Salt hydrolysis Questions in English

(B) The hydrolysis constant $(K_h)$ for a salt of a weak acid and a weak base is given by the formula:
$K_h = \frac{K_w}{K_a \times K_b}$
Substituting the given values:
$K_h = \frac{1 \times 10^{-14}}{(1.75 \times 10^{-5}) \times (4.6 \times 10^{-10})}$
$K_h = \frac{1 \times 10^{-14}}{8.05 \times 10^{-15}}$
$K_h = \frac{10}{8.05} \approx 1.242$

(A) $NaOCl$ is a salt of a strong base and a weak acid.
The formula for the degree of hydrolysis $(h)$ is $h = \sqrt{\frac{K_h}{C}}$.
Given $K_h = 10^{-6}$ and $C = 0.01 \ M = 10^{-2} \ M$.
$h = \sqrt{\frac{10^{-6}}{10^{-2}}} = \sqrt{10^{-4}} = 10^{-2}$.
Percentage of hydrolysis = $h \times 100 = 10^{-2} \times 100 = 1\%$.

(C) The degree of hydrolysis $(h)$ for a salt of a weak acid and a weak base is given by the formula $h = \sqrt{K_h}$.
Since $K_h = \frac{K_w}{K_a \times K_b}$,the expression for $h$ does not contain the concentration term $(C)$.
Therefore,the degree of hydrolysis for salts formed from a weak acid and a weak base is independent of dilution.
Both $NH_4CN$ and $CH_3COONH_4$ are salts of a weak acid and a weak base.

(C) Ammonium benzoate $(C_6H_5COONH_4)$ is a salt formed from a weak acid (benzoic acid,$K_a$) and a weak base (ammonium hydroxide,$K_b$).
For a salt of a weak acid and a weak base,the degree of hydrolysis $(h)$ is independent of the concentration $(C)$ of the salt solution.
The hydrolysis constant $(K_h)$ is given by $K_h = \frac{K_w}{K_a \times K_b}$.
The degree of hydrolysis $(h)$ is given by the formula $h = \sqrt{K_h}$.
Substituting the value of $K_h$,we get $h = \sqrt{\frac{K_w}{K_a \times K_b}}$.

(B) For the salt of a weak acid and strong base,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{\frac{K_w}{K_a \times C}}$
Given: $K_w = 10^{-14}$,$K_a = 1.4 \times 10^{-9}$,and $C = 1/100 = 10^{-2} \ M$.
Substituting the values: $h = \sqrt{\frac{10^{-14}}{1.4 \times 10^{-9} \times 10^{-2}}}$
$h = \sqrt{\frac{10^{-14}}{1.4 \times 10^{-11}}} = \sqrt{\frac{10^{-3}}{1.4}} = \sqrt{0.714 \times 10^{-3}} = \sqrt{7.14 \times 10^{-4}}$
$h \approx 2.67 \times 10^{-2} \approx 2.7 \times 10^{-2}$.

(C) $FeCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$.
When dissolved in water,the $Fe^{3+}$ ion undergoes hydrolysis.
The reaction is: $Fe^{3+} + 3H_2O \rightleftharpoons Fe(OH)_3 + 3H^+$.
Due to the production of $H^+$ ions,the solution becomes acidic.

(C) The degree of hydrolysis $h = \frac{4.88}{100} = 0.0488$.
The concentration $C = \frac{n}{V} = \frac{1 \ mol}{99.2 \ L} = 0.01008 \ M$.
For the salt of a weak base and strong acid,the hydrolysis constant is given by $K_h = C h^2$.
$K_h = 0.01008 \times (0.0488)^2$.
$K_h = 0.01008 \times 0.00238144 \approx 2.4 \times 10^{-5}$.

(D) When $CH_3COONa$ is dissolved in water,it undergoes complete dissociation into $CH_3COO^-$ and $Na^+$ ions.
Since $CH_3COO^-$ is the conjugate base of a weak acid $(CH_3COOH)$,it reacts with water to form $CH_3COOH$ and $OH^-$ ions.
This reaction of an ion with water to change the $pH$ of the solution is known as salt hydrolysis.
The overall process involves both the dissociation of the salt and the hydrolysis of the anion.

