.75$.

(C) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation:
$pH = pKa + \log \frac{[Salt]}{[Acid]}$
Given:
$pKa = 9.3$
$[Salt] = [NaCN] = 0.2 \ M$
$[Acid] = [HCN] = 0.1 \ M$
Substituting the values:
$pH = 9.3 + \log \frac{0.2}{0.1}$
$pH = 9.3 + \log 2$
$pH = 9.3 + 0.3010$
$pH = 9.6010 \approx 9.6$

(C) buffer solution is classified based on the $pH$ of the solution.
$1$. Acidic buffers consist of a weak acid and its salt with a strong base (e.g.,$CH_3COOH$ and $CH_3COONa$).
$2$. Basic buffers consist of a weak base and its salt with a strong acid (e.g.,$NH_4OH$ and $NH_4Cl$).
$3$. $A$ neutral buffer is typically formed by the mixture of a weak acid and a weak base,such as $CH_3COOH$ and $NH_4OH$,which results in a solution with a $pH$ near $7$.

(D) The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation: $pH = p K_{a} + \log \frac{[salt]}{[acid]}$.
Initially,the solution contains $0.1 \ mol$ of sodium acetate in $1000 \ cm^{3}$ $(1 \ L)$ of $0.1 \ M$ acetic acid.
After adding an additional $0.1 \ mol$ of sodium acetate,the total amount of salt becomes $0.1 + 0.1 = 0.2 \ mol$.
The concentration of salt $[salt] = \frac{0.2 \ mol}{1 \ L} = 0.2 \ M$.
The concentration of acid $[acid] = 0.1 \ M$.
Substituting these values into the equation: $pH = p K_{a} + \log \frac{0.2}{0.1}$.
$pH = p K_{a} + \log 2$.

(D) Initially,the amount of salt $[CH_{3}COONa] = 0.1 \ mol$ and the amount of acid $[CH_{3}COOH] = 0.1 \ mol$.
When an additional $0.1 \ mol$ of $CH_{3}COONa$ is added,the total amount of salt becomes $0.1 + 0.1 = 0.2 \ mol$.
The $pH$ of an acidic buffer is given by the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[salt]}{[acid]}$.
Substituting the values: $pH = pK_{a} + \log \frac{0.2}{0.1}$.
Therefore,$pH = pK_{a} + \log 2$.

(C) The mixture of a weak acid $(HA)$ and its salt with a strong base $(NaA)$ forms an acidic buffer solution.
Using the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$
Given: $K_{a} = 10^{-5}$,so $pK_{a} = -\log(10^{-5}) = 5$.
Since the volumes are equal,the ratio of concentrations is equal to the ratio of moles: $\frac{[Salt]}{[Acid]} = \frac{0.2 \ M}{0.1 \ M} = 2$.
Substituting the values: $pH = 5 + \log(2)$.
Given $\log 2 = 0.3$,so $pH = 5 + 0.3 = 5.3$.

(D) The reaction between $NaOH$ and acetic acid $(CH_3COOH)$ is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
Initial moles of $CH_3COOH = 0.06 \ M \times 0.050 \ L = 0.003 \ mol$.
Initial moles of $NaOH = 0.02 \ M \times 0.050 \ L = 0.001 \ mol$.
After the reaction,$0.001 \ mol$ of $CH_3COONa$ is formed and $0.002 \ mol$ of $CH_3COOH$ remains.
Since a buffer solution is formed,we use the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
$pH = 4.76 + \log \frac{0.001}{0.002} = 4.76 + \log (0.5)$.
$pH = 4.76 - 0.30 = 4.46$.

(D) The given solution is a basic buffer consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.
For a basic buffer,the $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
First,calculate the $pOH$ from the given $pH$: $pOH = 14 - pH = 14 - 8.2 = 5.8$.
Since the volumes of both solutions are equal $(30 \ mL)$,the ratio of concentrations $\frac{[Salt]}{[Base]}$ is equal to the ratio of their molarities: $\frac{[NH_4Cl]}{[NH_4OH]} = \frac{2 \ M}{0.2 \ M} = 10$.
Substitute these values into the Henderson-Hasselbalch equation: $5.8 = pK_b + \log(10)$.
Since $\log(10) = 1$,we have $5.8 = pK_b + 1$.
Therefore,$pK_b = 5.8 - 1 = 4.8$.

