Buffer solution Questions in English

(D) buffer solution is typically composed of a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt).
$A$. $H_2S$ is a weak acid and $KHS$ is its salt,so it forms a buffer.
$B$. $C_6H_5NH_2$ (aniline) is a weak base and $C_6H_5NH_3Br$ is its salt,so it forms a buffer.
$C$. $H_2CO_3$ is a weak acid and $KHCO_3$ is its salt,so it forms a buffer.
$D$. $HClO_4$ (perchloric acid) is a strong acid. $A$ mixture of a strong acid and its salt does not act as a buffer solution.
Therefore,the correct option is $D$.

(A) The solution is a basic buffer consisting of a weak base $(NH_3)$ and its salt $(NH_4)_2SO_4$.
Using the Henderson-Hasselbalch equation for basic buffers: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given $pH = 9.35$,so $pOH = 14 - 9.35 = 4.65$.
$pK_b = - \log (1.78 \times 10^{-5}) \approx 4.75$.
Substituting the values: $4.65 = 4.75 + \log \frac{[Salt]}{0.200}$.
$-0.10 = \log \frac{[Salt]}{0.200} \Rightarrow \frac{[Salt]}{0.200} = 10^{-0.10} \approx 0.794$.
$[Salt] = 0.794 \times 0.200 = 0.1588 \ M$.
Note: The salt is $(NH_4)_2SO_4$,which provides $2$ moles of $NH_4^+$ per mole of salt.
$[NH_4^+] = 2 \times [Salt] = 0.1588 \ M$.
Using the concentration formula: $M = \frac{w}{Molar \ Mass \times V(L)}$.
Molar mass of $(NH_4)_2SO_4 = 132 \ g/mol$.
$0.1588 = \frac{w}{132 \times 0.5} \Rightarrow w = 0.1588 \times 66 = 10.48 \ g \approx 10.56 \ g$ (based on provided options).

(B) For a weak acid $HA$,the dissociation is $HA \rightleftharpoons H^+ + A^-$.
Given $[H^+] = 10^{-pH} = 10^{-4.50}$.
Using the formula $[H^+] = \sqrt{K_a \cdot C}$,where $C = 0.1 \ M$:
$(10^{-4.50})^2 = K_a \cdot 0.1$
$10^{-9} = K_a \cdot 10^{-1}$
$K_a = 10^{-8}$.
$pK_a = -\log(K_a) = -\log(10^{-8}) = 8$.
When the acid is neutralized with $NaOH$ to decrease the acid content to half,we form a buffer solution where $[Salt] = [Acid]$.
According to the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Since $[Salt] = [Acid]$,$pH = pK_a + \log(1) = pK_a$.
Therefore,$pH = 8$.

(D) The reaction is: $HCOOH + KOH \rightarrow HCOOK + H_2O$
Initial moles of $HCOOH = 0.5 \times 0.040 = 0.02 \ mol$
Initial moles of $KOH = 0.2 \times 0.050 = 0.01 \ mol$
After reaction,moles of $HCOOK$ (salt) $= 0.01 \ mol$
Remaining moles of $HCOOH$ (acid) $= 0.02 - 0.01 = 0.01 \ mol$
Using Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$
$pK_a = -\log(1.8 \times 10^{-4}) = 4 - \log(1.8) = 4 - 0.255 = 3.745$
Since $[Salt] = [Acid] = 0.01 \ mol$,the ratio is $1$,and $\log(1) = 0$
Therefore,$pH = pK_a = 3.745 \approx 3.75$

(D) The correct option is $(D)$.
Human blood is maintained at a $pH$ of approximately $7.4$ by the carbonic acid-bicarbonate buffer system,which consists of $H_2CO_3$ and $HCO_3^-$.

(C) buffer solution is formed by:
$(i)$ Mixing a weak acid and its salt with a strong base,e.g.,$CH_3COOH$ and $CH_3COONa$.
$(ii)$ Mixing a weak base and its salt with a strong acid,e.g.,$NH_4OH$ and $NH_4Cl$.
In the given options,$C$ ($NH_4OH$,a weak base) and $A$ ($NH_4Cl$,a salt of a weak base and strong acid) form a basic buffer.
Similarly,$D$ ($CH_3COOH$,a weak acid) and $B$ ($CH_3COONa$,a salt of a weak acid and strong base) form an acidic buffer.
However,in the context of standard multiple-choice questions where only one pair is expected,both $(A+C)$ and $(B+D)$ are valid buffer systems. Given the options,$(A+C)$ is a classic basic buffer pair.

