Buffer solution Questions in English

(D) buffer solution is formed by a weak acid and its conjugate base,or a weak base and its conjugate acid.
In option $A$,$CH_3COOH$ (weak acid) and $NaOH$ (strong base) react in a $1:1$ molar ratio ($5 \ mmol$ each) to form $CH_3COONa$ (salt),leaving no weak acid behind.
In option $B$,$HCl$ (strong acid) and $NaOH$ (strong base) neutralize completely to form $NaCl$ and water.
In option $C$,$NH_3$ (weak base) and $HCl$ (strong acid) react in a $1:1$ molar ratio ($10 \ mmol$ each) to form $NH_4Cl$ (salt),leaving no weak base behind.
In option $D$,$NH_3$ $(10 \ mmol)$ reacts with $HCl$ $(5 \ mmol)$ to form $5 \ mmol$ of $NH_4Cl$ and leaves $5 \ mmol$ of unreacted $NH_3$. This mixture of a weak base $(NH_3)$ and its conjugate acid $(NH_4^+)$ acts as a basic buffer.

(B) An acidic buffer is prepared by mixing a weak acid and its salt with a strong base.
Since the solution contains a weak acid,the concentration of $H^+$ ions is greater than $10^{-7} \ M$.
Therefore,the $pH$ of an acidic buffer mixture is always less than $7$ $(pH < 7)$.

(A) The given solution is an acidic buffer.
Meq of acetic acid $(CH_3COOH)$ $= 50 \, mL \times 2 \, N = 100 \, meq$.
Meq of sodium acetate $(CH_3COONa)$ $= 10 \, mL \times 1 \, N = 10 \, meq$.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
$pK_a = -\log(K_a) = -\log(10^{-5}) = 5$.
$pH = 5 + \log \frac{10}{100} = 5 + \log(0.1) = 5 - 1 = 4$.

(C) Given $pH = 9.3$ and $K_a = 5 \times 10^{-10}$.
$pK_a = -\log(K_a) = -\log(5 \times 10^{-10}) = 10 - \log 5 = 10 - 0.7 = 9.3$.
Using the Henderson-Hasselbalch equation for an acidic buffer:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
$9.3 = 9.3 + \log \frac{[NaCN]}{[HCN]}$
$\log \frac{[NaCN]}{[HCN]} = 0$
$\frac{[NaCN]}{[HCN]} = 10^0 = 1$.

(C) The solution is a buffer containing $CH_3COOH$ (acid) and $CH_3COONa$ (conjugate base).
Initially,$[Acid] = 0.15 \ mol$ and $[Salt] = 0.15 \ mol$.
When $0.05 \ mol$ of $NaOH$ is added,it reacts with the acid $(CH_3COOH)$:
$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$
New moles of acid = $0.15 - 0.05 = 0.10 \ mol$.
New moles of salt = $0.15 + 0.05 = 0.20 \ mol$.
Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
$pH = 4.8 + \log \frac{0.20}{0.10}$
$pH = 4.8 + \log (2)$
$pH = 4.8 + 0.301 = 5.101 \approx 5.1$.

(A) The reaction is: $CH_3COOH + NaOH \to CH_3COONa + H_2O$
Initial millimoles of $CH_3COOH = 30 \ mL \times 0.2 \ M = 6 \ mmol$
Initial millimoles of $NaOH = 20 \ mL \times 0.1 \ M = 2 \ mmol$
After the reaction,$2 \ mmol$ of $NaOH$ reacts with $2 \ mmol$ of $CH_3COOH$ to form $2 \ mmol$ of $CH_3COONa$.
Remaining $CH_3COOH = 6 - 2 = 4 \ mmol$
Formed $CH_3COO^- = 2 \ mmol$
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[CH_3COO^-]}{[CH_3COOH]}$
$pH = 4.74 + \log \frac{2}{4} = 4.74 + \log(0.5)$
$pH = 4.74 - 0.30 = 4.44$

(B) The $pH$ of an acidic buffer is given by the Henderson-Hasselbalch equation: $pH = pKa + \log \frac{[Salt]}{[Acid]}$.
Given: $pH = 4.8$,$pKa = 4.8$,$V_{acid} = 50 \ mL$,$M_{acid} = 0.2 \ M$,$V_{salt} = 250 \ mL$.
Substituting the values: $4.8 = 4.8 + \log \frac{[Salt]}{[Acid]}$.
This implies $\log \frac{[Salt]}{[Acid]} = 0$,so $\frac{[Salt]}{[Acid]} = 10^0 = 1$.
Therefore,$[Salt] = [Acid]$.
Calculate the moles of acid: $n_{acid} = M \times V = 0.2 \ M \times 50 \ mL = 10 \ mmol$.
Since $[Salt] = [Acid]$,the moles of salt must also be $10 \ mmol$ in the final mixture.
Concentration of salt $[Salt] = \frac{n_{salt}}{V_{total}} = \frac{10 \ mmol}{(50 + 250) \ mL} = \frac{10}{300} \ M = 0.0333 \ M$.
Wait,re-evaluating the question context: If the question implies the concentration of the stock solution of $CH_3COONa$ added,let $M_{salt}$ be the concentration. Then $n_{salt} = M_{salt} \times 250 \ mL$. Since $[Salt] = [Acid]$ in the mixture,$\frac{M_{salt} \times 250}{300} = \frac{0.2 \times 50}{300}$.
Thus,$M_{salt} \times 250 = 10$,so $M_{salt} = \frac{10}{250} = 0.04 \ M$.

