pH of weak Acids and weak Bases Questions in English

(A) For a weak acid,the concentration of $H_3O^+$ is given by $[H_3O^+] = \sqrt{K_a \times C}$.
Substituting the values: $[H_3O^+] = \sqrt{1.4 \times 10^{-5} \times 0.1} = \sqrt{1.4 \times 10^{-6}} = 1.183 \times 10^{-3} \ M$.
Now,$pH = -\log[H_3O^+] = -\log(1.183 \times 10^{-3}) = 3 - \log(1.183) = 3 - 0.073 = 2.927 \approx 2.92$.

(A) For a weak acid,the concentration of $H^+$ ions is given by $[H^+] = \sqrt{K_a \times C}$.
Given $K_a = 1.74 \times 10^{-5}$ and $C = 0.1$ $M$.
$[H^+] = \sqrt{1.74 \times 10^{-5} \times 0.1} = \sqrt{1.74 \times 10^{-6}} = 1.319 \times 10^{-3}$ $M$.
$pH = -\log[H^+] = -\log(1.319 \times 10^{-3}) = 3 - \log(1.319) = 3 - 0.12 = 2.88$.

(A) For a weak base $NH_3$,the dissociation is $NH_3 + H_2O \rightleftharpoons NH_4^+ + OH^-$.
The concentration of $OH^-$ ions is given by $[OH^-] = \sqrt{K_b \times C}$.
$[OH^-] = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \ M$.
$pOH = -\log[OH^-] = -\log(1.34 \times 10^{-3}) = 3 - \log(1.34) \approx 3 - 0.127 = 2.873$.
Since $pH + pOH = 14$ at $298 \ K$,$pH = 14 - 2.873 = 11.127 \approx 11.13$.

For a weak base,the equilibrium is: $(CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^{+} + OH^{-}$.
Using the approximation $[OH^{-}] = \sqrt{K_b \times C}$,where $K_b = 5.4 \times 10^{-5}$ and $C = 0.2 \ M$.
$[OH^{-}] = \sqrt{5.4 \times 10^{-5} \times 0.2} = \sqrt{1.08 \times 10^{-5}} = \sqrt{10.8 \times 10^{-6}} \approx 3.286 \times 10^{-3} \ M$.
$pOH = -\log[OH^{-}] = -\log(3.286 \times 10^{-3}) \approx 2.48$.
$pH = 14 - pOH = 14 - 2.48 = 11.52$.
$[H_3O^{+}] = 10^{-pH} = 10^{-11.52} \approx 3.02 \times 10^{-12} \ M$.

(B) For a weak base $BOH$,the dissociation is $BOH \rightleftharpoons B^{+} + OH^{-}$.
The dissociation constant $K_b = \frac{[B^{+}][OH^{-}]}{[BOH]} = \frac{c \alpha^2}{1-\alpha} \approx c \alpha^2$.
Given $K_b = 1.0 \times 10^{-12}$ and $c = 0.01 \ M = 10^{-2} \ M$.
$1.0 \times 10^{-12} = 10^{-2} \times \alpha^2 \implies \alpha^2 = 10^{-10} \implies \alpha = 10^{-5}$.
$[OH^{-}] = c \alpha = 10^{-2} \times 10^{-5} = 1.0 \times 10^{-7} \ M$.

(B) For a weak diprotic acid like $H_2CO_3$,the first dissociation step is $H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$.
Since $K_1 \gg K_2$,the concentration of $H^+$ ions is primarily determined by the first dissociation step.
In the first step,one molecule of $H_2CO_3$ produces one $H^+$ ion and one $HCO_3^-$ ion.
However,a small amount of $HCO_3^-$ further dissociates into $H^+$ and $CO_3^{2-}$ according to the second dissociation step: $HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}$.
This second step produces additional $H^+$ ions,making the total $[H^+]$ slightly greater than $[HCO_3^-]$.

