Buffer solution Questions in English

(B) The reaction is: $CH_3NH_2 + HCl \rightarrow CH_3NH_3^+ + Cl^-$
Initial moles: $CH_3NH_2 = 0.1 \ mol$,$HCl = 0.08 \ mol$.
After reaction: $CH_3NH_2 = 0.02 \ mol$,$CH_3NH_3^+ = 0.08 \ mol$.
This forms a basic buffer solution.
Using the Henderson-Hasselbalch equation for a basic buffer:
$pOH = pK_b + \log \frac{[salt]}{[base]}$
$pK_b = -\log(5 \times 10^{-4}) = 4 - \log(5) = 4 - 0.699 = 3.301$.
$pOH = 3.301 + \log \frac{0.08}{0.02} = 3.301 + \log(4) = 3.301 + 0.602 = 3.903$.
$pH = 14 - pOH = 14 - 3.903 = 10.097$.
$[H^{+}] = 10^{-pH} = 10^{-10.097} \approx 8 \times 10^{-11} \ M$.

(A) buffer solution is typically a mixture of a weak acid and its conjugate base (salt of the weak acid with a strong base).
In option $A$,$CH_3COOH$ is a weak acid and $CH_3COONa$ is its salt with a strong base $(NaOH)$,which forms an acidic buffer.
Therefore,the correct option is $A$.

(B) The Henderson-Hasselbalch equation for an acidic buffer 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: $6 = 5 + \log \frac{[Salt]}{[Acid]}$.
Rearranging gives: $\log \frac{[Salt]}{[Acid]} = 6 - 5 = 1$.
Taking the antilog on both sides: $\frac{[Salt]}{[Acid]} = 10^1 = 10$.
Therefore,the ratio of the concentration of salt to acid is $10 : 1$.

(D) buffer solution is defined as a solution that resists changes in $pH$ upon the addition of small amounts of acid or base.
$(a)$ $A$ buffer consisting of a weak acid and its conjugate base is known as an acidic buffer.
$(b)$ $A$ buffer consisting of a weak base and its conjugate acid is known as a basic buffer.
$(c)$ The property of resisting $pH$ change is the fundamental characteristic of a buffer solution.
Since all statements $(a)$,$(b)$,and $(c)$ are correct,the correct choice is $(d)$.

(A) buffer solution of strong acidic nature is formed by a strong acid and its conjugate base.
Among the given options,$HCOOH$ (formic acid) is a relatively stronger acid compared to acetic acid $(CH_3COOH)$,boric acid $(H_3BO_3)$,and oxalic acid $(H_2C_2O_4)$ in the context of typical buffer systems used in chemistry problems.
Therefore,the mixture of $HCOOH + HCOO^-$ acts as a buffer solution with a lower $pH$ value,indicating a stronger acidic nature.

(A) buffer solution is defined as a solution that resists changes in $pH$ upon the addition of small amounts of acid or base. It typically consists of a weak acid and its conjugate base,or a weak base and its conjugate acid.
$HCl$ is a strong acid and $NaCl$ is a salt of a strong acid and a strong base.
Since $HCl$ is a strong acid,it dissociates completely in water,making the solution highly acidic with a $pH < 7$.
Because there is no weak acid-conjugate base pair present,this mixture does not act as a buffer solution.
Therefore,the correct description is that it is not a buffer solution and has a $pH < 7$.

(A) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \left( \frac{[Salt]}{[Acid]} \right)$
Given:
$[Acid] = [HCN] = 0.1 \ M$
$[Salt] = [NaCN] = 0.2 \ M$
$pK_a = 9.30$
Substituting the values:
$pH = 9.30 + \log \left( \frac{0.2}{0.1} \right)$
$pH = 9.30 + \log(2)$
Since $\log(2) \approx 0.3010$:
$pH = 9.30 + 0.3010 = 9.6010 \approx 9.60$
Therefore,the correct option is $A$.