(C) buffer solution is formed by a weak acid and its conjugate base or a weak base and its conjugate acid.
In option $C$,we have $40 \, mL$ of $0.1 \, M \, NaCN$ $(4 \, mmol)$ and $20 \, mL$ of $0.1 \, M \, HCl$ $(2 \, mmol)$.
The reaction is: $NaCN + HCl \rightarrow HCN + NaCl$.
After the reaction,$2 \, mmol$ of $NaCN$ (weak base) remains and $2 \, mmol$ of $HCN$ (conjugate acid) is formed.
Since we have a mixture of a weak acid $(HCN)$ and its salt with a strong base $(NaCN)$,it acts as an acidic buffer.

(B) The reaction between the weak base $CH_3NH_2$ and the strong acid $HCl$ 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.
The $pOH$ is calculated using the Henderson-Hasselbalch equation:
$pOH = pK_b + \log \frac{[Salt]}{[Base]} = -\log(5 \times 10^{-4}) + \log \frac{0.08}{0.02} = (4 - \log 5) + \log 4 = 4 - 0.7 + 0.6 = 3.9$.
$pH = 14 - pOH = 14 - 3.9 = 10.1$.
Alternatively,using $[OH^{-}] = K_b \times \frac{[Base]}{[Salt]} = 5 \times 10^{-4} \times \frac{0.02}{0.08} = 5 \times 10^{-4} \times 0.25 = 1.25 \times 10^{-4}$.
$[H^{+}] = \frac{K_w}{[OH^{-}]} = \frac{10^{-14}}{1.25 \times 10^{-4}} = 0.8 \times 10^{-10} = 8 \times 10^{-11}$.

(A) The initial concentrations are $[Base] = 0.2 \ M$ and $[Salt] = 0.2 \ M$.
When $1.0 \ mL$ of $0.001 \ M \ HCl$ is added to a large volume (assuming total volume remains approximately constant at $V$),the amount of $HCl$ added is negligible compared to the moles of $NH_4OH$ and $NH_4Cl$ present.
Since $NH_4OH$ is a weak base and $NH_4Cl$ is its salt,the solution acts as a basic buffer.
The Henderson-Hasselbalch equation for a basic buffer is: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given $K_b = 2 \times 10^{-5}$,$pK_b = -\log(2 \times 10^{-5}) = 5 - \log 2 = 5 - 0.301 = 4.699$.
Since the amount of $HCl$ added is extremely small $(10^{-6} \ mol)$,the change in concentrations of $NH_4OH$ and $NH_4Cl$ is negligible.
Thus,$[OH^-] = K_b \times \frac{[Base]}{[Salt]} = 2 \times 10^{-5} \times \frac{0.2}{0.2} = 2 \times 10^{-5} \ M$.

(C) buffer solution is defined as a solution that resists changes in its $pH$ value upon the addition of small amounts of acid or base.
Therefore,its primary function is to maintain a constant $pH$ value.

(A) buffer solution is a mixture of a weak base and its salt with a strong acid,or a weak acid and its salt with a strong base.
$NH_4OH$ is a weak base and $NH_4Cl$ is its salt with the strong acid $HCl$.
Therefore,the mixture of $NH_4OH$ and $NH_4Cl$ forms a basic buffer solution.
Other options involve strong acids or strong bases which do not form buffer solutions.

(C) $H_3PO_4$ is a triprotic acid $(K_{a1}, K_{a2}, K_{a3})$.
During titration with $NaOH$,buffers are formed in the regions where the acid and its conjugate base coexist.
$1$. The first buffer region is formed between $H_3PO_4$ and $H_2PO_4^-$.
$2$. The second buffer region is formed between $H_2PO_4^-$ and $HPO_4^{2-}$.
Therefore,a total of $2$ buffer regions are formed during the titration.

