Buffer solution Questions in English

(A) Given that the concentration of the base $B^-$ and its conjugate acid $HB$ are equal,i.e.,$[B^-] = [HB]$.
For a basic buffer,the $pOH$ is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \left( \frac{[salt]}{[base]} \right)$.
Since $[B^-] = [HB]$,the ratio $\frac{[salt]}{[base]} = 1$,and $\log(1) = 0$.
Therefore,$pOH = pK_b = -\log(K_b)$.
Given $K_b = 10^{-10}$,we have $pOH = -\log(10^{-10}) = 10$.
We know that $pH + pOH = 14$ at $25^{\circ}C$.
Thus,$pH = 14 - pOH = 14 - 10 = 4$.

(A) For a basic buffer,the formula is $pOH = pK_b + \log \left( \frac{[Salt]}{[Base]} \right)$.
Given: $[Salt] = [NH_4^+] = 0.20 \ M$,$[Base] = [NH_3] = 0.30 \ M$,and $K_b = 1.8 \times 10^{-5}$.
First,calculate $pK_b = -\log(K_b) = -\log(1.8 \times 10^{-5}) \approx 4.74$.
Now,$pOH = 4.74 + \log \left( \frac{0.20}{0.30} \right) = 4.74 + \log(0.667) = 4.74 - 0.176 = 4.564$.
Finally,$pH = 14 - pOH = 14 - 4.564 = 9.436 \approx 9.43$.

(A) Buffer solutions resist changes in $pH$ upon the addition of small amounts of acid or base. This is because they contain a reserve of acid and base that reacts with the added $H^+$ or $OH^-$ ions to form unionized acid or base,thereby maintaining the $pH$ constant.

(A) Given: $[Acid] = 0.25 \ M$,$[Salt] = 0.125 \ M$,$K_a = 1.8 \times 10^{-5}$.
First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
$pH = 4.745 + \log \frac{0.125}{0.25} = 4.745 + \log(0.5)$.
Since $\log(0.5) = -0.301$,$pH = 4.745 - 0.301 = 4.444$.
Rounding to two decimal places,the $pH$ is $4.44$.

(A) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Given: $pK_a = 4.76$,$[Salt] = [CH_3COONa] = 6.1 \times 10^{-2} \ M$,and $[Acid] = [CH_3COOH] = 0.1 \ M$.
Substituting the values: $pH = 4.76 + \log \frac{6.1 \times 10^{-2}}{0.1}$.
$pH = 4.76 + \log(0.61)$.
Since $\log(0.61) \approx -0.2147$,we get $pH = 4.76 - 0.2147 = 4.5453 \approx 4.55$.
Rounding to the nearest provided option,the answer is $4.52$.

(A) $1$. Calculate the millimoles of reactants:
$n(CH_3COOH) = 100 \ mL \times 0.1 \ M = 10 \ mmol$
$n(NaOH) = 30 \ mL \times 0.1 \ M = 3 \ mmol$
$2$. Reaction: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
After reaction:
$n(CH_3COOH)_{remaining} = 10 - 3 = 7 \ mmol$
$n(CH_3COONa)_{formed} = 3 \ mmol$
$3$. Use Henderson-Hasselbalch equation:
$pH = pK_a + \log(\frac{[Salt]}{[Acid]}) = 4.76 + \log(\frac{3}{7})$
$pH = 4.76 + \log(0.428) = 4.76 - 0.368 = 4.392 \approx 4.39$

(A) The given solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its conjugate base $(CH_3COONa)$.
According to the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Given: $pH = 4.74$,$[Salt] = 0.1 \ mol / 0.5 \ L = 0.2 \ M$,and $[Acid] = 0.1 \ mol / 0.5 \ L = 0.2 \ M$.
Substituting the values: $4.74 = pK_a + \log \frac{0.2}{0.2}$.
Since $\log(1) = 0$,we get $pK_a = 4.74$.
We know that $pK_a = -\log(K_a)$,so $K_a = 10^{-4.74}$.
$K_a \approx 1.82 \times 10^{-5}$.