(B) $MgCl_2$ is a salt formed from a weak base $(Mg(OH)_2)$ and a strong acid $(HCl)$.
For a salt of a strong acid and a weak base,the hydrolysis constant $(K_h)$ is related to the dissociation constant of the weak base $(K_b)$ by the expression: $K_h = \frac{K_w}{K_b}$.

(C) To determine the $pH$,we analyze the nature of the salts:
$1$. $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,so its solution is neutral $(pH \approx 7)$.
$2$. $NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$,so its solution is acidic $(pH < 7)$.
$3$. $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,so its solution is basic $(pH > 7)$.
$4$. $CH_3COONH_4$ is a salt of a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$,so its solution is approximately neutral $(pH \approx 7)$.
Therefore,the solution of $CH_3COONa$ has the highest $pH$ value.

(A) Given: Concentration $C = 0.1 \, M$,Degree of hydrolysis $h = 1 \% = 0.01 = 10^{-2}$.
For the salt of a weak acid and strong base,the hydrolysis constant $K_h$ is given by $K_h = C h^2$.
$K_h = 0.1 \times (10^{-2})^2 = 0.1 \times 10^{-4} = 10^{-5}$.
We know $K_h = \frac{K_w}{K_a}$,so $K_a = \frac{K_w}{K_h} = \frac{10^{-14}}{10^{-5}} = 10^{-9}$.
The concentration of $CH_3COOH$ formed is $[CH_3COOH] = C \times h = 0.1 \times 10^{-2} = 10^{-3} \, M$.
Thus,the values are $10^{-5}, 10^{-9}, 10^{-3}$.

(A) For a salt of a weak acid and a weak base,the $pH$ is given by the formula:
$pH = 7 + \frac{1}{2}pK_a - \frac{1}{2}pK_b$
Given that $pK_b > pK_a$,the term $(\frac{1}{2}pK_a - \frac{1}{2}pK_b)$ will be negative.
Therefore,$pH = 7 - \text{positive value}$,which implies $pH < 7$.

(C) $NaCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HCN)$.
When dissolved in water,it undergoes hydrolysis:
$CN^- + H_2O \rightleftharpoons HCN + OH^-$
Since $OH^-$ ions are produced in the solution,the resulting solution becomes basic in nature.

(C) The given salt $CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
For such a salt,the concentration of $[H^{+}]$ is given by the formula:
$[H^{+}] = \sqrt{\frac{K_w \times K_a}{C}}$
Given:
$K_w = 10^{-14}$
$K_a = 10^{-5}$
$C = 0.1 \, M = 10^{-1} \, M$
Substituting the values:
$[H^{+}] = \sqrt{\frac{10^{-14} \times 10^{-5}}{10^{-1}}} = \sqrt{10^{-19} \times 10^{1}} = \sqrt{10^{-18}} = 10^{-9} \, M$

(B) $Na_2SO_4$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$,so its solution is neutral $(pH \approx 7)$.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so its solution is acidic $(pH < 7)$.
$KNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(KOH)$,so its solution is neutral $(pH \approx 7)$.
$Na_3PO_4$ is a salt of a weak acid $(H_3PO_4)$ and a strong base $(NaOH)$,so its solution is basic $(pH > 7)$.
Therefore,$NH_4Cl$ has the minimum $pH$.

(D) basic salt is formed by the partial neutralization of a polyacidic base by an acid.
$2PbCO_3 \cdot Pb(OH)_2$ (White lead) is a basic salt because it contains the hydroxide ion $(OH^-)$ along with the carbonate ion $(CO_3^{2-})$.

(D) The given reaction $X^{-} + H_2O \rightleftharpoons HX + OH^{-}$ represents the hydrolysis of an anion $X^{-}$.
This reaction produces $OH^{-}$ ions,which makes the solution basic.
This type of hydrolysis occurs when a salt is formed from a weak acid $(HX)$ and a strong base $(MOH)$.
In such a salt $(MX)$,the anion $X^{-}$ reacts with water to form the weak acid $HX$ and releases $OH^{-}$ ions into the solution.