(D) 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 $D$,$HNO_3$ is a strong acid and $NH_4OH$ is a weak base.
The reaction is: $HNO_3 + NH_4OH \rightleftharpoons NH_4NO_3 + H_2O$.
Initial moles $(n_0)$: $HNO_3 = 0.2 \ mol$,$NH_4OH = 0.3 \ mol$.
After the reaction $(n_f)$: $HNO_3 = 0 \ mol$,$NH_4OH = 0.1 \ mol$,$NH_4NO_3 = 0.2 \ mol$.
Since we have a mixture of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4NO_3)$,this forms a basic buffer solution.

(A) For the titration of a weak acid $(CH_3COOH)$ with a strong base $(NaOH)$,the $pH$ at any stage of neutralization is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
At $1/4$ neutralization,the concentration of salt is $1/4$ and the concentration of remaining acid is $3/4$. Thus,$pH_1 = pK_a + \log \frac{1/4}{3/4} = pK_a + \log \frac{1}{3}$.
At $3/4$ neutralization,the concentration of salt is $3/4$ and the concentration of remaining acid is $1/4$. Thus,$pH_2 = pK_a + \log \frac{3/4}{1/4} = pK_a + \log 3$.
The difference in $pH$ is $pH_2 - pH_1 = (pK_a + \log 3) - (pK_a + \log \frac{1}{3}) = \log 3 - \log \frac{1}{3} = \log 3 - (-\log 3) = 2 \log 3$.

(A) An acidic buffer is prepared by mixing a weak acid and its salt with a strong base.
Acetic acid $(CH_3COOH)$ is a weak acid and sodium acetate $(CH_3COONa)$ is its salt with a strong base $(NaOH)$.
Therefore,a mixture of acetic acid and sodium acetate acts as an acidic buffer.
Its $pH$ is typically around $4.75$.

(C) The reaction between $CH_3NH_2$ and $HCl$ is: $CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^-$.
Initial moles: $CH_3NH_2 = 0.1$,$HCl = 0.08$.
After reaction: $CH_3NH_2 = 0.1 - 0.08 = 0.02 \ mol$,$CH_3NH_3^+ = 0.08 \ mol$.
This forms a basic buffer solution.
Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pK_b + \log(\frac{[Salt]}{[Base]})$.
$pK_b = -\log(5 \times 10^{-4}) = 4 - \log(5) = 4 - 0.7 = 3.3$.
$pOH = 3.3 + \log(\frac{0.08}{0.02}) = 3.3 + \log(4) = 3.3 + 0.6 = 3.9$.
$[OH^-] = 10^{-pOH} = 10^{-3.9} = 1.258 \times 10^{-4} \ M$.
Using $[H^+][OH^-] = 10^{-14}$,$[H^+] = \frac{10^{-14}}{1.258 \times 10^{-4}} \approx 7.95 \times 10^{-11} \ M \approx 8 \times 10^{-11} \ M$.
Wait,re-evaluating: $[H^+] = \frac{K_w}{[OH^-]} = \frac{10^{-14}}{1.258 \times 10^{-4}} \approx 0.795 \times 10^{-10} = 7.95 \times 10^{-11} \ M$.
Checking options,$1.6 \times 10^{-11}$ is closest if $K_b$ calculation or ratio differs,but let's re-calculate: $[H^+] = K_a \times \frac{[Salt]}{[Base]} = \frac{K_w}{K_b} \times \frac{[Salt]}{[Base]} = \frac{10^{-14}}{5 \times 10^{-4}} \times \frac{0.08}{0.02} = 2 \times 10^{-11} \times 4 = 8 \times 10^{-11} \ M$.
Since $8 \times 10^{-11}$ is not an option,let's re-check the math: $0.08/0.02 = 4$. $K_a = 2 \times 10^{-11}$. $[H^+] = 2 \times 10^{-11} \times 4 = 8 \times 10^{-11}$. Given the options,$1.6 \times 10^{-11}$ might be intended if $K_b$ was $2.5 \times 10^{-4}$ or similar. However,based on the provided values,the answer is $8 \times 10^{-11} \ M$. Assuming a typo in options,$C$ is the closest magnitude.