(A) The given solution is a buffer solution consisting of a weak acid $(HCN)$ and its salt with a strong base $(KCN)$.
For an acidic buffer,the concentration of hydrogen ions is given by the Henderson-Hasselbalch equation:
$[H^{+}] = K_{a} \times \frac{[Acid]}{[Salt]}$
Given:
$K_{a} = 4 \times 10^{-10}$
$[Acid] = [HCN] = 0.2 \ M$
$[Salt] = [KCN] = 1 \ M$
Substituting the values:
$[H^{+}] = 4 \times 10^{-10} \times \frac{0.2}{1}$
$[H^{+}] = 4 \times 10^{-10} \times 0.2 = 0.8 \times 10^{-10} = 8 \times 10^{-11} \ M$

(A) Using the Henderson-Hasselbalch equation for an acidic buffer: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Given $pH = 4.0$ and $K_a = 1.8 \times 10^{-5}$,we find $pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74$.
Substituting the values: $4.0 = 4.74 + \log \frac{[CH_3COONa]}{0.1}$.
$-0.74 = \log \frac{[CH_3COONa]}{0.1}$.
$\frac{[CH_3COONa]}{0.1} = 10^{-0.74} \approx 0.182$.
$[CH_3COONa] = 0.182 \times 0.1 = 0.0182 \ M$.
Mass of $CH_3COONa = \text{Molarity} \times \text{Volume} \times \text{Molar Mass}$. Assuming $1 \ L$ solution: $0.0182 \ mol \times 82 \ g \ mol^{-1} = 1.4924 \ g$.

(A) For a basic buffer solution,the $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given: $K_b = 1.8 \times 10^{-5}$,$[Base] = 0.15 \ M$,$[Salt] = 0.25 \ M$.
$pK_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
$pOH = 4.745 + \log \frac{0.25}{0.15} = 4.745 + \log(1.667) = 4.745 + 0.222 = 4.967$.
Since $pH + pOH = 14$,$pH = 14 - 4.967 = 9.033$.

(A) An acidic buffer is a mixture of a weak acid and its salt with a strong base.
$CH_3COOH + CH_3COONa$ is a mixture of a weak acid and its salt.
$HCN + NaCN$ is a mixture of a weak acid and its salt.
$H_3BO_3 + Na_2B_4O_7$ acts as a buffer system.
$HClO_4$ is a strong acid and $NaClO_4$ is its salt with a strong base $(NaOH)$. $A$ mixture of a strong acid and its salt does not form a buffer solution. Therefore,$HClO_4 + NaClO_4$ is not an acidic buffer.

(N/A) buffer solution is a solution that resists changes in $pH$ upon the addition of small amounts of acid or base.
Types of buffer solutions:
$1$. Acidic Buffer: $A$ mixture of a weak acid and its salt with a strong base. Example: $CH_3COOH + CH_3COONa$ $(pH < 7)$.
$2$. Basic Buffer: $A$ mixture of a weak base and its salt with a strong acid. Example: $NH_4OH + NH_4Cl$ $(pH > 7)$.
Biological Importance:
$1$. The $pH$ of human blood is maintained at approximately $7.4$ due to the $H_2CO_3/HCO_3^-$ buffer system.
$2$. Many biochemical reactions in cells occur only within a specific $pH$ range,which is maintained by intracellular buffer systems.

(C) An acidic buffer is formed by a mixture of a weak acid and its salt with a strong base.
In option $C$,we have $100 \ mL$ of $0.1 \ M \ HCl$ $(10 \ mmol)$ and $200 \ mL$ of $0.1 \ M \ CH_3COONa$ $(20 \ mmol)$.
The reaction is: $HCl + CH_3COONa \rightarrow CH_3COOH + NaCl$
Initial moles: $10 \ mmol \ HCl$ and $20 \ mmol \ CH_3COONa$.
After reaction: $0 \ mmol \ HCl$,$10 \ mmol \ CH_3COONa$ remaining,and $10 \ mmol \ CH_3COOH$ formed.
Since the final mixture contains a weak acid $(CH_3COOH)$ and its conjugate base ($CH_3COO^-$ from $CH_3COONa$),it acts as an acidic buffer.