(A) $CH_3COONa$ is a salt of a strong base $(NaOH)$ and a weak acid $(CH_3COOH)$.
The $pH$ of a salt of a weak acid and strong base is given by the formula: $pH = \frac{1}{2} [pK_w + pK_a + log C]$.
On dilution,the concentration $C$ decreases.
As $C$ decreases,$log C$ becomes more negative,which leads to a decrease in the $pH$ value.
Therefore,the $pH$ of the solution decreases.

(A) For the salt of a weak acid and a strong base,the hydrolysis constant is given by $K_h = \frac{K_w}{K_a}$.
Taking the negative logarithm on both sides:
$-\log K_h = -\log \left( \frac{K_w}{K_a} \right)$
$-\log K_h = -(\log K_w - \log K_a)$
$-\log K_h = -\log K_w + \log K_a$
Rearranging the terms:
$\log K_a - \log K_h = \log K_w$ is incorrect,rather:
$pK_h = pK_w - pK_a$
Therefore,$pK_a + pK_h = pK_w$.

(A) $CuSO_4$ is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Cu(OH)_2)$.
In water,$Cu^{2+}$ ions undergo hydrolysis:
$Cu^{2+} + 2H_2O ⇌ Cu(OH)_2 + 2H^+$
Since $H^+$ ions are produced in the solution,the aqueous solution of copper sulfate becomes acidic.

(A) The degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{\frac{K_h}{C}}$
Given: $K_h = 2.5 \times 10^{-10}$ and $C = 0.1 \ M = 10^{-1} \ M$
Substituting the values: $h = \sqrt{\frac{2.5 \times 10^{-10}}{10^{-1}}}$
$h = \sqrt{2.5 \times 10^{-9}}$
$h = \sqrt{25 \times 10^{-10}}$
$h = 5 \times 10^{-5}$

(A) Salt hydrolysis is the reaction of an ion with water to change the $pH$ of the solution.
$NaCN$ is a salt of a strong base $(NaOH)$ and a weak acid $(HCN)$. The anion $CN^-$ undergoes hydrolysis: $CN^- + H_2O \rightleftharpoons HCN + OH^-$.
$NH_4Cl$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(HCl)$. The cation $NH_4^+$ undergoes hydrolysis: $NH_4^+ + H_2O \rightleftharpoons NH_4OH + H^+$.
$CuSO_4$ and $FeCl_3$ are salts of weak bases and strong acids,where the metal cations undergo hydrolysis.
Therefore,$NaCN$ is the salt containing an anion that undergoes hydrolysis.

(C) Sodium acetate $(CH_3COONa)$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
When dissolved in water,it undergoes anionic hydrolysis:
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
Due to the production of $OH^-$ ions,the solution becomes basic.
Therefore,the $pH$ of the aqueous solution of sodium acetate is greater than $7$.

(C) Let us analyze the nature of the salts in aqueous solution:
$(1)$ $FeCl_3$ is a salt of a strong acid $(HCl)$ and a weak base $(Fe(OH)_3)$. It undergoes cationic hydrolysis to form an acidic solution. Thus,statement $(1)$ is incorrect.
$(2)$ $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$. It undergoes cationic hydrolysis to form an acidic solution. Thus,statement $(2)$ is correct.
$(3)$ $NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$. It undergoes anionic hydrolysis to form a basic solution. Thus,statement $(3)$ is incorrect.
$(4)$ $Na_2CO_3$ is a salt of a weak acid $(H_2CO_3)$ and a strong base $(NaOH)$. It undergoes anionic hydrolysis to form a basic solution. Thus,statement $(4)$ is correct.
Since statements $(1)$ and $(3)$ are incorrect,the correct option is $C$.

(C) The $pH$ of a salt solution formed from a weak acid and a weak base (e.g.,$CH_3COONH_4$) is given by the formula:
$pH = 7 + \frac{1}{2}pK_a - \frac{1}{2}pK_b$
Since this expression does not contain any concentration term,the $pH$ of such a salt solution is independent of dilution.
Therefore,the $pH$ remains constant.

(B) $NaHSO_4$ is an acidic salt because it contains a replaceable hydrogen atom and is derived from a strong acid $(H_2SO_4)$ and a weak base $(NaOH)$. $NaH_2PO_3$ is also an acidic salt. $HCOONa$ (Sodium formate) is a salt of a weak acid $(HCOOH)$ and a strong base $(NaOH)$,making it a basic salt. Therefore,$HCOONa$ is not an acidic salt.