(C) The solution is a buffer solution containing a weak acid (propanoic acid) and its salt with a strong base (sodium propanoate).
For a buffer solution,the $pH$ is calculated using the Henderson-Hasselbalch equation:
$pH = pKa + \log \frac{[Salt]}{[Acid]}$
Given that equal volumes are mixed,the concentrations of both the salt and the acid are halved,but their ratio remains the same:
$[Salt] = 0.05 \ M$,$[Acid] = 0.05 \ M$
$pKa = -\log(Ka) = -\log(1.3 \times 10^{-5}) = 5 - \log(1.3) \approx 5 - 0.1139 = 4.8861$
$pH = 4.8861 + \log \frac{0.05}{0.05} = 4.8861 + \log(1) = 4.8861 + 0 = 4.8861$
Rounding to two decimal places,the $pH$ is approximately $4.89$.

(D) The given solution is a buffer solution consisting of a weak acid $(HCN)$ and its salt $(KCN)$.
Using the Henderson-Hasselbalch equation for an acidic buffer:
$pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right)$
Given: $pH = 3$,$pK_a = 6$,$[HCN] = 0.1 \, M$,$V_{HCN} = 200 \, mL$,$[KCN] = 0.2 \, M$.
Let the volume of $KCN$ be $V \, mL$.
$3 = 6 + \log \left( \frac{0.2 \times V}{0.1 \times 200} \right)$
$-3 = \log \left( \frac{0.2V}{20} \right)$
$-3 = \log \left( \frac{V}{100} \right)$
$10^{-3} = \frac{V}{100}$
$V = 100 \times 10^{-3} = 0.1 \, mL$.

(D) buffer solution is formed when a weak acid reacts with a limited amount of a strong base,or a weak base reacts with a limited amount of a strong acid,resulting in a mixture of the weak electrolyte and its conjugate salt.
$(A)$ $100 \ mL, 0.1 \ M \ CH_3COOH$ $(10 \ mmol)$ + $100 \ mL, 0.05 \ M \ NaOH$ $(5 \ mmol)$: This results in $5 \ mmol$ of $CH_3COOH$ and $5 \ mmol$ of $CH_3COONa$,forming an acidic buffer.
$(B)$ $200 \ mL, 0.1 \ M \ NH_4OH$ $(20 \ mmol)$ + $200 \ mL, 0.08 \ M \ HCl$ $(16 \ mmol)$: This results in $4 \ mmol$ of $NH_4OH$ and $16 \ mmol$ of $NH_4Cl$,forming a basic buffer.
$(C)$ $300 \ mL, 0.1 \ M \ NaOH$ $(30 \ mmol)$ + $500 \ mL, 0.1 \ M \ C_6H_5COOH$ $(50 \ mmol)$: This results in $20 \ mmol$ of $C_6H_5COOH$ and $30 \ mmol$ of $C_6H_5COONa$,forming an acidic buffer.
Since all options result in a mixture of a weak acid/base and its salt,all are buffer solutions.

(A) Given that the concentration of $X^{-}$ and $HX$ are equal,$[X^{-}] = [HX]$.
For the conjugate acid-base pair,$K_a \times K_b = K_w = 10^{-14}$.
$K_a = \frac{K_w}{K_b} = \frac{10^{-14}}{10^{-10}} = 10^{-4}$.
$pK_a = -\log(K_a) = -\log(10^{-4}) = 4$.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[X^{-}]}{[HX]}$.
Since $[X^{-}] = [HX]$,$\log \frac{[X^{-}]}{[HX]} = \log(1) = 0$.
Therefore,$pH = pK_a = 4$.

(C) The solution is a buffer solution containing a weak acid $(CH_3COOH)$ and its salt with a strong base $(CH_3COONa)$.
$pH$ is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$.
$1$. Calculate moles of acetic acid $(CH_3COOH)$: Molar mass = $60 \ g/mol$. Moles = $2 \ g / 60 \ g/mol = 0.0333 \ mol$.
$2$. Calculate moles of sodium acetate $(CH_3COONa)$: Molar mass = $82 \ g/mol$. Moles = $3 \ g / 82 \ g/mol = 0.0366 \ mol$.
$3$. Since the volume is the same $(100 \ mL)$,the ratio of concentrations is equal to the ratio of moles: $\frac{[Salt]}{[Acid]} = \frac{0.0366}{0.0333} \approx 1.1$.
$4$. Calculate $pK_a$: $pK_a = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
$5$. Calculate $pH$: $pH = 4.745 + \log(1.1) = 4.745 + 0.041 = 4.786$.
Thus,the $pH$ is approximately $4.78$.