(A) For the first solution: $CH_3COONa + HCl \rightarrow CH_3COOH + NaCl$.
Initially,we have $1 \ mol$ of $CH_3COONa$ and $0.5 \ mol$ of $HCl$.
After the reaction,$0.5 \ mol$ of $CH_3COONa$ remains and $0.5 \ mol$ of $CH_3COOH$ is formed.
Using the Henderson-Hasselbalch equation: $pH_1 = pK_a + \log(\frac{[Salt]}{[Acid]}) = pK_a + \log(\frac{0.5}{0.5}) = pK_a$.
For the second solution: $1 \ mol$ $CH_3COONa$ and $1 \ mol$ $CH_3COOH$ are present.
$pH_2 = pK_a + \log(\frac{1}{1}) = pK_a$.
Therefore,the ratio of $pH_1 : pH_2 = pK_a : pK_a = 1 : 1$.

(C) For a buffer solution,the Henderson-Hasselbalch equation is: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Since $50\%$ of the acid is ionized,the concentration of the salt (conjugate base) is equal to the concentration of the remaining unionized acid.
Therefore,$[Salt] = [Acid]$,and $\log(1) = 0$.
$pH = pK_a = 4.5$.
We know that $pH + pOH = 14$.
$pOH = 14 - pH = 14 - 4.5 = 9.5$.

(A) The given solution is a buffer solution consisting of a weak acid $(HCN)$ and its salt with a strong base $(KCN)$.
We use the Henderson-Hasselbalch equation for an acidic buffer:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Given:
$pK_a = 9.30$
$Moles \ of \ Salt \ (KCN) = 2.5 \ mol$
$Moles \ of \ Acid \ (HCN) = 2.5 \ mol$
Since the volume is the same for both,the ratio of concentrations is equal to the ratio of moles:
$pH = 9.30 + \log \left( \frac{2.5}{2.5} \right)$
$pH = 9.30 + \log(1)$
Since $\log(1) = 0$,
$pH = 9.30 + 0 = 9.30$

(B) Ammonium hydroxide $(NH_4OH)$ is a weak base and ammonium chloride $(NH_4Cl)$ is its salt with a strong acid,forming a basic buffer solution.
For a basic buffer,the concentration of $OH^-$ ions is given by the formula: $[OH^-] = K_b \times \frac{[Base]}{[Salt]}$
Given: $K_b = 1.8 \times 10^{-5}$,$[Base] = 0.05 \, M$,$[Salt] = 0.001 \, M$.
Substituting the values: $[OH^-] = 1.8 \times 10^{-5} \times \frac{0.05}{0.001}$
$[OH^-] = 1.8 \times 10^{-5} \times 50$
$[OH^-] = 90 \times 10^{-5} = 9.0 \times 10^{-4} \, M$.

(A) The resulting solution is a buffer solution. The $pH$ is calculated using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
First,calculate $pK_a$:
$pK_a = -\log(1.5 \times 10^{-5}) = 5 - \log(1.5) = 5 - 0.176 = 4.824$
Since the volumes are equal,the ratio of concentrations is equal to the ratio of moles:
$pH = 4.824 + \log \left( \frac{0.30}{0.20} \right)$
$pH = 4.824 + \log(1.5) = 4.824 + 0.176 = 5.0$
Thus,the $pH$ of the solution is $5$.

(B) The $pH$ of human blood is maintained at a constant value of approximately $7.4$ by a buffer system consisting of carbonic acid $(H_2CO_3)$ and bicarbonate ions $(HCO_3^-)$. This ability of a solution to resist changes in $pH$ upon the addition of small amounts of acid or base is known as $Buffer \ action$.

(A) The reaction is $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
Since $60\%$ of the acid is neutralized,the concentration of salt $[Salt] = 0.6$ and the remaining acid $[Acid] = 0.4$.
Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
$pH = 4.7 + \log \frac{0.6}{0.4}$
$pH = 4.7 + \log(1.5)$
Since $\log(1.5) \approx 0.176$,
$pH = 4.7 + 0.176 = 4.876$.
This value is greater than $4.7$ and less than $5.0$.