(D) basic buffer solution is formed by mixing a weak base with its salt of a strong acid,or by the partial neutralization of a weak base with a strong acid.
In option $D$,we have $100 \ mL$ of $0.1 \ M$ $HCl$ $(10 \ mmol)$ and $200 \ mL$ of $0.1 \ M$ $NH_4OH$ $(20 \ mmol)$.
The reaction is: $NH_4OH + HCl \rightarrow NH_4Cl + H_2O$.
After the reaction,$10 \ mmol$ of $NH_4OH$ (weak base) remains,and $10 \ mmol$ of $NH_4Cl$ (salt of weak base and strong acid) is formed.
This mixture of a weak base and its salt acts as a basic buffer.

(C) The mixture of acetic acid and sodium acetate acts as an acidic buffer.
The $pH$ of the solution can be calculated using the Henderson-Hasselbalch equation:
$pH = pK_{a} + \log \frac{[\text{salt}]}{[\text{acid}]} = pK_{a} + \log \frac{[\text{sodium acetate}]}{[\text{acetic acid}]}$
Since equal volumes of solutions with equal normality are mixed,the concentrations of the salt and acid in the final mixture remain equal,i.e.,$[\text{sodium acetate}] = [\text{acetic acid}]$.
Therefore,$pH = pK_{a} + \log(1) = pK_{a} + 0$.
Given $pK_{a} = 4.75$,the $pH$ of the resultant solution is $4.75$.

(B) An acidic buffer is a solution that resists changes in $pH$ and is prepared by mixing a weak acid with its salt of a strong base.
$HCOOH$ (weak acid) and $HCOOK$ (salt) form an acidic buffer.
$C_6H_5COOH$ (weak acid) and $C_6H_5COONa$ (salt) form an acidic buffer.
$HCN$ (weak acid) and $KCN$ (salt) form an acidic buffer.
$HClO_4$ is a strong acid,and $NaClO_4$ is its corresponding salt. $A$ mixture of a strong acid and its salt does not act as a buffer solution because a buffer requires a weak acid to maintain the equilibrium $HA \rightleftharpoons H^+ + A^-$. Therefore,$HClO_4 \& NaClO_4$ is not an acidic buffer.

(C) buffer solution consists of a mixture of a weak acid and its salt with a strong base,or a weak base and its salt with a strong acid.
$LiOH$ is a strong base,while $HNO_3$ and $HBr$ are strong acids.
$HNO_2$ is a weak acid and $NaNO_2$ is its salt formed by the combination of the weak acid $(HNO_2)$ and a strong base $(NaOH)$.
Therefore,the pair $(HNO_2 + NaNO_2)$ constitutes an acidic buffer.

(B) The Henderson-Hasselbalch equation for an acidic buffer is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Given: $K_a = 10^{-6}$,so $pK_a = -\log(10^{-6}) = 6$.
Given: $pH = 6$.
Substituting the values into the equation: $6 = 6 + \log \frac{[Salt]}{[Acid]}$.
This simplifies to: $\log \frac{[Salt]}{[Acid]} = 0$.
Taking the antilog on both sides: $\frac{[Salt]}{[Acid]} = 10^0 = 1$.
Therefore,the molar ratio of the weak acid $[Acid]$ to its salt $[Salt]$ is $\frac{[Acid]}{[Salt]} = \frac{1}{1} = 1$.

(D) The number of millimoles of acetic acid $(CH_3COOH)$ $= 10 \ mL \times 1.0 \ M = 10 \ mmol$.
The number of millimoles of sodium acetate $(CH_3COONa)$ $= 20 \ mL \times 0.5 \ M = 10 \ mmol$.
According to the Henderson-Hasselbalch equation:
$pH = pK_a + \log \left( \frac{[salt]}{[acid]} \right)$.
Since both components are in the same total volume $(100 \ mL)$,the ratio of their concentrations is equal to the ratio of their millimoles:
$pH = 4.76 + \log \left( \frac{10 \ mmol}{10 \ mmol} \right)$.
$pH = 4.76 + \log(1)$.
Since $\log(1) = 0$,we get $pH = 4.76$.
Therefore,option $(D)$ is correct.