(D) The Henderson-Hasselbalch equation for an acidic buffer is given by:
$pH = pKa + \log \frac{[Salt]}{[Acid]}$
Given values are:
$pH = 5.74$
$pKa = 4.74$
$[Acid] = 1.0 \ M$
Substituting these values into the equation:
$5.74 = 4.74 + \log \frac{[Salt]}{1.0}$
$5.74 - 4.74 = \log [Salt]$
$1 = \log [Salt]$
Taking the antilog on both sides:
$[Salt] = 10^1 = 10 \ M$
Therefore,the concentration of sodium acetate is $10 \ M$.

(A) The mixture consists of a weak base $(NH_{3})$ and its salt $(NH_{4}Cl)$,forming a basic buffer.
First,calculate the moles of each component:
$n(NH_{4}Cl) = 0.0504 \ M \times 5.0 \ mL = 0.252 \ mmol$
$n(NH_{3}) = 0.0210 \ M \times 2.0 \ mL = 0.042 \ mmol$
Using the Henderson-Hasselbalch equation for a basic buffer:
$[OH^{-}] = K_{b} \times \frac{[Base]}{[Salt]} = K_{b} \times \frac{n(NH_{3})}{n(NH_{4}Cl)}$
$[OH^{-}] = 1.8 \times 10^{-5} \times \frac{0.042}{0.252}$
$[OH^{-}] = 1.8 \times 10^{-5} \times \frac{1}{6} = 0.3 \times 10^{-5} = 3 \times 10^{-6} \ M$
Comparing this with $x \times 10^{-6} \ M$,we get $x = 3$.

(D) The mixture of a weak acid $(CH_{3}COOH)$ and its salt with a strong base $(CH_{3}COONa)$ forms an acidic buffer solution.
The concentration of salt $[Salt] = 0.10 \ M$.
The concentration of acid $[Acid] = 0.01 \ M$.
The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation:
$pH = pK_{a} + \log \left( \frac{[Salt]}{[Acid]} \right)$
Substituting the given values:
$pH = 4.57 + \log \left( \frac{0.10}{0.01} \right)$
$pH = 4.57 + \log(10)$
Since $\log(10) = 1$,we get:
$pH = 4.57 + 1 = 5.57$

(C) Initial moles of $CH_3COOH = 0.1 \, M \times 50 \, mL = 5 \, mmol$.
Moles of $NaOH$ added = $0.1 \, M \times 25 \, mL = 2.5 \, mmol$.
The reaction is: $CH_3COOH + NaOH \longrightarrow CH_3COONa + H_2O$.
After the reaction,$2.5 \, mmol$ of $CH_3COOH$ remains and $2.5 \, mmol$ of $CH_3COONa$ is formed.
Since we have a weak acid and its conjugate base,a buffer solution is formed.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \left(\frac{[CH_3COO^-]}{[CH_3COOH]}\right)$.
Since the volume is the same for both,the ratio of concentrations is equal to the ratio of moles: $pH = 4.76 + \log \left(\frac{2.5}{2.5}\right) = 4.76 + \log(1) = 4.76$.
Thus,$pH = 476 \times 10^{-2}$.

(B) The Henderson-Hasselbalch equation for an acidic buffer is given by:
$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$
First,calculate $pK_{a}$:
$pK_{a} = -\log(K_{a}) = -\log(1.3 \times 10^{-5}) = 5 - \log(1.3) \approx 5 - 0.1139 = 4.8861$
Substitute the values into the equation:
$4 = 4.8861 + \log \frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]}$
$\log \frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]} = 4 - 4.8861 = -0.8861$
Taking the antilog:
$\frac{[CH_{3}CH_{2}COO^{-}]}{[CH_{3}CH_{2}COOH]} = 10^{-0.8861} \approx 0.13$