(B) 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: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given: $pH = 9.25$,$[Salt] = 0.1 \ N$,$[Base] = 0.1 \ N$.
First,calculate $pOH$: $pOH = 14 - pH = 14 - 9.25 = 4.75$.
Substitute the values into the equation: $4.75 = pK_b + \log \frac{0.1}{0.1}$.
Since $\log(1) = 0$,we get $4.75 = pK_b + 0$.
Therefore,$pK_b = 4.75$.

(C) buffer solution is a mixture of a weak acid and its conjugate base,or a weak base and its conjugate acid.
$NaH_2PO_4 + H_3PO_4$ is a mixture of a weak acid and its conjugate base.
$CH_3COOH + CH_3COONa$ is a mixture of a weak acid and its conjugate base.
$B(OH)_3 + \text{borax}$ acts as a buffer system.
$HCl + NH_4OH$ consists of a strong acid $(HCl)$ and a weak base $(NH_4OH)$. When mixed,they react to form a salt $(NH_4Cl)$ and water. Since the strong acid is completely neutralized by the weak base,it does not form a buffer solution.

(A) The given mixture is an acidic buffer consisting of a weak acid $(CH_3COOH)$ and its salt with a strong base $(CH_3COONa)$.
According to the Henderson-Hasselbalch equation for an acidic buffer:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Here,the salt is sodium acetate $(CH_3COONa)$ and the acid is acetic acid $(CH_3COOH)$.
When more sodium acetate is added,the concentration of the salt $[Salt]$ increases.
Since $pH$ is directly proportional to $\log [Salt]$,an increase in the concentration of the salt leads to an increase in the $pH$ of the solution.

(D) buffer solution with a $pH > 7$ is a basic buffer,which consists of a weak base and its salt with a strong acid.
$CH_3COOH + CH_3COONa$ is an acidic buffer $(pH < 7)$.
$HCOOH + HCOOK$ is an acidic buffer $(pH < 7)$.
$CH_3COONH_4 + CH_3COOH$ is an acidic buffer $(pH < 7)$.
$NH_4OH + NH_4Cl$ consists of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$,forming a basic buffer with $pH > 7$.

(A) The $pH$ of blood is maintained by the bicarbonate buffer system: $CO_2 + H_2O \rightleftharpoons H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$.
When a person breathes rapidly,$CO_2$ is expelled from the body,causing the concentration of $CO_2$ to decrease.
According to Le Chatelier's principle,the equilibrium shifts to the left to produce more $CO_2$.
This consumption of $H^+$ ions leads to a decrease in the concentration of $H^+$ ions in the blood.
Since $pH = -\log[H^+]$,a decrease in $[H^+]$ results in an increase in the $pH$ of the blood.

(A) The acidity of a buffer solution is determined by the $pK_a$ value of the weak acid used. $A$ lower $pK_a$ value indicates a stronger acid.
Comparing the $pK_a$ values of the conjugate acids:
$1$. $HCOOH$: $pK_a \approx 3.75$
$2$. $CH_3COOH$: $pK_a \approx 4.76$
$3$. $HC_2O_4^-$: $pK_a \approx 4.19$
$4$. $H_3BO_3$: $pK_a \approx 9.24$
Since $HCOOH$ has the lowest $pK_a$ value among the given options,the buffer solution containing $HCOOH + HCOO^-$ will be the most acidic.

(C) buffer solution with $pH = 9$ is basic in nature.
For a basic buffer,$pOH = 14 - pH = 14 - 9 = 5$.
The $pKb$ of $NH_4OH$ is approximately $4.75$.
Using the Henderson-Hasselbalch equation for a basic buffer: $pOH = pKb + log \frac{[Salt]}{[Base]}$.
Since $NH_4Cl$ (salt) and $NH_4OH$ (weak base) form a basic buffer with a $pOH$ close to $5$,this mixture is the correct choice.
$CH_3COONa$ and $CH_3COOH$ form an acidic buffer $(pH < 7)$.
$NaCl$ and $NaOH$ do not form a buffer.
$KH_2PO_4$ and $K_2HPO_4$ form a buffer with $pH$ near $7.2$.

(B) For an acidic buffer,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Given: $pH = 6$ and $K_a = 10^{-5}$.
First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(10^{-5}) = 5$.
Substitute the values into the equation: $6 = 5 + \log \frac{[Salt]}{[Acid]}$
$6 - 5 = \log \frac{[Salt]}{[Acid]}$
$1 = \log \frac{[Salt]}{[Acid]}$
Taking the antilog on both sides: $\frac{[Salt]}{[Acid]} = 10^1 = 10$.
Therefore,the ratio of salt to acid is $10 : 1$.

(D) buffer solution is formed by a weak acid and its conjugate base,or a weak base and its conjugate acid.
$H_2S$ is a weak acid and $KHS$ is its conjugate base,so option $A$ is a buffer.
$C_6H_5NH_2$ (aniline) is a weak base and $C_6H_5NH_3Br$ is its conjugate acid,so option $B$ is a buffer.
$H_2CO_3$ is a weak acid and $KHCO_3$ is its conjugate base,so option $C$ is a buffer.
$HClO_4$ is a strong acid. $A$ mixture of a strong acid and its salt does not form a buffer solution. Therefore,option $D$ is not a buffer.