(A) For a basic buffer,the Henderson-Hasselbalch equation is given by: $pOH = pK_b + \log \frac{[\text{salt}]}{[\text{base}]}$.
Given $pH = 10.5$,we calculate $pOH$ as: $pOH = 14 - 10.5 = 3.5$.
Substituting the values into the equation: $3.5 = pK_b + \log \frac{0.01}{0.1}$.
$3.5 = pK_b + \log(10^{-1})$.
$3.5 = pK_b - 1$,which gives $pK_b = 4.5$.
Using the relation $pK_a + pK_b = 14$,we find the $pK_a$ of the conjugate acid: $pK_a = 14 - 4.5 = 9.5$.

(B) Given: $pH = 8.6$.
$pOH = 14 - pH = 14 - 8.6 = 5.4$.
For a basic buffer: $pOH = pK_b + \log\frac{[\text{Salt}]}{[\text{Base}]}$.
$[\text{Salt}] = \frac{1 \times 30}{60} = 0.5 \text{ M}$ and $[\text{Base}] = \frac{0.1 \times 30}{60} = 0.05 \text{ M}$.
$5.4 = pK_b + \log\frac{0.5}{0.05} = pK_b + \log(10) = pK_b + 1$.
$pK_b = 5.4 - 1 = 4.4$.

(C) The given solution is a basic buffer consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.
Using the Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
First,calculate the number of millimoles:
$n(NH_4OH) = 50 \ mL \times 0.1 \ M = 5 \ mmol$
$n(NH_4Cl) = 25 \ mL \times 2.0 \ M = 50 \ mmol$
Since the total volume is $75 \ mL$,the concentrations are $[Salt] = \frac{50}{75} \ M$ and $[Base] = \frac{5}{75} \ M$.
Substituting these into the equation:
$pOH = 4.8 + \log \frac{50/75}{5/75} = 4.8 + \log(10) = 4.8 + 1 = 5.8$
Now,calculate $pH$ using $pH + pOH = 14$:
$pH = 14 - 5.8 = 8.2$

(B) For a buffer solution,the $pH$ is given by the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[\text{salt}]}{[\text{acid}]}$.
Since the concentration of the acid and salt are proportional to their volumes (as molarity is the same),the $pH$ is lowest when the ratio $\frac{[\text{salt}]}{[\text{acid}]}$ is the smallest.
Calculating the ratio $\frac{V_{\text{salt}}}{V_{\text{acid}}}$ for each:
$I: \frac{4.0}{4.0} = 1.0$
$II: \frac{40.0}{4.0} = 10.0$
$III: \frac{4.0}{40.0} = 0.1$
$IV: \frac{10.0}{0.1} = 100.0$
Comparing the ratios,$III$ $(0.1)$ and $I$ $(1.0)$ have the lowest values. However,looking at the options provided,we re-evaluate the question. The $pH$ is lowest when the acid concentration is highest relative to the salt.
Comparing the ratio $\frac{[\text{acid}]}{[\text{salt}]}$,the highest ratio corresponds to the lowest $pH$.
$I: \frac{4}{4} = 1$
$II: \frac{4}{40} = 0.1$
$III: \frac{40}{4} = 10$
$IV: \frac{0.1}{10} = 0.01$
The lowest $pH$ occurs for the solutions with the highest acid-to-salt ratio,which are $III$ and $I$. Given the options,$III$ and $IV$ is not correct,but $III$ and $I$ is not an option. Re-checking the question: the lowest $pH$ occurs when the ratio $\frac{[\text{salt}]}{[\text{acid}]}$ is minimum. The values are $III$ $(0.1)$ and $I$ $(1.0)$. If we look for the lowest $pH$ among the sets,$III$ is the lowest. The question asks for a set. Based on the ratio $\frac{[\text{salt}]}{[\text{acid}]}$,$III$ and $I$ are the lowest. Since $I$ and $III$ is option $B$,that is the correct choice.

(B) The reaction is: $CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^-$
Initial moles: $0.1 \ mol$ of $CH_3NH_2$ and $0.08 \ mol$ of $HCl$.
After reaction: $0.02 \ mol$ of $CH_3NH_2$ remains and $0.08 \ mol$ of $CH_3NH_3Cl$ is formed.
This forms a basic buffer solution.
The $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \left( \frac{[Salt]}{[Base]} \right)$.
$pK_b = -\log(5 \times 10^{-4}) = 4 - \log 5 = 4 - 0.7 = 3.3$.
$pOH = 3.3 + \log \left( \frac{0.08}{0.02} \right) = 3.3 + \log(4) = 3.3 + 0.6 = 3.9$.