(C) The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(5 \times 10^{-10}) = 10 - \log(5) = 10 - 0.699 = 9.301 \approx 9.3$.
Given $pH = 9$,$[Acid] = [HCN] = 2 \, M$,$V_{acid} = 10 \, mL$,$[Salt] = [KCN] = 5 \, M$,and $V_{salt} = V$.
The concentration of salt in the mixture is $\frac{5V}{10+V}$ and acid is $\frac{2 \times 10}{10+V}$.
Substituting into the equation: $9 = 9.3 + \log \left( \frac{5V}{20} \right)$.
$-0.3 = \log \left( \frac{V}{4} \right)$.
$10^{-0.3} = \frac{V}{4}$.
$0.5 = \frac{V}{4} \implies V = 2 \, mL$.

(A) Buffer capacity is defined as the number of moles of acid or base added per liter to change the $pH$ by one unit.
The formula is: $\beta = \frac{\Delta n}{\Delta pH}$.
Given: $\Delta n = 0.02 \ mol/L$ and $\Delta pH = 5.80 - 5.75 = 0.05$.
Therefore,$\beta = \frac{0.02}{0.05} = 0.4$.

(A) buffer solution is a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.
$H_3BO_3$ (Boric acid) is a weak acid.
Borax $(Na_2B_4O_7 \cdot 10H_2O)$ in water hydrolyzes to form $H_3BO_3$ and $Na^+ + OH^-$.
The mixture of $H_3BO_3$ and its conjugate base (borate ion) acts as an acidic buffer.
Therefore,the correct answer is $A$.

(D) buffer solution is a mixture of a weak acid and its conjugate base,or a weak base and its conjugate acid.
$NaCl + NaOH$ is a mixture of a strong salt and a strong base,which does not form a buffer.
$NaOH + NH_4OH$ is a mixture of a strong base and a weak base,which does not form a buffer.
$CH_3COONH_4 + HCl$ results in the formation of $CH_3COOH$ (weak acid) and $NH_4Cl$ (salt of weak base and strong acid),which can act as a buffer,but the mixture itself is not a standard buffer system.
Since none of the given options represent a standard buffer system,the correct answer is $D$.

(A) The reaction is: $HCOOH + KOH \rightarrow HCOOK + H_2O$.
Since the acid is half-neutralized,the concentration of the salt $[HCOOK]$ is equal to the concentration of the remaining acid $[HCOOH]$.
Using the Henderson-Hasselbalch equation for an acidic buffer: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Given $K_a = 2 \times 10^{-4}$,so $pK_a = -\log(2 \times 10^{-4}) = 4 - \log 2 = 4 - 0.3010 = 3.6990$.
Since $[Salt] = [Acid]$,the ratio $\frac{[Salt]}{[Acid]} = 1$.
Therefore,$pH = 3.6990 + \log(1) = 3.6990 + 0 = 3.6990$.

(C) The Henderson-Hasselbalch equation for a basic buffer is given by:
$pOH = pK_b + log \frac{[\text{Conjugate Acid}]}{[\text{Base}]}$
Given the equation:
$pOH - pK_b = 1$
Substituting this into the Henderson-Hasselbalch equation:
$1 = log \frac{[\text{Conjugate Acid}]}{[\text{Base}]}$
Taking the antilog on both sides:
$10^1 = \frac{[\text{Conjugate Acid}]}{[\text{Base}]}$
Therefore,$\frac{[\text{Conjugate Acid}]}{[\text{Base}]} = 10$
This implies the ratio $[\text{Conjugate Acid}] : [\text{Base}] = 10 : 1$.

(A) For a buffer solution,the Henderson-Hasselbalch equation for a basic component is: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given that $[Salt] = [Base]$ (concentrations of $X^-$ and $HX$ are equal),the equation becomes: $pOH = pK_b + \log(1)$.
Since $K_b = 10^{-10}$,$pK_b = -\log(10^{-10}) = 10$.
Thus,$pOH = 10 + 0 = 10$.
The $pH$ is calculated as: $pH = 14 - pOH = 14 - 10 = 4$.