(A) $1$. Calculate the molar mass of acetic acid $(CH_3COOH)$: $M = (2 \times 12) + (4 \times 1) + (2 \times 16) = 60 \ g/mol$.
$2$. Calculate the number of moles of acetic acid: $n = \frac{6.0 \ g}{60 \ g/mol} = 0.1 \ mol$.
$3$. Calculate the molarity $(C)$ of the solution: $C = \frac{0.1 \ mol}{0.250 \ L} = 0.4 \ M$.
$4$. For a weak acid,$[H^+] = \sqrt{K_a \times C} = \sqrt{1.8 \times 10^{-5} \times 0.4} = \sqrt{0.72 \times 10^{-5}} = \sqrt{7.2 \times 10^{-6}} \approx 2.68 \times 10^{-3} \ M$.
$5$. Calculate $pH$: $pH = -\log[H^+] = -\log(2.68 \times 10^{-3}) = 3 - \log(2.68) \approx 3 - 0.428 = 2.572$.

(A) $1$. Calculate the molarity $(M)$ of the solution:
$M = \frac{\text{mass}}{\text{molar mass} \times \text{volume in L}} = \frac{7 \ g}{35 \ g \ mol^{-1} \times 0.5 \ L} = 0.4 \ M$.
$2$. For a weak base,$[OH^-] = \sqrt{K_b \times C} = \sqrt{1.8 \times 10^{-5} \times 0.4} = \sqrt{0.72 \times 10^{-5}} = \sqrt{7.2 \times 10^{-6}} \approx 2.68 \times 10^{-3} \ M$.
$3$. Calculate $pOH$:
$pOH = -\log[OH^-] = -\log(2.68 \times 10^{-3}) = 3 - \log(2.68) \approx 3 - 0.43 = 2.57$.
$4$. Calculate $pH$:
$pH = 14 - pOH = 14 - 2.57 = 11.43$.

(A) For a weak acid,the concentration of $H^{+}$ ions is given by $[H^{+}] = \sqrt{K_a \times C}$.
Given $K_a = 6.5 \times 10^{-5}$ and $C = 0.15 \ M$.
$[H^{+}] = \sqrt{6.5 \times 10^{-5} \times 0.15} = \sqrt{9.75 \times 10^{-6}} \approx 3.12 \times 10^{-3} \ M$.
$pH = -\log[H^{+}] = -\log(3.12 \times 10^{-3}) = 3 - \log(3.12) \approx 3 - 0.494 = 2.506 \approx 2.5$.

(A) For a weak base $(CH_3)_2NH$,the dissociation is: $(CH_3)_2NH + H_2O \rightleftharpoons (CH_3)_2NH_2^+ + OH^-$.
The concentration of $OH^-$ is given by $[OH^-] = \sqrt{K_b \times C}$.
$[OH^-] = \sqrt{5.4 \times 10^{-4} \times 0.25} = \sqrt{1.35 \times 10^{-4}} = 1.16 \times 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(1.16 \times 10^{-2}) = 2 - 0.064 = 1.936$.
Since $pH + pOH = 14$,$pH = 14 - 1.936 = 12.064$.

(A) For a weak acid $HCN$,the dissociation is given by: $HCN \rightleftharpoons H^+ + CN^-$.
The concentration of $H^+$ ions is calculated from $pH$ as: $[H^+] = 10^{-pH} = 10^{-5.2}$.
$[H^+] = 6.31 \times 10^{-6} \ M$.
The dissociation constant $K_a$ is given by the formula: $K_a = \frac{[H^+][CN^-]}{[HCN]}$.
Since $[H^+] = [CN^-]$,we have $K_a = \frac{[H^+]^2}{[HCN]}$.
Substituting the values: $K_a = \frac{(6.31 \times 10^{-6})^2}{0.1} = \frac{3.98 \times 10^{-11}}{0.1} = 3.98 \times 10^{-10}$.