(C) For a basic buffer,the $pH$ is related to $pOH$ by the equation: $pH + pOH = 14$.
Given $pH = 8.26$,so $pOH = 14 - 8.26 = 5.74$.
The Henderson-Hasselbalch equation for a basic buffer is: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Substituting the given values: $5.74 = 4.74 + \log \frac{[NH_4^+]}{0.2}$.
$1 = \log \frac{[NH_4^+]}{0.2}$,which implies $\frac{[NH_4^+]}{0.2} = 10^1 = 10$.
Therefore,$[NH_4^+] = 10 \times 0.2 = 2 \, M$.
Since the volume is $1 \, L$,the number of moles of $NH_4Cl$ required is $2 \, mol$.
Mass of $NH_4Cl = \text{moles} \times \text{molar mass} = 2 \, mol \times 53.5 \, g \, mol^{-1} = 107 \, g$.

(A) buffer solution is a mixture that resists changes in $pH$ upon the addition of small amounts of acid or base. It typically consists of a weak base and its salt,or a weak acid and its salt.
In option $A$,mixing equal volumes of $0.2 \, M \ NH_{4}OH$ (weak base) and $0.1 \, M \ HCl$ (strong acid) results in a reaction where $0.1 \, M$ of $NH_{4}OH$ reacts with $0.1 \, M$ of $HCl$ to form $0.1 \, M \ NH_{4}Cl$ (salt). The remaining concentration of $NH_{4}OH$ is $0.1 \, M$. Since the final mixture contains both the weak base $(NH_{4}OH)$ and its salt $(NH_{4}Cl)$,it forms a basic buffer solution.

(B) Given that the mixture is a basic buffer consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.
$pH = 8$,so $pOH = 14 - 8 = 6$.
Using the Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
Here,$[Salt] = [NH_4Cl] = 1 \, M$ and $[Base] = [NH_4OH] = 0.1 \, M$.
Substituting the values:
$6 = pK_b + \log \left( \frac{1}{0.1} \right)$
$6 = pK_b + \log(10)$
$6 = pK_b + 1$
$pK_b = 6 - 1 = 5$.
Therefore,the correct option is $B$.

(D) The number of millimoles $(mmol)$ of acetic acid $(CH_3COOH)$ is $100 \, mL \times 0.1 \, mol \, L^{-1} = 10 \, mmol$.
The number of millimoles $(mmol)$ of sodium acetate $(CH_3COONa)$ is $50 \, mL \times 0.2 \, mol \, L^{-1} = 10 \, mmol$.
According to the Henderson-Hasselbalch equation for an acidic buffer:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Since the volume is the same for both components in the mixture,the ratio of concentrations is equal to the ratio of millimoles:
$pH = pK_a + \log \frac{mmol \, of \, CH_3COO^-}{mmol \, of \, CH_3COOH}$
Substituting the values:
$pH = 4.76 + \log \frac{10}{10}$
$pH = 4.76 + \log 1$
Since $\log 1 = 0$,we get:
$pH = 4.76$

(C) The mixture of $CH_3COOH$ and $CH_3COONa$ forms an acidic buffer solution.
Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Calculate the concentrations in the final $50 \ mL$ mixture:
$[CH_3COONa] = \frac{25 \ mL \times 0.2 \ M}{50 \ mL} = 0.1 \ M$
$[CH_3COOH] = \frac{25 \ mL \times 0.02 \ M}{50 \ mL} = 0.01 \ M$
Substitute the values into the equation:
$5 = pK_a + \log \frac{0.1}{0.01}$
$5 = pK_a + \log(10)$
$5 = pK_a + 1$
$pK_a = 4$
Since $pK_a = -\log(K_a)$:
$K_a = 10^{-4} = 10 \times 10^{-5}$
Comparing this with $x \times 10^{-5}$,we get $x = 10$.

(B) The initial moles are $n_{NH_3} = 0.1 \ mol$ and $n_{NH_4^+} = 0.1 \ mol$.
When $0.02 \ mol$ of $HCl$ is added,it reacts with $NH_3$ as follows:
$NH_3 + HCl \rightarrow NH_4^+ + Cl^-$
After the reaction:
$n_{NH_3} = 0.1 - 0.02 = 0.08 \ mol$
$n_{NH_4^+} = 0.1 + 0.02 = 0.12 \ mol$
Using the Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]}$
$pOH = 4.745 + \log \frac{0.12}{0.08}$
$pOH = 4.745 + \log(1.5)$
$pOH = 4.745 + (\log 3 - \log 2)$
$pOH = 4.745 + (0.477 - 0.301) = 4.745 + 0.176 = 4.921$
Since $pH + pOH = 14$ at $298 \ K$:
$pH = 14 - 4.921 = 9.079$
Expressing as $9079 \times 10^{-3}$,the value is $9079$.