(C) The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Given:
$pH = 4.0$,$pK_a = 3.7$
$[Acid] = 0.05 \, M$,$Volume_{acid} = 50 \, mL$
$[Salt] = 0.10 \, M$,$Volume_{salt} = V$
Substituting the values:
$4.0 = 3.7 + \log \frac{0.10 \times V}{0.05 \times 50}$
$0.3 = \log \frac{0.10 \times V}{2.5}$
Since $\log 2 \approx 0.3$,we have:
$2 = \frac{0.10 \times V}{2.5}$
$0.10 \times V = 5.0$
$V = 50 \, mL$

(B) For a basic buffer,the $pOH$ is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Initially,the ratio $\frac{[Salt]}{[Base]} = \frac{1}{1} = 1$,so $pOH = pK_b + \log(1) = pK_b$.
Since $pH + pOH = 14$,the initial $pH = 14 - pK_b$.
When the ratio becomes $\frac{2}{1} = 2$,the new $pOH' = pK_b + \log(2)$.
Since $\log(2) \approx 0.301$,the $pOH$ increases.
As $pH = 14 - pOH$,an increase in $pOH$ results in a decrease in the $pH$ value.

(D) buffer solution is a solution that resists changes in $pH$ upon the addition of small amounts of acid or base.
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 a buffer solution.
Therefore,the correct option is $D$.

(D) For a basic buffer,$pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given $pH = 9$,so $pOH = 14 - 9 = 5$.
$pK_b = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$.
Substituting the values: $5 = 4.745 + \log \frac{[NH_4Cl]}{[NH_4OH]}$.
$0.255 = \log \frac{[NH_4Cl]}{1.0}$.
$[NH_4Cl] = 10^{0.255} \approx 1.8 \, M$.
Since the volume is $1.0 \, L$,the number of moles of $NH_4Cl$ is $1.8 \, mol$.

(B) The $pH$ of human blood is maintained at approximately $7.4$ by the action of buffer systems present in the blood. The primary buffer system in blood is the carbonic acid-bicarbonate buffer system $(H_2CO_3 / HCO_3^-)$. This system resists changes in $pH$ upon the addition of small amounts of acids or bases,thereby maintaining homeostasis.

(A) The reaction is: $CH_3COOH + NaOH \rightarrow CH_3COONa + H_2O$.
Since this is a titration of a weak acid with a strong base,the $pH$ is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$.
Given $K_a = 10^{-5}$,so $pK_a = -\log(10^{-5}) = 5$.
At $25\%$ titration: $[Salt] = 25$,$[Acid] = 75$. $pH = 5 + \log(25/75) = 5 + \log(1/3)$.
At $50\%$ titration: $[Salt] = 50$,$[Acid] = 50$. $pH = 5 + \log(50/50) = 5 + 0 = 5$.
At $75\%$ titration: $[Salt] = 75$,$[Acid] = 25$. $pH = 5 + \log(75/25) = 5 + \log(3)$.

(D) The added $H^+$ ions from $HCl$ react with $CH_3COO^-$ to form $CH_3COOH$.
Initial concentrations: $[CH_3COOH] = 1 \, M$,$[CH_3COO^-] = 1 \, M$.
After adding $0.2 \, mol$ of $HCl$:
$[CH_3COOH]_{new} = 1 + 0.2 = 1.2 \, M$
$[CH_3COO^-]_{new} = 1 - 0.2 = 0.8 \, M$
Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
$pK_a = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) = 5 - 0.255 = 4.745$
$pH = 4.745 + \log \frac{0.8}{1.2} = 4.745 + \log(0.667)$
$pH = 4.745 - 0.176 = 4.569 \approx 4.56$

(A) The Henderson-Hasselbalch equation for a basic buffer is given by: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Initially,$5 = pK_b + \log \frac{[Salt]}{[Base]}$.
When the concentration of the salt is tripled,the new concentration becomes $3 \times [Salt]$.
The new $pOH$ is: $pOH_{new} = pK_b + \log \frac{3 \times [Salt]}{[Base]}$.
This can be written as: $pOH_{new} = (pK_b + \log \frac{[Salt]}{[Base]}) + \log 3$.
Substituting the initial values: $pOH_{new} = 5 + 0.48 = 5.48$.