(A) The solution is a buffer containing a weak acid (acetic acid) and its salt (sodium acetate).
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Number of milliequivalents of salt = $10 \ mL \times 1 \ N = 10 \ meq$.
Number of milliequivalents of acid = $50 \ mL \times 2 \ N = 100 \ meq$.
Total volume = $10 \ mL + 50 \ mL = 60 \ mL$.
Since the volume cancels out in the ratio,we use milliequivalents: $\frac{[Salt]}{[Acid]} = \frac{10}{100} = 0.1$.
$pK_a = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) \approx 5 - 0.255 = 4.745$.
$pH = 4.745 + \log(0.1) = 4.745 - 1 = 3.745$.
The value is approximately $4$.

(A) buffer solution is defined as a solution that resists changes in $pH$ upon the addition of small amounts of a strong acid or a strong base.
Therefore,when a small amount of a strong acid is added to a buffer solution,the $pH$ remains essentially constant.

(B) The $pH$ of an acidic buffer is calculated using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Given: $pK_a = 4.75$,$[Salt] = 0.1 \ M$,and $[Acid] = 0.1 \ M$.
Substituting the values:
$pH = 4.75 + \log \frac{0.1}{0.1}$
$pH = 4.75 + \log(1)$
Since $\log(1) = 0$,
$pH = 4.75 + 0 = 4.75$.

(D) An acidic buffer is prepared by mixing a weak acid and its salt with a strong base.
Such buffers are typically used to maintain a $pH$ in the acidic range,specifically $pH$ $4 - 5$.
Therefore,the correct mixture is a weak acid and its salt with a strong base.

(D) An acidic buffer solution 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 $CH_3COOH$ and $CH_3COONa$ forms an acidic buffer.

(B) An acidic buffer solution is prepared by mixing 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 $CH_3COOH + CH_3COONa$ forms an acidic buffer.

(A) For a basic buffer,the formula is $pOH = pK_b + \log \frac{[salt]}{[base]}$.
Given that the volumes are equal,the concentration ratio is equal to the mole ratio.
$pOH = 5 + \log \frac{0.02}{0.2} = 5 + \log(0.1) = 5 - 1 = 4$.
Since $pH + pOH = 14$,we have $pH = 14 - 4 = 10$.

(B) Given: $[Salt] = 0.1 \, M$,$[Acid] = 0.1 \, M$ and $K_a = 1.8 \times 10^{-5}$.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) \approx 5 - 0.255 = 4.745$.
Substituting the values: $pH = 4.745 + \log \frac{0.1}{0.1} = 4.745 + \log(1) = 4.745 + 0 = 4.745$.
Therefore,the $pH$ of the solution is about $4.7$.

(A) buffer solution is a mixture of a weak acid and its conjugate base,or a weak base and its conjugate acid.
$NH_4OH$ is a weak base and $NH_4Cl$ is its salt with a strong acid $(HCl)$.
Therefore,the mixture of $NH_4Cl + NH_4OH$ acts as a basic buffer solution.

(B) buffer solution is defined as a solution that resists changes in $pH$ upon the addition of small amounts of acid or base.
Therefore,when a small amount of dilute $HCl$ is added to the buffer solution,the $pH$ remains effectively unchanged.

(A) For the base $X^{-}$,the equilibrium is: $X^{-} + H_2O \rightleftharpoons OH^{-} + HX$.
Given $K_b = 10^{-10}$.
We know that $K_a \times K_b = K_w = 10^{-14}$.
Therefore,$K_a = \frac{10^{-14}}{10^{-10}} = 10^{-4}$.
For a buffer solution,the Henderson-Hasselbalch equation is: $pH = pK_a + \log\frac{[salt]}{[acid]}$.
Since the concentrations of $X^{-}$ (salt) and $HX$ (acid) are equal,$[X^{-}] = [HX]$.
Thus,$pH = pK_a + \log(1) = pK_a$.
$pH = -\log(K_a) = -\log(10^{-4}) = 4$.