(D) The reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
Initial moles of $CH_3COOH = 50 \ mL \times 0.1 \ M = 5 \ mmol$.
Initial moles of $NaOH = 20 \ mL \times 0.1 \ M = 2 \ mmol$.
After reaction,moles of $CH_3COONa$ (salt) $= 2 \ mmol$ and remaining moles of $CH_3COOH$ (acid) $= 5 - 2 = 3 \ mmol$.
Using the Henderson-Hasselbalch equation: $pH = pKa + \log_{10} (\frac{[salt]}{[acid]})$
$pH = 4.76 + \log_{10} (\frac{2}{3})$
$pH = 4.76 + (\log 2 - \log 3) = 4.76 + (0.30 - 0.48) = 4.76 - 0.18 = 4.58$.
Thus,$pH = 4.58 = 458 \times 10^{-2}$.

(A) The buffer solution consists of benzoic acid (weak acid) and sodium benzoate (its conjugate base).
Using the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[\text{salt}]}{[\text{acid}]}$.
Given $pH = 4.5$,$pK_{a} = 4.2$,$[\text{salt}] = [\text{sodium benzoate}]$,and $[\text{acid}] = [\text{benzoic acid}]$.
$4.5 = 4.2 + \log \frac{[\text{salt}]}{[\text{acid}]}$
$0.3 = \log \frac{[\text{salt}]}{[\text{acid}]}$
Since $10^{0.3} \approx 2$,we have $\frac{[\text{salt}]}{[\text{acid}]} = 2$.
Let $V_{a}$ be the volume of benzoic acid and $V_{s}$ be the volume of sodium benzoate.
Since the molarities are equal $(1 \ M)$,the ratio of moles is equal to the ratio of volumes: $\frac{V_{s}}{V_{a}} = 2$,so $V_{s} = 2V_{a}$.
Given the total volume $V_{s} + V_{a} = 300 \ mL$.
Substituting $V_{s} = 2V_{a}$ into the equation: $2V_{a} + V_{a} = 300 \ mL$.
$3V_{a} = 300 \ mL \implies V_{a} = 100 \ mL$.

(A) buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid in specific proportions,not just any quantity.
Statement $I$ is false because it implies any quantity of salt and acid/base works,whereas buffer action depends on specific ratios.
Blood is a naturally occurring buffer system maintained by the $H_2CO_3 / HCO_3^{-}$ pair,which keeps the $pH$ constant.
Statement $II$ is true.
Therefore,Statement $I$ is false but Statement $II$ is true.

(C) buffer solution is formed by a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.
$1$. In option $C$,$HNO_3$ (strong acid) reacts with $CH_3COONa$ (salt of weak acid) to form $CH_3COOH$ (weak acid) and $NaNO_3$. If $HNO_3$ is the limiting reagent,the resulting mixture contains $CH_3COOH$ and $CH_3COONa$,which acts as an acidic buffer.
$2$. In option $D$,the mixture contains $CH_3COOH$ (weak acid) and $CH_3COONa$ (conjugate base),which is a classic acidic buffer system.
Therefore,both $C$ and $D$ can form a buffer.