(A) For a basic buffer solution,the $pOH$ is given by the Henderson-Hasselbalch equation:
$pOH = pK_{b} + \log \frac{[Salt]}{[Base]}$
Given that the molar concentrations of the base and its conjugate acid (salt) are the same,we have $[Salt] = [Base]$.
Substituting this into the equation:
$pOH = pK_{b} + \log(1)$
Since $\log(1) = 0$,we get:
$pOH = pK_{b}$
Therefore,the $pOH$ of the buffer solution is the same as the $pK_{b}$ of the base.

(A) The solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its conjugate base $(CH_3COO^-)$.
The Henderson-Hasselbalch equation is: $pH = pK_{a} + \log \frac{[\text{Salt}]}{[\text{Acid}]}$
Calculate the concentration of acid: $[\text{Acid}] = \frac{10 \ mL \times 1.0 \ M}{100 \ mL} = 0.1 \ M$
Calculate the concentration of salt: $[\text{Salt}] = \frac{20 \ mL \times 0.5 \ M}{100 \ mL} = 0.1 \ M$
Substitute the values into the equation: $pH = 4.76 + \log \frac{0.1}{0.1}$
Since $\log(1) = 0$,we get $pH = 4.76 + 0 = 4.76$.

(C) buffer solution is formed by a mixture of a weak acid and its conjugate base,or a weak base and its conjugate acid.
In option $A$,$NH_3$ (weak base) and $HCl$ (strong acid) in a $2:1$ ratio result in $NH_4Cl$ and $NH_3$,forming a basic buffer.
In option $B$,$CH_3COOH$ (weak acid) and $NaOH$ (strong base) in a $2:1$ ratio result in $CH_3COONa$ and $CH_3COOH$,forming an acidic buffer.
In option $C$,$NaOH$ and $CH_3COOH$ in a $1:1$ ratio undergo complete neutralization to form $CH_3COONa$ (a salt of a weak acid and strong base),which does not act as a buffer.
In option $D$,$NH_4Cl$ and $NH_3$ form a standard basic buffer.

(C) Given: Concentration of salt $[NH_4Cl] = 0.01 \ M$.
Concentration of base $[NH_4OH] = 0.1 \ M$.
$pK_b = 5$.
$A$ mixture of $NH_4OH$ and $NH_4Cl$ forms a basic buffer solution.
The Henderson-Hasselbalch equation for a basic buffer solution is:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Substituting the values into the equation:
$pOH = 5 + \log \frac{0.01}{0.1} = 5 + \log(0.1) = 5 - 1 = 4$.
Since $pH + pOH = 14$ at $25^{\circ}C$:
$pH = 14 - pOH = 14 - 4 = 10$.

(A) $H_2CO_3$ is a weak acid and dissociates in the following step:
$H_2CO_3{_{\text{(aq)}}} + H_2O_{\text{(l)}} \rightleftharpoons HCO_3^{-}{_{\text{(aq)}}} + H_3O^{+}{_{\text{(aq)}}}$
The $H_2CO_3 / HCO_3^{-}$ buffer system is the primary mechanism that helps to maintain the $pH$ of human blood within the physiological range of $7.26$ to $7.42$.

(B) The mixture of $NH_4OH$ (a weak base) and $NH_4Cl$ (its salt with a strong acid) forms a basic buffer solution.
For a basic buffer,the $pOH$ is calculated using the Henderson-Hasselbalch equation:
$pOH = pK_b + \log \frac{[\text{salt}]}{[\text{base}]}$
Given: $[\text{salt}] = 0.2 \ M$,$[\text{base}] = 0.02 \ M$,and $pK_b = 4.8$.
$pOH = 4.8 + \log \left( \frac{0.2}{0.02} \right) = 4.8 + \log(10) = 4.8 + 1 = 5.8$.
Now,calculate the $pH$ using the relation $pH + pOH = 14$ at $25^{\circ}C$:
$pH = 14 - 5.8 = 8.2$.

(B) Let the concentration of potassium acetate be $x \ M$. The mixture forms an acidic buffer. Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[\text{salt}]}{[\text{acid}]}$
Given: $pH = 4.8$,$K_a = 1.8 \times 10^{-5}$,$pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74$.
The number of moles of acid $= 20 \ mL \times 0.1 \ M = 2 \ mmol$.
The number of moles of salt $= 50 \ mL \times x \ M = 50x \ mmol$.
Substituting these into the equation:
$4.8 = 4.74 + \log \frac{50x}{2}$
$0.06 = \log(25x)$
$25x = 10^{0.06} \approx 1.148$
$x = \frac{1.148}{25} \approx 0.0459 \ M \approx 0.04 \ M$ (rounding to the nearest option).