(C) buffer solution is typically composed of a weak acid and its conjugate base or a weak base and its conjugate acid.
Specifically,the mixture of $H_2PO_4^-$ (dihydrogen phosphate,acting as a weak acid) and $HPO_4^{2-}$ (hydrogen phosphate,acting as its conjugate base) forms a well-known buffer system in biological fluids.
Therefore,the pair $[HPO_4^{2-}] , [H_2PO_4^{-}]$ constitutes a buffer solution.

(C) The $pH$ of a buffer solution is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
For the given buffer,$[Salt] = [Acid] = 1 \ M$,so $pH = pK_a$.
Adding a small amount of strong acid or base will change the ratio $\frac{[Salt]}{[Acid]}$,but the change is minimized by the buffer capacity.
Option $C$ ($5 \ mL$ of $5 \ M \ HCl$) adds $25 \ mmol$ of $H^+$ ions,which is a significant amount compared to the $25 \ mmol$ of $CH_3COONa$ present,causing a drastic change in the $pH$ compared to the other options.
Therefore,the $pH$ is most appreciably affected by $5 \ mL$ of $5 \ M \ HCl$.

(D) The Henderson-Hasselbalch equation is given by $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
For a buffer to be most efficient,the $pH$ of the solution should be as close as possible to the $pK_a$ of the weak acid used,i.e.,$pH \approx pK_a$.
Given the target $pH = 3.58$,we look for an acid with a $pK_a$ value closest to $3.58$.
Comparing the given options:
$(A)$ $pK_a = 3.98$
$(B)$ $pK_a = 4.41$
$(C)$ $pK_a = 2.97$
$(D)$ $pK_a = 3.58$
Since Acetoacetic acid has a $pK_a$ of $3.58$,which is exactly equal to the desired $pH$,it provides the maximum buffer capacity at this $pH$.

(A) For a buffer solution containing a weak acid and its salt,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Let the initial ratio be $R_1 = \frac{[Salt]}{[Acid]}$. Then $pH_1 = pK_a + \log(R_1)$.
The problem states that the ratio of acid to salt is increased ten-fold,which means the new ratio $\frac{[Acid]}{[Salt]}$ is $10 \times \frac{[Acid]_{initial}}{[Salt]_{initial}}$.
Therefore,the new ratio of salt to acid is $R_2 = \frac{[Salt]}{[Acid]} = \frac{1}{10} \times R_1$.
The new $pH$ is $pH_2 = pK_a + \log(R_2) = pK_a + \log(\frac{R_1}{10}) = pK_a + \log(R_1) - \log(10)$.
Since $\log(10) = 1$,we get $pH_2 = pH_1 - 1$.
Thus,the $pH$ decreases by $1$.

(A) buffer solution is defined as a solution that resists changes in its $pH$ upon the addition of small amounts of an acid or a base.
Therefore,when a small amount of acid or alkali is added to a buffer solution,the $pH$ remains practically unchanged or changes only negligibly.

(C) The Henderson-Hasselbalch equation for an acidic buffer is: $pH = pK_a + \log\frac{[Salt]}{[Acid]}$
Given $pH = 5.5$,$pK_a = 4.5$,and $[Acid] = 0.1 \ M$.
Substituting the values: $5.5 = 4.5 + \log\frac{[Salt]}{0.1}$
$\log\frac{[Salt]}{0.1} = 5.5 - 4.5 = 1$
Taking antilog on both sides: $\frac{[Salt]}{0.1} = 10^1 = 10$
$[Salt] = 10 \times 0.1 = 1 \ M$.
Since the volume is $1 \ L$,the number of moles of sodium acetate required is $1 \ mol$.

(C) buffer solution is one that resists changes in $pH$ upon the addition of small amounts of acid or base.
$(C)$ $CH_3COONH_4$ (ammonium acetate) is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.
Such salts act as simple buffers because they can neutralize both added $H^+$ and $OH^-$ ions.

(D) 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.
In option $D$,$NH_4OH$ is a weak base and $CH_3COONH_4$ is its salt with a weak acid $(CH_3COOH)$. This mixture acts as a buffer solution.
Therefore,the correct option is $D$.