(A) The initial moles are $0.01 \ mol$ $(10 \ mmol)$ each of $H_2CO_3$,$NaHCO_3$,$Na_2CO_3$,and $NaOH$.
Step $1$: Reaction between $H_2CO_3$ and $NaOH$:
$H_2CO_3 + NaOH \longrightarrow NaHCO_3 + H_2O$
Initial: $10 \ mmol, 10 \ mmol, 10 \ mmol$
Final: $0 \ mmol, 0 \ mmol, 20 \ mmol$
Step $2$: The final mixture contains $20 \ mmol$ of $NaHCO_3$ (from initial $10 \ mmol$ + $10 \ mmol$ produced) and $10 \ mmol$ of $Na_2CO_3$.
Step $3$: This forms a buffer solution of $NaHCO_3$ (acid) and $Na_2CO_3$ (salt).
Using the Henderson-Hasselbalch equation:
$pH = pK_{a2} + \log \left( \frac{[Na_2CO_3]}{[NaHCO_3]} \right)$
$pH = 10.32 + \log \left( \frac{10}{20} \right)$
$pH = 10.32 - \log 2$
$pH = 10.32 - 0.30 = 10.02$
Rounding to the nearest option,the answer is $10.1$.

(A) Initial $pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]} = 4.745 + \log \frac{0.10}{0.10} = 4.745$.
Since $pH + pOH = 14$,initial $pH = 14 - 4.745 = 9.255$.
On adding $0.05 \ mol$ of $HCl$,the reaction is: $NH_3 + H^+ \rightarrow NH_4^+$.
New moles: $NH_3 = 0.10 - 0.05 = 0.05 \ mol$,$NH_4^+ = 0.10 + 0.05 = 0.15 \ mol$.
New $pOH' = 4.745 + \log \frac{0.15}{0.05} = 4.745 + \log 3 = 4.745 + 0.477 = 5.222$.
New $pH' = 14 - 5.222 = 8.778$.
Change in $pH = |pH' - pH| = |8.778 - 9.255| = 0.477$.
$0.477 = 47.7 \times 10^{-2} \approx 48 \times 10^{-2}$.

(D) The reaction forms an acidic buffer solution.
$pK_a = -\log(K_a) = -\log(2 \times 10^{-5}) = 5 - \log(2) = 5 - 0.3 = 4.7$.
$CH_3COOH + KOH \rightarrow CH_3COOK + H_2O$.
Initial moles of $CH_3COOH = 50 \times 0.1 = 5 \ mmol$.
Initial moles of $KOH = 12.5 \times 0.1 = 1.25 \ mmol$.
After reaction,remaining $CH_3COOH = 5 - 1.25 = 3.75 \ mmol$.
Formed $CH_3COOK = 1.25 \ mmol$.
Using Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
$pH = 4.7 + \log \frac{1.25}{3.75} = 4.7 + \log(\frac{1}{3})$.
$pH = 4.7 - 0.477 \approx 4.22$.

(D) Given $pH = 9.80$,so $pOH = 14 - 9.80 = 4.20$.
Using the Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_b + \log \frac{[NH_4^+]}{[NH_3]}$
$4.20 = 4.74 + \log \frac{[NH_4^+]}{[NH_3]}$
$-0.54 = \log \frac{[NH_4^+]}{[NH_3]}$
Given $\log 0.28 = -0.54$,so $\frac{[NH_4^+]}{[NH_3]} = 0.28$.
Let $V$ be the volume of $1 \ M \ NH_4Cl$ in $1000 \ mL$ of solution. Then the volume of $1 \ M \ NH_3$ is $(1000 - V) \ mL$.
Since molarities are equal,the ratio of concentrations is equal to the ratio of volumes:
$\frac{V}{1000 - V} = 0.28$
$V = 280 - 0.28V$
$1.28V = 280$
$V = \frac{280}{1.28} \approx 218.75 \ mL \approx 218 \ mL$.

(C) buffer solution is formed by a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt).
$HCl$ is a strong acid and $NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$.
The combination of a strong acid and a salt does not form a buffer solution.
Therefore,$HCl + NH_4Cl$ does not act as a buffer.

(B) For a basic buffer solution,the $pOH$ is calculated using the Henderson-Hasselbalch equation:
$pOH = pK_b + \log_{10} \frac{[Salt]}{[Base]}$
Given $pK_b = 4.744$,$[Salt] = [NH_4Cl] = 1 \ M$,and $[Base] = [NH_4OH] = 0.2 \ M$.
$pOH = 4.744 + \log_{10} \frac{1}{0.2} = 4.744 + \log_{10} (5)$
$pOH = 4.744 + 0.699 = 5.443$
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - 5.443 = 8.557 \approx 8.56$.