(N/A) For a weak acid $ClCH_2COOH$,the concentration $C = 0.02 \ M$ and $K_a = 1.36 \times 10^{-3}$.
First,calculate the degree of dissociation $\alpha = \sqrt{K_a / C} = \sqrt{1.36 \times 10^{-3} / 0.02} = \sqrt{0.068} \approx 0.2608$.
$[H^+] = C \times \alpha = 0.02 \times 0.2608 = 0.005216 \ M$.
$pH = -\log[H^+] = -\log(0.005216) \approx 2.283$.
To find $pK_b$,first calculate $pK_a = -\log(K_a) = -\log(1.36 \times 10^{-3}) \approx 2.866$.
Using the relation $pK_a + pK_b = 14$,we get $pK_b = 14 - 2.866 = 11.134$.

(A) The concentration of the pyridinium ion $[PyH^+]$ is given by the formula: $[PyH^+] = C \times \alpha$.
Given:
Concentration $C = 0.1 \ M$.
Degree of dissociation $\alpha = 0.013 \% = \frac{0.013}{100} = 1.3 \times 10^{-4}$.
Therefore,$[PyH^+] = 0.1 \times 1.3 \times 10^{-4} = 1.3 \times 10^{-5} \ M$.

(A) For a weak acid $HA$,the dissociation is given by $HA \rightleftharpoons H^+ + A^-$.
The degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.
Given $K_a = 1.4 \times 10^{-5}$ and $C = 0.1 \ M$.
$\alpha = \sqrt{\frac{1.4 \times 10^{-5}}{0.1}} = \sqrt{1.4 \times 10^{-4}} = 1.18 \times 10^{-2}$.
Percentage of dissociation = $\alpha \times 100 = 1.18 \times 10^{-2} \times 100 = 1.18 \%$.
The concentration of $H^+$ ions is $[H^+] = C \times \alpha = 0.1 \times 1.18 \times 10^{-2} = 1.18 \times 10^{-3} \ M$.
$pH = -\log[H^+] = -\log(1.18 \times 10^{-3}) = 3 - \log(1.18) = 3 - 0.0719 = 2.9281 \approx 2.93$.

(B) $H_{2}SO_{3}$ is a dibasic acid with concentration $c = 0.588 \ M$.
Since $Ka_{1} \gg Ka_{2}$,the $pH$ is primarily determined by the first dissociation step:
$H_{2}SO_{3(aq)} \rightleftharpoons H^{+}_{(aq)} + HS{O_{3}}^{-}_{(aq)}$
$Ka_{1} = \frac{[H^{+}][HSO_{3}^{-}]}{[H_{2}SO_{3}]} = 1.7 \times 10^{-2}$
Let $x$ be the concentration of $H^{+}$ at equilibrium:
$\frac{x^{2}}{0.588 - x} = 0.017$
$x^{2} + 0.017x - 0.009996 = 0$
Using the quadratic formula $x = \frac{-b + \sqrt{b^{2} - 4ac}}{2a}$:
$x = \frac{-0.017 + \sqrt{(0.017)^{2} - 4(1)(-0.009996)}}{2} \approx 0.09186 \ M$
$pH = -\log[H^{+}] = -\log(0.09186) \approx 1.036$
Rounding to the nearest integer,the $pH$ is $1$.

(C) For a weak acid,the $pH$ is given by the formula: $pH = \frac{1}{2} (pK_{a} - \log C)$.
Given $K_{a} = 2 \times 10^{-5}$,so $pK_{a} = -\log(2 \times 10^{-5}) = 5 - \log 2 = 5 - 0.30 = 4.7$.
Concentration $C = 0.2 \ M = 2 \times 10^{-1} \ M$.
$pH = \frac{1}{2} (4.7 - \log(2 \times 10^{-1}))$.
$pH = \frac{1}{2} (4.7 - (\log 2 + \log 10^{-1}))$.
$pH = \frac{1}{2} (4.7 - (0.30 - 1)) = \frac{1}{2} (4.7 - (-0.7)) = \frac{1}{2} (5.4) = 2.7$.
$pH = 2.7 = 27 \times 10^{-1}$.
Thus,the value is $27$.