(B) For a buffer solution,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Given that the concentrations of $X^-$ (salt) and $HX$ (acid) are equal,i.e.,$[X^-] = [HX]$.
Therefore,$\frac{[X^-]}{[HX]} = 1$.
Since $\log(1) = 0$,the equation simplifies to $pH = pK_a$.
Given $K_a = 10^{-8}$,we calculate $pK_a = -\log(K_a) = -\log(10^{-8}) = 8$.
Thus,$pH = 8$.

(C) The given solution is a buffer solution consisting of a weak acid (acetic acid) and its conjugate base (sodium acetate).
The $pH$ of such a buffer can be calculated using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \left( \frac{[\text{salt}]}{[\text{acid}]} \right)$
Given:
$pK_a = 4.57$
$[\text{salt}] = [\text{CH}_3\text{COONa}] = 0.10 \ M$
$[\text{acid}] = [\text{CH}_3\text{COOH}] = 0.03 \ M$
Substituting the values:
$pH = 4.57 + \log \left( \frac{0.10}{0.03} \right)$
$pH = 4.57 + \log(3.333)$
$pH = 4.57 + 0.5228$
$pH = 5.0928 \approx 5.09$

(C) The reaction is: $NH_3 + HCl \to NH_4Cl$.
Initial moles of $NH_3 = 40 \ mL \times 0.1 \ M = 4 \ mmol$.
Initial moles of $HCl = 20 \ mL \times 0.1 \ M = 2 \ mmol$.
After the reaction,$2 \ mmol$ of $NH_3$ remains and $2 \ mmol$ of $NH_4Cl$ is formed.
This forms a basic buffer solution.
Using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Since the volume is the same for both,we can use the ratio of moles: $pOH = 4.74 + \log \frac{2}{2} = 4.74 + 0 = 4.74$.
Finally,$pH = 14 - pOH = 14 - 4.74 = 9.26$.

(C) For the weak acid $HA$,the degree of ionization $\alpha = 0.1$ at concentration $C = 0.01 \ M$.
The dissociation constant $K_a$ is given by $K_a = \frac{C \alpha^2}{1 - \alpha} = \frac{0.01 \times (0.1)^2}{1 - 0.1} = \frac{0.0001}{0.9} = 1.11 \times 10^{-4}$.
Calculate $pK_a$: $pK_a = -\log(1.11 \times 10^{-4}) = 4 - \log(1.11) \approx 4 - 0.0458 = 3.9542$.
Using the Henderson-Hasselbalch equation for the buffer solution containing $HA$ $(0.1 \ M)$ and $NaA$ $(0.05 \ M)$:
$pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right) = 3.9542 + \log \left( \frac{0.05}{0.1} \right)$.
$pH = 3.9542 + \log(0.5) = 3.9542 - 0.3010 = 3.6532 \approx 3.653$.

(A) Blood maintains a constant $pH$ because it contains serum proteins and bicarbonate ions which act as a buffer system.
This buffer system resists changes in $pH$ upon the addition of small amounts of acid or base.

(D) An acidic buffer is a solution containing a mixture of a weak acid and its salt with a strong base.
$CH_3COOH$ is a weak acid,but $CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.
Therefore,the mixture of $CH_3COOH$ and $CH_3COONH_4$ does not form an acidic buffer.
Since the assertion is false and the definition provided in the reason is also incorrect (it should be a salt with a strong base),both the assertion and the reason are incorrect.

(A) In the titration of a weak acid $(HA)$ with a strong base $(NaOH)$,the reaction is $HA + OH^- \rightarrow A^- + H_2O$.
At the half equivalence point,half of the acid has been neutralized,meaning $[HA] = [A^-]$.
According to the Henderson-Hasselbalch equation,$pH = pK_a + \log(\frac{[salt]}{[acid]})$.
Since $[salt] = [acid]$ at this point,$\log(1) = 0$,so $pH = pK_a$.
This mixture acts as an acidic buffer,and the buffer capacity is indeed maximum when the concentration of the acid equals the concentration of its conjugate base (salt).

(D) The Assertion is incorrect because the buffer system of carbonic acid $(H_2CO_3)$ and sodium bicarbonate $(NaHCO_3)$ is a biological buffer found in blood,not a reagent used for the precipitation of group $III$ hydroxides (which typically uses $NH_4Cl$ and $NH_4OH$).
The Reason is correct as this specific buffer system does maintain the $pH$ of human blood at approximately $7.4$.