(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: $pK_{a} = 4.762$,$[\text{Salt}] = 0.01 \ M$,$[\text{Acid}] = 0.004 \ M$
Substituting the values:
$pH = 4.762 + \log_{10} \left( \frac{0.01}{0.004} \right)$
$pH = 4.762 + \log_{10} (2.5)$
Since $\log_{10} (2.5) \approx 0.398$:
$pH = 4.762 + 0.398 = 5.16$

(D) The solution is an acidic buffer consisting of a weak acid $(CH_3COOH)$ and its salt with a strong base $(CH_3COONa)$.
Using the Henderson-Hasselbalch equation:
$pH = pK_{a} + \log_{10} \frac{[Salt]}{[Acid]}$
Given:
$pK_{a} = 4.50$
$[Salt] = 0.1 \ M$
$[Acid] = 0.01 \ M$
Substituting the values:
$pH = 4.50 + \log_{10} \frac{0.1}{0.01}$
$pH = 4.50 + \log_{10} (10)$
Since $\log_{10} (10) = 1$,
$pH = 4.50 + 1 = 5.50$

(C) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$.
Given: $pK_{a} = 4.680$,$[Salt] = 0.02 \ M$,$[Acid] = 0.01 \ M$.
Substituting the values: $pH = 4.680 + \log \frac{0.02}{0.01}$.
$pH = 4.680 + \log(2)$.
Since $\log(2) \approx 0.301$,we get $pH = 4.680 + 0.301 = 4.981$.

(C) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right)$
Given:
$pK_a = 4.7447$
$[\text{Salt}] = 0.05 \ M$
$[\text{Acid}] = 0.01 \ M$
Substituting the values:
$pH = 4.7447 + \log \left( \frac{0.05}{0.01} \right)$
$pH = 4.7447 + \log(5)$
Since $\log(5) \approx 0.6990$:
$pH = 4.7447 + 0.6990 = 5.4437$
Rounding to two decimal places,the $pH$ is $5.44$.

(C) For an acidic buffer,the Henderson-Hasselbalch equation is given by:
$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$
Given:
$pH = 5.5$
$pK_{a} = 4.5$
$[Acid] = 0.1 \ M$
Substituting the values:
$5.5 = 4.5 + \log \frac{[Salt]}{0.1}$
$5.5 - 4.5 = \log \frac{[Salt]}{0.1}$
$1 = \log \frac{[Salt]}{0.1}$
Taking antilog on both sides:
$10^{1} = \frac{[Salt]}{0.1}$
$[Salt] = 10 \times 0.1 = 1.0 \ M$

(B) For a basic buffer,the Henderson-Hasselbalch equation is given by:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
Since equal volumes are mixed,the new concentrations are halved,but the ratio of $[Salt]/[Base]$ remains the same as the ratio of their initial concentrations.
$[Salt] = [NH_4Cl] = 0.5 \ M$
$[Base] = [NH_4OH] = 0.4 \ M$
$pOH = 4.730 + \log \left( \frac{0.5}{0.4} \right)$
$pOH = 4.730 + \log(1.25)$
$pOH = 4.730 + 0.0969 \approx 4.8269 \approx 4.83$

(B) The Henderson-Hasselbalch equation for an acidic buffer is given by: $pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$
Given $pH = 7.2$ and $pK_{a} = 6.2$.
Substituting the values: $7.2 = 6.2 + \log \frac{[Salt]}{[Acid]}$
$7.2 - 6.2 = \log \frac{[Salt]}{[Acid]}$
$1 = \log \frac{[Salt]}{[Acid]}$
Since $\log_{10} 10 = 1$,we have $\frac{[Salt]}{[Acid]} = 10$.