(B) buffer solution is defined as a solution that resists changes in its $pH$ upon the addition of small amounts of strong acid or base. Therefore,adding a small amount of dilute $HCl$ to the buffer solution will result in negligible change to the $pH$.

(A) The solution is a buffer solution consisting of a weak acid and its salt.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Given: $K_a = 10^{-5}$,so $pK_a = -\log(10^{-5}) = 5$.
Number of equivalents of salt = $1 \, N \times 10 \, mL = 10 \, meq$.
Number of equivalents of acid = $2 \, N \times 50 \, mL = 100 \, meq$.
Substituting the values: $pH = 5 + \log \frac{10}{100} = 5 + \log(0.1) = 5 + (-1) = 4$.

(A) When a small amount of acid $(H^{+})$ or base $(OH^{-})$ is added to a buffer solution,it reacts with the components of the buffer to form unionized weak acid or weak base,respectively. This prevents significant changes in the $pH$ of the solution.

(B) An acidic buffer is a mixture of a weak acid and its salt 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 $CH_3COOH + CH_3COONa$ acts as an acidic buffer.
$NH_4OH + NH_4Cl$ is a basic buffer (weak base + its salt with strong acid).
$NaOH + HCl$ is a strong acid-strong base mixture,not a buffer.

(C) The buffer is a basic buffer consisting of a weak base $(NH_3)$ and its salt $(NH_4Cl)$.
Using the Henderson-Hasselbalch equation for basic buffers:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
Given $K_b = 1.8 \times 10^{-5}$,so $pK_b = -\log(1.8 \times 10^{-5}) \approx 4.7447$.
Calculate the moles of salt and base:
$Moles \, of \, NH_4Cl = 500 \, mL \times 0.5 \, M = 250 \, mmol$.
$Moles \, of \, NH_3 = 300 \, mL \times 0.3 \, M = 90 \, mmol$.
$pOH = 4.7447 + \log \left( \frac{250}{90} \right) = 4.7447 + \log(2.777) \approx 4.7447 + 0.4437 = 5.1884$.
Since $pH + pOH = 14$,$pH = 14 - 5.1884 = 8.8116$.

(A) mixture of $1 \ M \ NaCl$ and $1 \ M \ HCl$ contains a strong acid $(HCl)$ and its salt with a strong base $(NaCl)$.
Since $HCl$ is a strong acid,it dissociates completely to provide a high concentration of $H^+$ ions,resulting in a $pH < 7$.
For a solution to act as a buffer,it must contain a weak acid and its conjugate base,or a weak base and its conjugate acid.
Since $HCl$ is a strong acid,this mixture does not resist changes in $pH$ upon the addition of small amounts of acid or base.
Therefore,it is not a buffer solution.

(D) For a basic buffer,the $pOH$ is calculated using the Henderson-Hasselbalch equation:
$pOH = pK_b + \log \frac{[Salt]}{[Base]}$
Given that the concentrations of $NH_4Cl$ (salt) and $NH_4OH$ (base) are equimolar,$[Salt] = [Base]$.
Therefore,$\log \frac{[Salt]}{[Base]} = \log(1) = 0$.
$pOH = pK_b + 0 = 4.74$.
Since $pH + pOH = 14$ at $25^{\circ}C$,
$pH = 14 - pOH = 14 - 4.74 = 9.26$.

(A) The given solution is a basic buffer solution consisting of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$.
Adding a small amount of water to a buffer solution causes dilution,which changes the concentration of the components,but the ratio of $[Salt]/[Base]$ remains constant.
According to the Henderson-Hasselbalch equation: $pOH = pK_b + \log(\frac{[Salt]}{[Base]})$.
Since the ratio $[Salt]/[Base]$ does not change significantly upon dilution,the $pH$ of the buffer solution remains unchanged.
Therefore,adding $1 \, mL$ of water will not change the $pH$.