(D) buffer solution is designed to resist changes in $pH$ upon the addition of small amounts of acid,base,or upon dilution.
The $pH$ of an acidic buffer is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Upon dilution,both the concentration of the salt $(CH_3COONa)$ and the acid $(CH_3COOH)$ decrease by the same factor.
Therefore,the ratio $\frac{[Salt]}{[Acid]}$ remains constant.
Consequently,the $pH$ and the $H^{+}$ ion concentration remain effectively unaltered.

(D) buffer solution is defined as a solution that resists changes in $pH$ upon the addition of small amounts of acid or base. $A$ common buffer solution consists of a weak acid and its conjugate base (e.g.,$CH_3COOH$ and $CH_3COO^-$) or a weak base and its conjugate acid (e.g.,$NH_4OH$ and $NH_4^+$). Therefore,the correct option is $D$.

(C) buffer solution is typically composed of a weak acid and its conjugate base,or a weak base and its conjugate acid.
$A$,$B$,and $D$ represent mixtures of weak acids and their conjugate bases,which form effective buffer systems.
$C$ consists of a strong acid $(HCl)$ and a salt of a weak base $(NH_4Cl)$. Since a strong acid does not form a buffer with its salt,this mixture cannot act as a buffer.

(C) The $pH$ of human blood is maintained at approximately $7.4$.
This slightly alkaline $pH$ is regulated by the bicarbonate buffer system $(HCO_3^- / H_2CO_3)$ present in the blood.

(C) buffer solution is a solution that resists changes in $pH$ when small amounts of an acid or a base are added to it.
It maintains a relatively constant $pH$ value due to the presence of a weak acid and its conjugate base or a weak base and its conjugate acid.

(B) buffer solution is typically formed by a weak acid and its conjugate base,or a weak base and its conjugate acid.
$NH_4Cl + NH_4OH$ is a mixture of a weak base $(NH_4OH)$ and its conjugate acid $(NH_4^+)$.
$Na_2HPO_4 + Na_3PO_4$ is a mixture of a weak acid $(HPO_4^{2-})$ and its conjugate base $(PO_4^{3-})$.
$CH_3COOH + CH_3COONa$ is a mixture of a weak acid $(CH_3COOH)$ and its conjugate base $(CH_3COO^-)$.
$NaCl + NaOH$ consists of a salt of a strong acid and strong base $(NaCl)$ and a strong base $(NaOH)$. Since $NaCl$ is not a weak acid or base,this mixture does not act as a buffer solution.

(B) For a buffer solution,the Henderson-Hasselbalch equation is given by: $pH = pK_a + \log\left(\frac{[Salt]}{[Acid]}\right)$.
Given that the concentration of the salt $[X^{-}]$ is equal to the concentration of the acid $[HX]$,we have $[Salt] = [Acid]$.
Substituting this into the equation: $pH = pK_a + \log(1) = pK_a$.
Since $pK_a = -\log(K_a)$,we calculate: $pK_a = -\log(10^{-8}) = 8$.
Therefore,the $pH$ of the buffer is $8$.

(C) The given solution is a buffer solution consisting of a weak acid $(HCN)$ and its salt with a strong base $(KCN)$.
Using the Henderson-Hasselbalch equation:
$pH = pK_a + \log \frac{[Salt]}{[Acid]}$
Given:
$K_a = 5 \times 10^{-10}$
$pK_a = - \log(5 \times 10^{-10}) = 10 - \log 5 = 10 - 0.698 = 9.302$
Concentration of salt $(KCN)$ = $\frac{0.15 \, mole}{0.5 \, dm^3} = 0.3 \, M$
Concentration of acid $(HCN)$ = $\frac{1.5 \, mole}{0.5 \, dm^3} = 3.0 \, M$
$pH = 9.302 + \log \left( \frac{0.3}{3.0} \right)$
$pH = 9.302 + \log(0.1)$
$pH = 9.302 - 1 = 8.302$

(D) buffer solution that has a $pH > 7$ is known as a basic buffer.
$A$. $CH_3COOH + CH_3COONa$ is an acidic buffer $(pH < 7)$.
$B$. $HCOOH + HCOOK$ is an acidic buffer $(pH < 7)$.
$C$. $CH_3COONH_4$ is a salt of a weak acid and a weak base,which typically results in a $pH$ near $7$.
$D$. $NH_4OH + NH_4Cl$ is a mixture of a weak base $(NH_4OH)$ and its salt with a strong acid $(NH_4Cl)$,which forms a basic buffer with $pH > 7$.