(A) For an acidic buffer solution,the Henderson-Hasselbalch equation is given by:
$pH = pK_{a} + \log_{10} \frac{[\text{Salt}]}{[\text{Acid}]}$
Given:
$pK_{a} = 4.2$
$[\text{Salt}] = 0.054 \ M$
$[\text{Acid}] = 0.027 \ M$
Substituting the values:
$pH = 4.2 + \log_{10} \left( \frac{0.054}{0.027} \right)$
$pH = 4.2 + \log_{10} (2)$
Since $\log_{10} 2 \approx 0.3010$,
$pH = 4.2 + 0.3010 = 4.5010$
Thus,the $pH$ is approximately $4.5$.

(C) For an acidic buffer,the Henderson-Hasselbalch equation is given by:
$pH = pK_{a} + \log_{10} \frac{[\text{salt}]}{[\text{acid}]}$
Given:
$pK_{a} = 4.56$
$[\text{salt}] = 0.70 \ M$
$[\text{acid}] = 0.35 \ M$
Substituting the values:
$pH = 4.56 + \log_{10} \frac{0.70}{0.35}$
$pH = 4.56 + \log_{10} (2)$
Since $\log_{10} (2) \approx 0.3010$:
$pH = 4.56 + 0.3010 = 4.861$
Thus,the $pH$ of the buffer solution is $4.86$.

(A) basic buffer is a mixture of a weak base and its salt with a strong acid.
In option $A$,$NH_4OH$ is a weak base and $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$.
Therefore,$NH_4OH + NH_4Cl$ acts as a basic buffer.
Options $B$,$C$,and $D$ consist of a weak acid and its salt with a strong base,which are examples of acidic buffers.

(B) The solution containing $0.04 \ M$ $NaF$ (salt) and $0.02 \ M$ $HF$ (weak acid) forms an acidic buffer.
Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log_{10} \frac{[Salt]}{[Acid]}$
Substituting the given values:
$pH = 3.142 + \log_{10} \frac{0.04}{0.02}$
$pH = 3.142 + \log_{10}(2)$
Since $\log_{10}(2) \approx 0.3010$:
$pH = 3.142 + 0.3010 = 3.443$
Rounding to one decimal place,we get $pH \approx 3.4$.

(C) The $pH$ of human blood is maintained naturally at $7.36-7.42$ by the $(HCO_3^{-} + H_2CO_3)$ buffer system.
This buffer system consists of carbonic acid $(H_2CO_3)$ and its conjugate base,the bicarbonate ion $(HCO_3^{-})$.

(A) For an acidic buffer,the Henderson-Hasselbalch equation is given by:
$pH = pK_{a} + \log \frac{[\text{Salt}]}{[\text{Acid}]}$
Given that the concentrations of the weak acid and its salt are equal,i.e.,$[\text{Salt}] = [\text{Acid}]$.
Therefore,the equation simplifies to:
$pH = pK_{a} = -\log K_{a}$
Given $K_{a} = 1.8 \times 10^{-5}$.
$pH = -\log (1.8 \times 10^{-5}) = -(\log 1.8 + \log 10^{-5})$
$pH = -(\log 1.8 - 5) = 5 - \log 1.8$
Using $\log 1.8 \approx 0.2553$:
$pH = 5 - 0.2553 = 4.7447$

(B) The concentration of $H^{+}$ ions in an acidic buffer can be calculated using the Henderson-Hasselbalch equation:
$[H^{+}] = K_{a} \times \frac{[Acid]}{[Salt]}$
Given:
$K_{a} = 6.6 \times 10^{-10}$
$[Acid] = [HCN] = 0.01 \ M$
$[Salt] = [NaCN] = 0.02 \ M$
Substituting the values:
$[H^{+}] = (6.6 \times 10^{-10}) \times \frac{0.01}{0.02}$
$[H^{+}] = (6.6 \times 10^{-10}) \times 0.5$
$[H^{+}] = 3.3 \times 10^{-10} \ M$

(C) An acidic buffer is prepared by mixing a weak acid and its salt with a strong base.
For example,a mixture of $CH_3COOH$ (weak acid) and $CH_3COONa$ (salt of weak acid and strong base) acts as an acidic buffer.
The weak acid and its conjugate base (provided by the salt) work together to resist changes in $pH$ upon the addition of small amounts of acid or base.