(A) The solution is a buffer containing a weak acid $(CH_3COOH)$ and its salt $((CH_3COO)_2Ba)$.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
First,calculate $pK_a = -\log(1.8 \times 10^{-5}) = 4.7447$.
Since $(CH_3COO)_2Ba$ dissociates as $(CH_3COO)_2Ba \rightarrow 2CH_3COO^- + Ba^{2+}$,the concentration of the conjugate base $[CH_3COO^-] = 2 \times 0.1 \, M = 0.2 \, M$.
The concentration of the acid $[CH_3COOH] = 0.1 \, M$.
Substituting the values: $pH = 4.7447 + \log \frac{0.2}{0.1} = 4.7447 + \log(2) = 4.7447 + 0.301 = 5.0457 \approx 5.046$.

(C) Let the volume of $NaHCO_3$ solution be $x \, mL$.
Moles of $NaHCO_3$ in $x \, mL$ of $5 \, M$ solution $= \frac{5x}{1000} = 0.005x \, mol$.
Moles of $H_2CO_3$ in $10 \, mL$ of $2 \, M$ solution $= \frac{2 \times 10}{1000} = 0.02 \, mol$.
Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
$pK_a = -\log(7.8 \times 10^{-7}) = 7 - \log(7.8) = 7 - 0.892 = 6.108$.
$7.4 = 6.108 + \log \left( \frac{0.005x}{0.02} \right)$.
$7.4 - 6.108 = \log(0.25x)$.
$1.292 = \log(0.25x)$.
$0.25x = 10^{1.292} \approx 19.59$.
$x = \frac{19.59}{0.25} = 78.36 \, mL$.

(B) The Henderson-Hasselbalch equation for an acidic buffer is:
$pH = pK_a + \log \frac{[\text{Conjugate Base}]}{[\text{Acid}]}$
Rearranging this,we get:
$pH - pK_a = \log \frac{[\text{Conjugate Base}]}{[\text{Acid}]}$
Given that $pH - pK_a = 5$,we substitute this into the equation:
$5 = \log \frac{[\text{Conjugate Base}]}{[\text{Acid}]}$
Taking the antilog on both sides:
$10^5 = \frac{[\text{Conjugate Base}]}{[\text{Acid}]}$
Therefore,$[\text{Conjugate Base}] = [\text{Acid}] \times 10^5$
Or,$[\text{Acid}] = [\text{Conjugate Base}] \times 10^{-5}$
Comparing this with the given options,option $B$ is correct.

(C) buffer solution is a mixture of a weak acid and its conjugate base (acidic buffer) or a weak base and its conjugate acid (basic buffer).
$H_3PO_4 + NaH_2PO_4$ is a mixture of a weak acid and its salt,which forms an acidic buffer.
$NaHCO_3 + H_2CO_3$ is a mixture of a weak acid and its salt,which forms an acidic buffer.
$CH_3COOH + CH_3COONa$ is a mixture of a weak acid and its salt,which forms an acidic buffer.
$NH_4Cl + HCl$ consists of a weak base salt $(NH_4Cl)$ and a strong acid $(HCl)$. $A$ mixture of a strong acid and a salt of a weak base does not form a buffer solution.

(B) Given: $[Salt] = 0.1 \ M$,$[Acid] = 0.1 \ M$,$K_a = 1.8 \times 10^{-5}$.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) \approx 5 - 0.26 = 4.74$.
Substitute the values into the equation: $pH = 4.74 + \log \frac{0.1}{0.1}$.
Since $\log(1) = 0$,$pH = 4.74 + 0 = 4.74$.

(A) The $pH$ of an acidic buffer is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
For the buffering action to be maximum,the buffer capacity is highest when $pH = pK_a$.
This occurs when $\log \frac{[Salt]}{[Acid]} = 0$,which implies $\frac{[Salt]}{[Acid]} = 10^0 = 1$.
Therefore,the ratio of salt to acid is $1$.

(D) The $H_2CO_3 / HCO_3^-$ buffer system is the primary buffer present in human blood.
It maintains the blood $\text{pH}$ within the range of $7.35$ to $7.45$.

(B) The blood $pH$ is maintained by the bicarbonate buffer system.
This system consists of carbonic acid $(H_2CO_3)$ and its conjugate base,the bicarbonate ion $(HCO_3^{-})$.
Therefore,the correct option is $(B)$.