(D) Buffer capacity,$\beta = \frac{dC_{HA}}{dpH}$.
Here,$dC_{HA}$ is the number of moles of acid added per liter of the solution.
$dC_{HA} = \frac{\text{moles of acetic acid}}{\text{volume in liters}} = \frac{0.12 / 60}{250 / 1000} = \frac{0.002}{0.25} = 0.008 \ M$.
The change in $pH$ is $dpH = 0.02$.
Therefore,$\beta = \frac{0.008}{0.02} = 0.4$.

(B) Both $(A)$ and $(R)$ are true statements,but $(R)$ is not the correct explanation of $(A)$.
The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[\text{salt}]}{[\text{acid}]}$
When the moles of salt and acid are equal,$[\text{salt}] = [\text{acid}]$,so:
$pH = pK_a + \log(1) = pK_a = 4.8$.
Thus,$(A)$ is true. The ionic product of water $(K_w)$ at $25^{\circ} C$ is indeed $10^{-14} \ mol^2 \ L^{-2}$,making $(R)$ a true statement,but it does not explain the $pH$ calculation of the buffer.

(B) For a buffer solution,the $pH$ is calculated using the Henderson-Hasselbalch equation:
$pH = pKa + \log \left( \frac{[salt]}{[acid]} \right)$
Since the molarity of the acid and the salt are the same $(0.1 \ M)$,the ratio $\frac{[salt]}{[acid]}$ is equivalent to the ratio of their volumes: $\frac{V_{salt}}{V_{acid}}$.
The $pH$ is lowest when the ratio $\frac{[salt]}{[acid]}$ is the smallest,which occurs when the volume of the acid is high and the volume of the salt is low.
Calculating the ratios for each:
$I: \frac{4.0}{4.0} = 1.0$
$II: \frac{40.0}{4.0} = 10.0$
$III: \frac{4.0}{40.0} = 0.1$
$IV: \frac{10.0}{0.1} = 100.0$
Comparing the ratios,$III$ $(0.1)$ and $I$ $(1.0)$ have the lowest values. However,looking at the options provided,$III$ has the lowest ratio,followed by $I$. Re-evaluating the question for the two sets with the lowest $pH$,$III$ and $I$ are the correct candidates.

(A) buffer solution is formed by mixing a weak acid and its salt with a strong base,or a weak base and its salt with a strong acid.
When $1 \ M \ CH_3COOH$ and $0.5 \ M \ NaOH$ are mixed in equal volumes,the reaction is:
$CH_3COOH + NaOH \longrightarrow CH_3COONa + H_2O$.
Since the concentration of $CH_3COOH$ $(1 \ M)$ is double that of $NaOH$ $(0.5 \ M)$,after the reaction,$0.5 \ M$ of $CH_3COOH$ remains unreacted,and $0.5 \ M$ of $CH_3COONa$ is formed.
This mixture of a weak acid $(CH_3COOH)$ and its salt $(CH_3COONa)$ acts as an acidic buffer.

(A) Using the Henderson-Hasselbalch equation for an acidic buffer:
$pH = pK_a + \log \frac{[\text{salt}]}{[\text{acid}]}$
Given $pH = 5.8$ and $pK_a = 4.8$:
$5.8 = 4.8 + \log \frac{[\text{salt}]}{[\text{acid}]}$
$\log \frac{[\text{salt}]}{[\text{acid}]} = 5.8 - 4.8 = 1.0$
Taking the antilog on both sides:
$\frac{[\text{salt}]}{[\text{acid}]} = 10^1 = 10$
Therefore,the ratio $\frac{[\text{acid}]}{[\text{salt}]} = \frac{1}{10} = 0.1$.