(C) basic buffer is formed by a mixture of a weak base and its salt with a strong acid.
In option $C$,we have $NH_{4}OH$ (weak base) and $HCl$ (strong acid).
Initial milli-moles of $HCl = 100 \times 0.1 = 10 \ mmol$.
Initial milli-moles of $NH_{4}OH = 200 \times 0.1 = 20 \ mmol$.
The reaction is: $HCl + NH_{4}OH \rightarrow NH_{4}Cl + H_{2}O$.
After the reaction,$10 \ mmol$ of $NH_{4}OH$ remains,and $10 \ mmol$ of $NH_{4}Cl$ (salt) is formed.
Since the mixture contains a weak base and its salt,it forms a basic buffer.

(D) $1$. Calculate the milliequivalents $(meq)$ in the $500 \; mL$ solution:
$meq$ of $HCl = 250 \; mL \times 0.1 \; M = 25 \; meq$.
$meq$ of $CH_3COOH = \frac{3 \; g}{60 \; g/mol} = 0.05 \; mol = 50 \; meq$.
$2$. Calculate the $meq$ in $20 \; mL$ of the solution:
$meq$ of $HCl = \frac{25 \; meq}{500 \; mL} \times 20 \; mL = 1 \; meq$.
$meq$ of $CH_3COOH = \frac{50 \; meq}{500 \; mL} \times 20 \; mL = 2 \; meq$.
$3$. Calculate the $meq$ of $NaOH$ added:
$meq$ of $NaOH = 5 \; M \times 0.5 \; mL = 2.5 \; meq$.
$4$. Reaction:
$HCl$ reacts first: $1 \; meq$ of $NaOH$ neutralizes $1 \; meq$ of $HCl$.
Remaining $NaOH = 2.5 - 1 = 1.5 \; meq$.
This $1.5 \; meq$ of $NaOH$ reacts with $CH_3COOH$:
$CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
Remaining $CH_3COOH = 2 - 1.5 = 0.5 \; meq$.
Formed $CH_3COONa = 1.5 \; meq$.
$5$. Calculate $pH$ using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]} = 4.75 + \log \frac{1.5}{0.5} = 4.75 + \log 3 = 4.75 + 0.4771 = 5.2271 \approx 5.23$.

(N/A) The given solution is a basic buffer containing a weak base $(NH_{3})$ and its salt with a strong acid $(NH_{4}Cl)$.
Using the Henderson-Hasselbalch equation for basic buffers:
$pOH = pK_{b} + \log \left( \frac{[Salt]}{[Base]} \right)$
Given:
$pK_{b} = 4.75$
$[Salt] = [NH_{4}Cl] = 0.2 \, M$
$[Base] = [NH_{3}] = 0.1 \, M$
Substituting the values:
$pOH = 4.75 + \log \left( \frac{0.2}{0.1} \right)$
$pOH = 4.75 + \log(2)$
$pOH = 4.75 + 0.301 = 5.051$
Since $pH + pOH = 14$ at $25^{\circ}C$:
$pH = 14 - 5.051 = 8.949 \approx 8.95$

(N/A) $NH_{3} + H_{2}O \rightleftharpoons NH_{4}^{+} + OH^{-}$
$K_{b} = \frac{[NH_{4}^{+}][OH^{-}]}{[NH_{3}]} = 1.77 \times 10^{-5}$
Before neutralization:
$[NH_{4}^{+}] = [OH^{-}] = x$
$[NH_{3}] = 0.10 - x \approx 0.10 \,M$
$\frac{x^{2}}{0.10} = 1.77 \times 10^{-5} \implies x = \sqrt{1.77 \times 10^{-6}} = 1.33 \times 10^{-3} \,M = [OH^{-}]$
$[H^{+}] = \frac{K_{w}}{[OH^{-}]} = \frac{10^{-14}}{1.33 \times 10^{-3}} = 7.52 \times 10^{-12} \,M$
$pH = -\log(7.52 \times 10^{-12}) = 11.12$
After adding $25 \,mL$ of $0.1 \,M$ $HCl$ to $50 \,mL$ of $0.1 \,M$ $NH_{3}$:
Initial $mmol$ of $NH_{3} = 50 \times 0.1 = 5 \,mmol$
Initial $mmol$ of $HCl = 25 \times 0.1 = 2.5 \,mmol$
$NH_{3} + HCl \rightarrow NH_{4}^{+} + Cl^{-}$
Remaining $NH_{3} = 5 - 2.5 = 2.5 \,mmol$
Formed $NH_{4}^{+} = 2.5 \,mmol$
Total volume = $75 \,mL$
Using Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_{b} + \log\left(\frac{[Salt]}{[Base]}\right)$
$pK_{b} = -\log(1.77 \times 10^{-5}) = 4.75$
$pOH = 4.75 + \log\left(\frac{2.5/75}{2.5/75}\right) = 4.75 + \log(1) = 4.75$
$pH = 14 - pOH = 14 - 4.75 = 9.25$

(B) The human blood $pH$ is tightly regulated by buffer systems,primarily the bicarbonate buffer system,to maintain a physiological range of $7.26$ to $7.42$.