(C) The $pH$ of a buffer solution is calculated using the Henderson-Hasselbalch equation: $pH = pK_a + \log\frac{[Salt]}{[Acid]}$.
Since the mixture is equimolar,the concentration of the salt (sodium acetate) is equal to the concentration of the acid (acetic acid),so $\frac{[Salt]}{[Acid]} = 1$.
Substituting the values: $pH = 4.74 + \log(1)$.
Since $\log(1) = 0$,we get $pH = 4.74 + 0 = 4.74$.

(C) Buffers work best when $pH = pK_a$.
From the Henderson-Hasselbalch equation:
$pH = pK_a + \log_{10} \frac{[A^{-}]}{[HA]}$
If $pH = pK_a$,then $\log_{10} \frac{[A^{-}]}{[HA]} = 0$,which implies $\frac{[A^{-}]}{[HA]} = 1$.
$A$ buffer exhibits the highest resistance to acid and base addition when the concentrations of the conjugate acid and base are equal (i.e.,when $pH = pK_a$).
Buffer capacity is highest when $pH$ is close to the $pK_a$ value. As the $pH$ deviates from the $pK_a$ value,the buffer capacity decreases.

(D) For a basic buffer,the relationship is given by the Henderson-Hasselbalch equation: $pOH = pK_b + \log \frac{[Salt]}{[Base]}$.
Given $pH = 9$,we calculate $pOH = 14 - 9 = 5$.
Given $K_b = 1.8 \times 10^{-5}$,we calculate $pK_b = -\log(1.8 \times 10^{-5}) \approx 4.74$.
Substituting the values into the equation: $5 = 4.74 + \log \frac{[NH_4Cl]}{1.0}$.
Rearranging gives $\log [NH_4Cl] = 5 - 4.74 = 0.26$.
Taking the antilog: $[NH_4Cl] = 10^{0.26} \approx 1.8 \ M$.
Since the volume is $1 \ L$,the number of moles of $NH_4Cl$ required is $1.8 \ mol \times 1 \ L = 1.8 \ mol$.

(B) The Henderson-Hasselbalch equation for an acidic buffer is: $pH = pK_a + \log \frac{[salt]}{[acid]}$
Given $K_a = 10^{-4}$,so $pK_a = - \log(10^{-4}) = 4$.
Substituting the values into the equation: $5 = 4 + \log \frac{[salt]}{[acid]}$
$\log \frac{[salt]}{[acid]} = 5 - 4 = 1$
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 consists of a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt).
$NH_4OH$ is a weak base and $NH_4Cl$ is its salt with a strong acid $(HCl)$.
Therefore,the pair ($NH_4OH$ and $NH_4Cl$) forms a basic buffer solution.
The correct option is $(D)$.

(C) buffer solution is typically a mixture of a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt).
$NH_4Cl + NH_4OH$ is a basic buffer (weak base + salt of strong acid).
$CH_3COOH + CH_3COONa$ is an acidic buffer (weak acid + salt of strong base).
$Borax + \text{Boric acid}$ acts as a buffer solution.
$CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$. It is a salt solution,not a buffer solution.

(A) The mixture forms an acidic buffer solution.
Using the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Given: $pK_a$ of acetic acid is approximately $4.74$ (or $5$ for approximation).
Number of milliequivalents of acid $= 50 \ mL \times 2 \ N = 100 \ meq$.
Number of milliequivalents of salt $= 10 \ mL \times 1 \ N = 10 \ meq$.
$pH = 4.74 + \log \frac{10}{100} = 4.74 + \log(0.1) = 4.74 - 1 = 3.74$.
The approximate value is $4$.