(C) To determine the lowest $pH$,we analyze the nature of each mixture:
$(A)$ $10 \ mL$ $0.05 \ N$ $CH_{3}COOH$ $(0.5 \ mmol)$ + $5 \ mL$ $0.1 \ N$ $NH_{4}OH$ $(0.5 \ mmol)$ forms a salt of weak acid and weak base $(CH_{3}COONH_{4})$,which is nearly neutral.
$(B)$ $5 \ mL$ $0.2 \ N$ $NH_{4}Cl$ + $5 \ mL$ $0.2 \ N$ $NH_{4}OH$ forms a basic buffer $(pH > 7)$.
$(C)$ $5 \ mL$ $0.1 \ N$ $CH_{3}COOH$ $(0.5 \ mmol)$ + $10 \ mL$ $0.05 \ N$ $CH_{3}COONa$ $(0.5 \ mmol)$ forms an acidic buffer where $pH = pK_{a} + \log(\frac{[Salt]}{[Acid]})$. Since $[Salt] = [Acid]$,$pH = pK_{a} \approx 4.76$.
$(D)$ $5 \ mL$ $0.1 \ N$ $CH_{3}COOH$ + $5 \ mL$ $0.1 \ N$ $NaOH$ forms $CH_{3}COONa$,which undergoes anionic hydrolysis,resulting in a basic solution $(pH > 7)$.
Comparing these,the acidic buffer in option $(C)$ has the lowest $pH$.

(B, C) buffer solution is formed by a mixture of a weak acid and its salt with a strong base,or a weak base and its salt with a strong acid.
Option $B$: $NH_4OH$ (weak base) + $HCl$ (strong acid) in $2:1$ mole ratio results in $NH_4Cl$ (salt) and remaining $NH_4OH$. This forms a basic buffer.
Option $C$: $CH_3COOH$ (weak acid) + $NaOH$ (strong base) in $2:1$ mole ratio results in $CH_3COONa$ (salt) and remaining $CH_3COOH$. This forms an acidic buffer.
Both $B$ and $C$ act as buffer solutions.

(C, D) buffer solution is formed by a weak acid and its conjugate base or a weak base and its conjugate acid.
$1$. $CH_{3}COOH$ (weak acid) and $CH_{3}COONa$ (salt of weak acid and strong base) form an acidic buffer. Thus,$(D)$ is correct.
$2$. When $HNO_{3}$ (strong acid) is mixed with $CH_{3}COONa$ (salt of weak acid) in a ratio such that the salt is in excess,the reaction $HNO_{3} + CH_{3}COO^{-} \rightarrow CH_{3}COOH + NO_{3}^{-}$ occurs. This results in a mixture of the weak acid $(CH_{3}COOH)$ and its conjugate base $(CH_{3}COO^{-})$,which acts as a buffer. Thus,$(C)$ is also correct.

(D) Given: $pH = 5.74$,$pK_a = 4.74$.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Substituting the values: $5.74 = 4.74 + \log \frac{[CH_3COONa]}{[CH_3COOH]}$.
$1 = \log \frac{[CH_3COONa]}{[CH_3COOH]}$.
Therefore,$\frac{[CH_3COONa]}{[CH_3COOH]} = 10^1 = 10$.
Since the concentrations of both solutions are equal $(0.1 \ N)$,the ratio of their volumes is equal to the ratio of their concentrations: $\frac{V_{acid}}{V_{salt}} = \frac{[CH_3COOH]}{[CH_3COONa]} = \frac{1}{10}$.

(B) The initial moles of acetic acid and sodium acetate in $1 \ \text{L}$ of solution are $0.01 \ \text{mole}$ each.
When $1 \times 10^{-3} \ \text{mole}$ of $HCl$ is added,it reacts with the salt (sodium acetate) to form acetic acid:
$CH_{3}COO^{-} + H^{+} \longrightarrow CH_{3}COOH$
New moles of salt = $0.01 - 0.001 = 0.009 \ \text{mole}$.
New moles of acid = $0.01 + 0.001 = 0.011 \ \text{mole}$.
Using the Henderson-Hasselbalch equation:
$pH = pK_{a} + \log \frac{[\text{salt}]}{[\text{acid}]} = 4.75 + \log \frac{0.009}{0.011} = 4.75 + \log(0.818) \approx 4.75 - 0.087 = 4.663$
Thus,the final $pH$ is approximately $4.66$.

(B) The reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
Initial millimoles of $CH_3COOH = 20 \ mL \times 0.1 \ N = 2 \ mmol$.
Initial millimoles of $NaOH = 10 \ mL \times 0.1 \ N = 1 \ mmol$.
After the reaction,$1 \ mmol$ of $CH_3COOH$ remains and $1 \ mmol$ of $CH_3COONa$ is formed.
This forms an acidic buffer solution.
Using Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
$pH = 4.74 + \log \frac{1 \ mmol}{1 \ mmol}$
$pH = 4.74 + \log(1) = 4.74 + 0 = 4.74$.