(N/A) Definition: The solutions which resist change in $pH$ on dilution or with the addition of small amounts of acid or alkali are called buffer solutions. Buffer solutions are of two types: acidic and basic.
$(A)$ Acidic buffer solutions: The $pH$ value of these solutions is less than $7.0$. The examples of acidic buffers are as under:

Acid $+$ Salt of acid	Approx $pH$
$CH_{3}COOH + CH_{3}COONa$	$4.75$
$C_{6}H_{5}COOH + C_{6}H_{5}COONa$	$3.7$
$C_{6}H_{4}(COOH)_{2} + C_{6}H_{4}(COOH)(COOK)$	$2.9$
$HCOOH + HCOONa$	$3.7$

$pH$ of an acidic buffer is calculated as follows:
$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$
$(B)$ Basic buffer solutions: The $pH$ value of these solutions is greater than $7.0$. These consist of a weak base and its salt with a strong acid. Example: $NH_{4}OH + NH_{4}Cl$ $(pH \approx 9.25)$.
$pOH = pK_{b} + \log \frac{[Salt]}{[Base]}$

(N/A) Buffer solutions are crucial in chemical and biochemical reactions,particularly in analytical chemistry.
$(i)$ In biological processes: For example,the $pH$ of human blood is maintained between $7.36$ and $7.42$ by the $(H_{2}CO_{3} + NaHCO_{3})$ buffer system. The human body also utilizes buffers containing $[HCO_{3}^{-}] / [CO_{3}^{2-}]$ and $[H_{2}PO_{4}^{-}] / [HPO_{4}^{2-}]$.
$(ii)$ In industrial processes: Buffers are essential in various industries,including electroplating,the manufacture of leather,photographic materials,dyes,and the preservation of food articles.
$(iii)$ In the preparation of medicines: Maintaining a specific $pH$ is critical for the efficacy of many medicines,making the use of buffers essential.
$(iv)$ In cosmetic items: Buffers are used to maintain the $pH$ of cosmetic products to ensure they are not harmful to the skin.
$(v)$ In agriculture: Soil $pH$ is often regulated by natural buffer systems involving carbonates,bicarbonates,phosphates,and organic acids to support crop growth.
$(vi)$ In analytical chemistry: Buffer solutions are used to maintain specific $pH$ levels for the selective precipitation of ions,such as in group-$II$ $(HCl + H_{2}S)$,group-$III$ $(NH_{4}Cl + NH_{4}OH)$,and group-$IV$ $(NH_{4}Cl + NH_{4}OH + (NH_{4})_{2}S)$ analysis.

(N/A) The Henderson-Hasselbalch equation relates the $pH$ of a buffer solution to the $pK_{a}$ of the weak acid and the ratio of the concentrations of the conjugate base and the acid.
Consider the ionization of a weak acid $HA$ in water:
$HA + H_{2}O ⇌ H_{3}O^{+} + A^{-}$
The acid dissociation constant $K_{a}$ is given by:
$K_{a} = \frac{[H_{3}O^{+}][A^{-}]}{[HA]}$
Rearranging for $[H_{3}O^{+}]$:
$[H_{3}O^{+}] = K_{a} \times \frac{[HA]}{[A^{-}]}$
Taking the negative logarithm $(-\log)$ on both sides:
$-\log [H_{3}O^{+}] = -\log K_{a} - \log \frac{[HA]}{[A^{-}]}$
Since $pH = -\log [H_{3}O^{+}]$ and $pK_{a} = -\log K_{a}$,we get:
$pH = pK_{a} - \log \frac{[HA]}{[A^{-}]}$
By inverting the fraction inside the logarithm,the sign changes:
$pH = pK_{a} + \log \frac{[A^{-}]}{[HA]}$
This is the Henderson-Hasselbalch equation,where $[A^{-}]$ is the concentration of the conjugate base and $[HA]$ is the concentration of the acid.

(N/A) An acidic buffer solution consists of a weak acid and its salt with a strong base,such as a mixture of acetic acid $(CH_3COOH)$ and sodium acetate $(CH_3COONa)$.
The equilibrium of ionization of a weak acid $HA$ in water is represented as:
$HA + H_2O \rightleftharpoons H_3O^+ + A^-$
The acid dissociation constant $K_a$ is given by:
$K_a = \frac{[H_3O^+][A^-]}{[HA]}$
Rearranging for $[H_3O^+]$:
$[H_3O^+] = K_a \times \frac{[HA]}{[A^-]}$
Taking the negative logarithm on both sides:
$-\log[H_3O^+] = -\log K_a - \log \frac{[HA]}{[A^-]}$
Since $pH = -\log[H_3O^+]$ and $pK_a = -\log K_a$,we get:
$pH = pK_a - \log \frac{[HA]}{[A^-]}$
By inverting the log term:
$pH = pK_a + \log \frac{[A^-]}{[HA]}$
This is the Henderson-Hasselbalch equation,where $[A^-]$ is the concentration of the conjugate base and $[HA]$ is the concentration of the acid.