(B) The $pH$ of human blood is maintained at a constant value of approximately $7.4$ by the action of buffer systems.
Specifically,the bicarbonate buffer system $(H_2CO_3 / HCO_3^-)$ and serum proteins act as buffers to resist changes in $pH$ upon the addition of small amounts of acids or bases.

(D) solution that resists change in $pH$ value upon addition of a small amount of strong acid or base (less than $1\%$) or when the solution is diluted is called a buffer solution.
An acidic buffer solution consists of a mixture of a weak acid and its salt with a strong base.
$A$ basic buffer solution consists of a mixture of a weak base and its salt with a strong acid.
Therefore,a buffer solution can be prepared by mixing a weak acid and its salt with a strong base.

(A) The $pH$ of an acidic buffer is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[Salt]}{[Acid]}$.
Initially,let the ratio be $R = \frac{[Salt]}{[Acid]}$. So,$pH_1 = pK_a + \log(R)$.
When the ratio is increased ten times,the new ratio becomes $10R$.
The new $pH$ is $pH_2 = pK_a + \log(10R)$.
$pH_2 = pK_a + \log(10) + \log(R) = pK_a + 1 + \log(R)$.
Comparing the two,$pH_2 = pH_1 + 1$.
Therefore,the $pH$ increases by $1$.

(D) The correct answer is $D$.
$A$ buffer solution is prepared by mixing a weak acid with its conjugate base (salt of a strong base) or a weak base with its conjugate acid (salt of a strong acid).
$1$. Sodium acetate $(CH_3COONa)$ and acetic acid $(CH_3COOH)$ form an acidic buffer because acetic acid is a weak acid and sodium acetate is its salt with a strong base $(NaOH)$.
$2$. Ammonia $(NH_3)$ and ammonium chloride $(NH_4Cl)$ form a basic buffer because ammonia is a weak base and ammonium chloride is its salt with a strong acid $(HCl)$.
$3$. Sodium acetate and hydrochloric acid $(HCl)$ react to form acetic acid and sodium chloride,which does not constitute a buffer system in the required proportions.
Therefore,both $A$ and $C$ are correct.

(D) buffer solution is typically composed of a weak acid and its conjugate base (salt) or a weak base and its conjugate acid (salt).
$A$. $NaCl$ (strong electrolyte) and $NaOH$ (strong base) do not form a buffer.
$B$. $NaOH$ (strong base) and $NH_4OH$ (weak base) do not form a buffer.
$C$. $CH_3COONH_4$ is a salt of a weak acid and a weak base,and adding $HCl$ (strong acid) does not create a standard buffer system.
Since none of the given combinations satisfy the criteria for a buffer solution,the correct answer is $D$.

(D) The solution is a buffer containing a weak acid $(HCN)$ and its salt $(NaCN)$.
Using the acid dissociation constant expression: $K_a = \frac{[H^{+}][CN^{-}]}{[HCN]}$
Given: $K_a = 6.2 \times 10^{-10}$,$[HCN] = 0.01 \ M$,and $[CN^{-}] = 0.02 \ M$.
Substituting the values: $6.2 \times 10^{-10} = \frac{[H^{+}](0.02)}{0.01}$
$[H^{+}] = \frac{6.2 \times 10^{-10} \times 0.01}{0.02}$
$[H^{+}] = 3.1 \times 10^{-10} \ M$.

(B) The solution is a buffer solution consisting of a weak acid $(CH_3COOH)$ and its salt with a strong base $(CH_3COONa)$.
We use the Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[salt]}{[acid]}$.
Given: $pK_a = 4.57$,$[salt] = 0.10\,M$,$[acid] = 0.03\,M$.
Substituting the values: $pH = 4.57 + \log \frac{0.10}{0.03} = 4.57 + \log(3.33)$.
Since $\log(3.33) \approx 0.52$,we get $pH = 4.57 + 0.52 = 5.09$.