(A) The salts are formed from a weak acid and a strong base.
The degree of hydrolysis ($h$ or $x$) is given by the formula: $x = \sqrt{\frac{K_h}{C}} = \sqrt{\frac{K_w}{K_a \times C}}$.
Assuming the concentration $(C)$ is the same for both salts,the relationship is $x \propto \sqrt{\frac{1}{K_a}}$.
For $KCN$,the acid is $HCN$ $(K_a = 7.2 \times 10^{-10})$.
For $CH_3COOK$,the acid is $CH_3COOH$ $(K_a = 1.8 \times 10^{-5})$.
Since $K_a(CH_3COOH) > K_a(HCN)$,it follows that $x_1 > x_2$ because $x$ is inversely proportional to the square root of $K_a$.

(D) Ammonium cyanide $(NH_4CN)$ is a salt of a weak acid $(HCN)$ and a weak base $(NH_4OH)$.
For such salts,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.
This formula shows that the $pH$ of the solution is independent of the concentration of the salt.
Therefore,adding water (dilution) does not change the $pH$ of the solution.
Thus,the $pH$ remains $7.02$.

(D) $KCN$ is a salt of a weak acid $(HCN)$ and a strong base $(KOH)$.
For such a salt,the $pH$ is given by: $pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C$.
Given $K_a = 10^{-7}$,so $pK_a = 7$. $C = 10^{-3} \ M$.
$pH = 7 + \frac{1}{2}(7) + \frac{1}{2}\log(10^{-3}) = 7 + 3.5 - 1.5 = 9$.
Degree of hydrolysis $(h)$ is given by: $h = \sqrt{\frac{K_w}{K_a \times C}} = \sqrt{\frac{10^{-14}}{10^{-7} \times 10^{-3}}} = \sqrt{10^{-4}} = 10^{-2}$.
Concentration of $[OH^-]$ is given by: $[OH^-] = C \times h = 10^{-3} \times 10^{-2} = 10^{-5} \ M$.

(D) For a salt of a weak acid and a weak base like $CH_3COONH_4$,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{K_h} = \sqrt{\frac{K_w}{K_a \times K_b}}$.
Since $K_w$,$K_a$,and $K_b$ are constants at a given temperature,the degree of hydrolysis $(h)$ for such salts is independent of the concentration of the salt solution.
Therefore,the degree of hydrolysis is the same for all the given concentrations.

(D) Alum (e.g.,potassium alum,$KAl(SO_4)_2 \cdot 12H_2O$) is a salt of a strong acid $(H_2SO_4)$ and a weak base $(Al(OH)_3)$.
When dissolved in water,it undergoes hydrolysis to produce $H^+$ ions.
Due to the presence of $H^+$ ions,the solution becomes acidic.
Therefore,the $pH$ of the aqueous solution of alum is $< 7$.

(C) $K_2S$ is a salt of a weak acid $(H_2S)$ and a strong base $(KOH)$.
In water,it undergoes anionic hydrolysis: $S^{2-} + H_2O \rightleftharpoons HS^- + OH^-$.
Due to the production of $OH^-$ ions,the solution becomes basic.
Therefore,the $pH$ of the $K_2S$ solution is greater than $7$.

(D) For a salt of a strong acid and a weak base,the hydrolysis constant $K_h$ is given by the formula:
$K_h = \frac{K_w}{K_b}$
Given:
$K_w = 10^{-12}$
$K_b = 10^{-5}$
Substituting the values:
$K_h = \frac{10^{-12}}{10^{-5}} = 10^{-7}$

(A) Cationic hydrolysis is directly proportional to the charge density of the cation.
Charge density is defined as the ratio of charge to size (ionic radius).
Among the given ions,$Al^{3+}$ has the smallest ionic radius and a high charge of $+3$.
Since $Al^{3+}$ has the highest charge-to-size ratio,it exhibits the maximum capacity for cationic hydrolysis.