(A) Given,dissociation constant,$K_{a} = 1.8 \times 10^{-5}$ and concentration $c = 0.1 \, M$.
For the dissociation of acetic acid: $CH_{3}COOH \rightleftharpoons CH_{3}COO^{-} + H^{+}$.
Since acetic acid is a weak acid,the concentration of $H^{+}$ ions is given by $[H^{+}] = \sqrt{K_{a} \times c}$.
$[H^{+}] = \sqrt{1.8 \times 10^{-5} \times 0.1} = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \, M$.
$pH = -\log[H^{+}] = -\log(1.34 \times 10^{-3}) = 3 - \log(1.34) \approx 3 - 0.127 = 2.873$.
Thus,the $pH$ is closest to $2.87$.

(C) The acidic strength of an acid depends on the stability of its conjugate base. Stronger acids have more stable conjugate bases.
The dissociation reactions are:
$HCl \rightleftharpoons H^{+} + Cl^{-}$
$HCOOH \rightleftharpoons H^{+} + HCOO^{-}$
$CH_{3}COOH \rightleftharpoons H^{+} + CH_{3}COO^{-}$
The stability of the conjugate bases follows the order: $Cl^{-} > HCOO^{-} > CH_{3}COO^{-}$.
Therefore,the acidic strength follows the order: $HCl > HCOOH > CH_{3}COOH$.
Since $pH = -\log[H^{+}]$,a higher concentration of $H^{+}$ ions results in a lower $pH$ value.
Thus,the order of $pH$ for $1\, N$ aqueous solutions is: $CH_{3}COOH > HCOOH > HCl$.

(B) The Henderson-Hasselbalch equation is given by: $pH = pK_a + \log \frac{[\text{conjugate base}]}{[\text{acid}]}$.
Given that the concentrations of the acid and its conjugate base are equal,we have $[\text{conjugate base}] = [\text{acid}]$,which implies $\frac{[\text{conjugate base}]}{[\text{acid}]} = 1$.
Substituting the values into the equation: $pH = 5.85 + \log(1)$.
Since $\log(1) = 0$,we get $pH = 5.85 + 0 = 5.85$.
Therefore,the correct option is $B$.

(A) For a weak acid,the degree of dissociation $\alpha$ is given by the formula $\alpha = \sqrt{\frac{K_a}{C}}$.
Given:
Concentration $C = 0.1 \ mol \ L^{-1} = 10^{-1} \ M$.
Dissociation constant $K_a = 10^{-5}$.
Substituting the values into the formula:
$\alpha = \sqrt{\frac{10^{-5}}{10^{-1}}}$
$\alpha = \sqrt{10^{-4}}$
$\alpha = 10^{-2} = 0.01$.
Thus,the degree of dissociation is $0.01$.

(C) For $CH_3COOH$ :
$[H^{+}] = C \cdot \alpha$
Given concentration $C = 0.01 \, M = 10^{-2} \, M$ and degree of ionisation $\alpha = 1 \% = 0.01 = 10^{-2}$.
$[H^{+}] = 10^{-2} \times 10^{-2} = 10^{-4} \, M$
$pH = -\log [H^{+}]$
$pH = -\log(10^{-4}) = 4$

(A) The dissociation reaction of ethylamine is: $C_2H_5NH_2 + H_2O \rightleftharpoons C_2H_5NH_3^+ + OH^-$
Given concentration $C = 1 \ mM = 10^{-3} \ M$.
Given $pH = 9$,so $pOH = 14 - 9 = 5$.
Thus,$[OH^-] = 10^{-pOH} = 10^{-5} \ M$.
Since $[OH^-] = [C_2H_5NH_3^+] = 10^{-5} \ M$ and $[C_2H_5NH_2] \approx C = 10^{-3} \ M$ (as degree of ionization is neglected),
$K_b = \frac{[C_2H_5NH_3^+][OH^-]}{[C_2H_5NH_2]} = \frac{10^{-5} \times 10^{-5}}{10^{-3}} = 10^{-7}$.
Comparing $10^{-7}$ with $10^{-x}$,we get $x = 7$.