(N/A) An acidic buffer solution is prepared by mixing a weak acid $(HA)$ and its salt with a strong base $(A^-)$.
The $pH$ of the solution is determined by the Henderson-Hasselbalch equation:
$pH = pK_{a} + \log \frac{[Salt]}{[Acid]}$
If the concentration of the weak acid and its conjugate base (salt) are equal,i.e.,$[HA] = [A^-]$,then:
$pH = pK_{a} + \log(1) = pK_{a} + 0 = pK_{a}$.
Therefore,to prepare a buffer with a desired $pH$,one should select a weak acid whose $pK_{a}$ value is close to the target $pH$.
Example: Acetic acid $(CH_3COOH)$ has a $pK_{a}$ of $4.76$. $A$ buffer solution prepared by mixing equimolar concentrations of acetic acid and sodium acetate $(CH_3COONa)$ will have a $pH$ approximately equal to $4.76$.

(A) basic buffer is prepared by mixing a weak base $(NH_3)$ and its salt with a strong acid $(NH_4Cl)$.
For a basic buffer,the $pOH$ is calculated using the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given that $[Salt] = [Base]$,we have $pOH = pK_b = 4.75$.
Since $pH + pOH = 14$ at $25^{\circ}C$,the $pH = 14 - 4.75 = 9.25$.

(N/A) $1$. The initial $pH$ of $0.1 \ M$ $CH_3COOH$ is calculated using the formula: $pH = \frac{1}{2}(pK_a - \log C)$.
$2$. Substituting the values: $pH = \frac{1}{2}(4.74 - \log 0.1) = \frac{1}{2}(4.74 - (-1)) = \frac{1}{2}(5.74) = 2.87$.
$3$. After adding $0.1 \ M$ $CH_3COONa$,the solution becomes a buffer. The $pH$ is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$.
$4$. Substituting the values: $pH = 4.74 + \log(\frac{0.1}{0.1}) = 4.74 + \log(1) = 4.74 + 0 = 4.74$.
$5$. The change in $pH$ is: $\Delta pH = 4.74 - 2.87 = 1.87$.

(N/A) $1$. Calculate the $pH$ of $0.1 \ M \ NH_4OH$ (weak base):
$[OH^-] = \sqrt{K_b \times C} = \sqrt{1.77 \times 10^{-5} \times 0.1} = \sqrt{1.77 \times 10^{-6}} \approx 1.33 \times 10^{-3} \ M$.
$pOH = -\log(1.33 \times 10^{-3}) \approx 2.876$.
$pH = 14 - 2.876 = 11.124$.
$2$. Calculate the $pH$ of the buffer solution formed by adding $0.1 \ M \ NH_4Cl$ to $0.1 \ M \ NH_4OH$:
Using Henderson-Hasselbalch equation: $pOH = pK_b + \log(\frac{[Salt]}{[Base]})$.
$pK_b = -\log(1.77 \times 10^{-5}) \approx 4.752$.
$pOH = 4.752 + \log(\frac{0.1}{0.1}) = 4.752 + 0 = 4.752$.
$pH = 14 - 4.752 = 9.248$.
$3$. Change in $pH = 11.124 - 9.248 = 1.876$.

(B) For a weak acid $HA$,the Henderson-Hasselbalch equation is given by: $pH = pKa + \log \frac{[A^-]}{[HA]}$.
Since the acid is $50\%$ ionized,the concentration of the conjugate base $[A^-]$ is equal to the concentration of the undissociated acid $[HA]$.
Therefore,$[A^-] = [HA]$,which implies $\log \frac{[A^-]}{[HA]} = \log(1) = 0$.
Thus,$pH = pKa = 4.5$.
Using the relation $pH + pOH = 14$ at $25^{\circ}C$,we get $pOH = 14 - pH = 14 - 4.5 = 9.5$.

(A) The solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its salt with a strong base $(CH_3COONa)$.
Using the Henderson-Hasselbalch equation for $[H^+]$:
$[H^+] = K_a \times \frac{[Acid]}{[Salt]}$
Given:
$K_a = 1.8 \times 10^{-5}$
$[Acid] = [CH_3COOH] = 0.10 \ M$
$[Salt] = [CH_3COONa] = 0.20 \ M$
Substituting the values:
$[H^+] = 1.8 \times 10^{-5} \times \frac{0.10}{0.20}$
$[H^+] = 1.8 \times 10^{-5} \times 0.5$
$[H^+] = 0.9 \times 10^{-5} = 9.0 \times 10^{-6} \ M$