(C) $NaCN$ is a salt of a strong base and a weak acid.
For the hydrolysis of such a salt,the hydrolysis constant $K_h$ is given by the formula $K_h = C h^2$,where $C$ is the concentration and $h$ is the degree of hydrolysis.
Given: $C = 0.01 \ M$ and $h = 3.37 \% = 0.0337$.
Substituting the values: $K_h = 0.01 \times (0.0337)^2$.
$K_h = 0.01 \times 0.00113569 = 1.13569 \times 10^{-5} \approx 1.13 \times 10^{-5}$.

(D) Salt hydrolysis occurs for salts derived from a weak acid and a strong base,or a strong acid and a weak base,or a weak acid and a weak base.
$Na_3PO_4$ is a salt of a weak acid $(H_3PO_4)$ and a strong base $(NaOH)$,so it undergoes anionic hydrolysis.
$CH_3COONa$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$,so it also undergoes anionic hydrolysis.
$NaNO_3$ is a salt of a strong acid $(HNO_3)$ and a strong base $(NaOH)$,so it does not undergo hydrolysis.
Therefore,both $(A)$ and $(B)$ undergo hydrolysis.

(A) The hydrolysis constant $(K_h)$ for a salt of a weak acid and a strong base is given by the formula: $K_h = \frac{K_w}{K_a}$.
Given: $K_w = 1.0 \times 10^{-14}$ and $K_a (HNO_2) = 4.5 \times 10^{-4}$.
Substituting the values: $K_h = \frac{1.0 \times 10^{-14}}{4.5 \times 10^{-4}}$.
$K_h = 0.222 \times 10^{-10} = 2.22 \times 10^{-11}$.

(A) mixed salt is a salt that contains more than one type of cation or anion.
$Ca(OH)Cl$ is a hydroxy salt (a type of mixed salt containing $OH^-$ and $Cl^-$ anions).
$NaKSO_4$ is a double salt containing two different cations ($Na^+$ and $K^+$).
$NaHSO_4$ is an acid salt containing a replaceable hydrogen atom.
Since both $Ca(OH)Cl$ and $NaKSO_4$ are types of mixed salts,the question implies identifying the most representative example of a mixed salt. $Ca(OH)Cl$ is a classic example of a mixed salt containing more than one type of anion.

(C) The hydrolysis constant $K_h$ is related to the dissociation constants of the acid $(K_a)$ and base $(K_b)$ and the ionic product of water $(K_w)$.
For a salt of a weak acid and a strong base (e.g.,$CH_3COONa$),the hydrolysis constant is given by $K_h = K_w / K_a$.
For a salt of a strong acid and a weak base (e.g.,$NH_4Cl$ or $FeCl_3$),the formula is $K_h = K_w / K_b$.
For a salt of a weak acid and a weak base (e.g.,$CH_3COONH_4$),the formula is $K_h = K_w / (K_a \times K_b)$.
Thus,$K_h = K_w / K_a$ applies to $CH_3COONa$.

(A) Salt hydrolysis occurs when a salt is formed from a weak acid and/or a weak base.
$1$. $Na_2SO_4$ is a salt of a strong acid $(H_2SO_4)$ and a strong base $(NaOH)$. Salts of strong acids and strong bases do not undergo hydrolysis in water.
$2$. $(NH_4)_2SO_4$ is a salt of a weak base $(NH_4OH)$ and a strong acid $(H_2SO_4)$,which undergoes cationic hydrolysis.
$3$. $NH_4CN$ is a salt of a weak base $(NH_4OH)$ and a weak acid $(HCN)$,which undergoes both cationic and anionic hydrolysis.
Therefore,$Na_2SO_4$ does not undergo hydrolysis.

(B) For a salt of a weak acid and a weak base,the hydrolysis constant $(K_h)$ is given by the formula:
$K_h = \frac{K_w}{K_a \times K_b}$
Therefore,$K_h$ is proportional to $(K_a \times K_b)^{-1}$.