(B) Given $K_a = 10^{-5}$,so $pK_a = -\log(10^{-5}) = 5$.
At one-third neutralization,the amount of salt formed is $1/3$ and the amount of acid remaining is $2/3$ of the initial concentration.
According to the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[\text{Salt}]}{[\text{Acid}]}$
$pH = 5 + \log \frac{1/3}{2/3}$
$pH = 5 + \log \frac{1}{2}$
$pH = 5 - \log 2$.

(C) For a buffer solution of a weak base and its salt,the Henderson-Hasselbalch equation is: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given $pH = 9$,so $pOH = 14 - 9 = 5$.
Given $pK_b = 5.699$,so $5 = 5.699 + \log \frac{[Salt]}{[Base]}$.
$\log \frac{[Salt]}{[Base]} = 5 - 5.699 = -0.699$.
Since $\log 5 = 0.699$,then $\log \frac{[Salt]}{[Base]} = -\log 5 = \log \frac{1}{5}$.
Therefore,$\frac{[Salt]}{[Base]} = \frac{1}{5}$.
Reaction: $B + HCl \rightarrow BH^+ + Cl^-$.
Initial moles: $n_B = 0.02y$,$n_{HCl} = 0.02x$.
After reaction: $n_{BH^+} = 0.02x$,$n_{B, remaining} = 0.02y - 0.02x$.
Ratio: $\frac{0.02x}{0.02y - 0.02x} = \frac{1}{5}$ $\Rightarrow \frac{x}{y-x} = \frac{1}{5}$ $\Rightarrow 5x = y - x$ $\Rightarrow y = 6x$.
Total volume: $x + y = 100 \ mL$.
Substituting $y$: $x + 6x = 100$ $\Rightarrow 7x = 100$ $\Rightarrow x = 14.28 \approx 14.3 \ mL$.
$y = 100 - 14.3 = 85.7 \ mL$.

(B) For a basic buffer,$pOH = pK_{b} + \log \frac{[Salt]}{[Base]}$.
Given $pH = 9.25$,so $pOH = 14 - 9.25 = 4.75$.
Substituting values: $4.75 = 4.75 + \log \frac{[Salt]}{[Base]}$,which implies $\log \frac{[Salt]}{[Base]} = 0$,so $[Salt] = [Base]$.
This means the number of millimoles of $NH_{4}Cl$ (salt) formed must equal the remaining millimoles of $NH_{4}OH$ (base).
Checking Option $(B)$:
$NH_{4}OH + HCl \rightarrow NH_{4}Cl + H_{2}O$
Initial millimoles of $NH_{4}OH = 0.2 \ M \times 500 \ mL = 100 \ mmol$.
Initial millimoles of $HCl = 0.1 \ M \times 500 \ mL = 50 \ mmol$.
After reaction,$50 \ mmol$ of $NH_{4}Cl$ is formed and $50 \ mmol$ of $NH_{4}OH$ remains.
Since $[Salt] = [Base]$,the $pH$ will be $9.25$.

(B) $1$. Calculate the initial moles of acetic acid $(CH_3COOH)$: Since $20 \text{ mL}$ of acetic acid is neutralized by $28.4 \text{ mL}$ of $0.1 \text{ M } NaOH$,the moles of acid = moles of base = $0.1 \text{ M} \times 28.4 \text{ mL} = 2.84 \text{ mmol}$.
$2$. Calculate the composition of solution $(X)$: Solution $(X)$ is formed by mixing $2.84 \text{ mmol}$ of $CH_3COOH$ and $1.42 \text{ mmol}$ of $NaOH$ $(0.1 \text{ M} \times 14.2 \text{ mL} = 1.42 \text{ mmol})$.
$3$. Reaction: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
$4$. After reaction: Remaining $CH_3COOH = 2.84 - 1.42 = 1.42 \text{ mmol}$. Formed $CH_3COONa$ (salt) = $1.42 \text{ mmol}$.
$5$. Apply Henderson-Hasselbalch equation: $pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right)$.
$6$. Since $[Salt] = [Acid] = 1.42 \text{ mmol}$,the ratio is $1$. Thus,$pH = 4.75 + \log(1) = 4.75 + 0 = 4.75$.