(C) The $pH$ of a salt of a weak acid and a weak base is given by the formula:
$pH = 7 + \frac{1}{2}(pK_a - pK_b)$
Given:
$pK_a = 4.80$
$pK_b = 4.78$
Substituting the values into the formula:
$pH = 7 + \frac{1}{2}(4.80 - 4.78)$
$pH = 7 + \frac{1}{2}(0.02)$
$pH = 7 + 0.01$
$pH = 7.01$

(B) The given reaction represents the hydrolysis of the formate ion $(HCOO^-)$,which is the anion of a weak acid $(HCOOH)$ and a strong base $(NaOH)$.
For the hydrolysis of a salt of a weak acid and a strong base,the degree of hydrolysis $(h)$ is related to the hydrolysis constant $(K_h)$ and the concentration $(C)$ of the salt solution by the formula:
$K_h = C h^2$
Rearranging this for $h$,we get:
$h = \sqrt{\frac{K_h}{C}}$
Thus,the correct relation is $h = \sqrt{\frac{K_h}{C}}$.

(D) salt formed from a weak acid and a strong base undergoes anionic hydrolysis,resulting in an alkaline solution $(pH > 7)$.
However,if the question states $pH = 7$ and alkaline properties,it implies a contradiction or a specific buffer condition.
Given the standard classification,a salt of a weak acid and a strong base is the correct choice for an alkaline solution.

(B) Methylamine hydrochloride $(CH_3NH_3Cl)$ is a salt of a strong acid $(HCl)$ and a weak base $(CH_3NH_2)$.
For such a salt,the hydrolysis constant $(K_h)$ is given by the formula:
$K_h = \frac{K_w}{K_b}$
Given $K_w = 10^{-14}$ and $K_b = 5 \times 10^{-4}$.
Substituting the values:
$K_h = \frac{10^{-14}}{5 \times 10^{-4}} = 0.2 \times 10^{-10}$.

(A) The salt $CH_3COONH_4$ is formed from a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$.
For a salt of a weak acid and a weak base,the degree of hydrolysis $(h)$ is given by the formula: $h = \sqrt{K_h} = \sqrt{\frac{K_w}{K_a \times K_b}}$.
Since $K_w$,$K_a$,and $K_b$ are constants at a given temperature,the degree of hydrolysis $(h)$ is independent of the concentration of the salt solution.
Therefore,$x_1 = x_2$.

(B) Ammonium acetate is a salt of a weak acid and a weak base.
For such salts,the degree of hydrolysis $(h)$ is independent of the concentration of the solution and is given by the formula:
$h = \sqrt{K_h}$
Given $K_h = 3.175 \times 10^{-5}$.
$h = \sqrt{3.175 \times 10^{-5}} = \sqrt{31.75 \times 10^{-6}} \approx 5.63 \times 10^{-3}$.

(A) Sodium acetate $(CH_3COONa)$ is a salt of a weak acid $(CH_3COOH)$ and a strong base $(NaOH)$.
For such salts,the $pH$ is given by the formula: $pH = 7 + \frac{1}{2}pK_a + \frac{1}{2}\log C$.
Given: $C = 0.018 \, M = 1.8 \times 10^{-2} \, M$ and $K_a = 1.8 \times 10^{-5}$.
First,calculate $pK_a$: $pK_a = -\log(K_a) = -\log(1.8 \times 10^{-5}) = 5 - \log(1.8) \approx 5 - 0.255 = 4.745 \approx 4.75$.
Now,substitute the values into the formula:
$pH = 7 + \frac{1}{2}(4.75) + \frac{1}{2}\log(1.8 \times 10^{-2})$
$pH = 7 + 2.375 + \frac{1}{2}(\log 1.8 + \log 10^{-2})$
$pH = 7 + 2.375 + \frac{1}{2}(0.255 - 2)$
$pH = 7 + 2.375 + \frac{1}{2}(-1.745)$
$pH = 7 + 2.375 - 0.8725 = 8.5025 \approx 8.5$.

(B) Sodium acetate $(CH_3COONa)$ undergoes hydrolysis in water as follows:
$CH_3COO^{-} + H_2O ⇌ CH_3COOH + OH^{-}$
According to the stoichiometry of the hydrolysis reaction,for every mole of $CH_3COOH$ produced,one mole of $OH^{-}$ is also produced.
Therefore,in an aqueous solution of sodium acetate,the concentration of acetic acid formed is equal to the concentration of hydroxide ions produced:
$[CH_3COOH] = [OH^{-}]$