(D) For a weak acid $HA$,the dissociation equilibrium is: $HA \rightleftharpoons H^{+} + A^{-}$.
At $t=0$,concentration is $C$ (or $a$).
At equilibrium,concentrations are: $[HA] = C(1-x)$,$[H^{+}] = Cx$,$[A^{-}] = Cx$.
The dissociation constant $K_{a}$ is given by: $K_{a} = \frac{[H^{+}][A^{-}]}{[HA]} = \frac{(Cx)(Cx)}{C(1-x)} = \frac{Cx^{2}}{1-x}$.
Taking negative logarithm on both sides: $-\log K_{a} = -\log\left(\frac{Cx^{2}}{1-x}\right)$.
$pK_{a} = -\log(Cx) - \log\left(\frac{x}{1-x}\right)$.
Since $pH = -\log[H^{+}] = -\log(Cx)$,we have $pK_{a} = pH - \log\left(\frac{x}{1-x}\right)$.
Therefore,$pH - pK_{a} = \log\left(\frac{x}{1-x}\right)$.

(D) For a weak acid $HX$,the dissociation is $HX \rightleftharpoons H^{+} + X^{-}$.
Given $C = 0.01 \ M$ and $pH = 5$,so $[H^{+}] = 10^{-5} \ M$.
Using $[H^{+}] = \sqrt{K_{a} \times C}$,we check: $\sqrt{4 \times 10^{-10} \times 10^{-2}} = \sqrt{4 \times 10^{-12}} = 2 \times 10^{-6} \ M$.
Since the given $[H^{+}] = 10^{-5} \ M$ does not match the calculated value,the problem statement is internally inconsistent.
However,assuming the dilution follows the relation $[H^{+}] = \sqrt{K_{a} \times C'}$ for the new concentration $C'$:
$10^{-6} = \sqrt{4 \times 10^{-10} \times C'}$
$10^{-12} = 4 \times 10^{-10} \times C'$
$C' = \frac{10^{-12}}{4 \times 10^{-10}} = 0.25 \times 10^{-2} = 25 \times 10^{-4} \ M$.
Thus,$x = 25$.

(B) The acidity of substituted benzoic acids depends on the stability of the conjugate base (carboxylate ion) formed after the loss of a proton.
In $o$-hydroxybenzoic acid (Salicylic acid),the hydroxyl group at the ortho position forms an intramolecular hydrogen bond with the carboxylate group after the loss of a proton,which significantly stabilizes the conjugate base.
In $2,6$-dihydroxybenzoic acid,there are two hydroxyl groups at ortho positions. The intramolecular hydrogen bonding is even more pronounced,and the inductive effect of the two hydroxyl groups further stabilizes the carboxylate anion.
Therefore,$2,6$-dihydroxybenzoic acid has the highest $K_{a}$ value among the given options due to the strong stabilization of its conjugate base by intramolecular hydrogen bonding and inductive effects.

(D) Given: Percent dissociation $= 1.42 \%$,Concentration $C = 0.05 \ M$.
Degree of dissociation $\alpha = \frac{1.42}{100} = 0.0142$.
For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = \alpha^2 C$ (assuming $\alpha << 1$).
$K_b = (0.0142)^2 \times 0.05$.
$K_b = 0.00020164 \times 0.05$.
$K_b = 1.0082 \times 10^{-5} \approx 1.0 \times 10^{-5}$.

(B) For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given,$C = 0.05 \ M$ and percentage dissociation $= 0.001 \%$.
Therefore,$\alpha = \frac{0.001}{100} = 10^{-5}$.
Substituting the values into the formula:
$K_a = 0.05 \times (10^{-5})^2$
$K_a = 0.05 \times 10^{-10}$
$K_a = 5 \times 10^{-12}$.

(A) For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the molar concentration and $\alpha$ is the degree of dissociation.
Given: $C = 0.1 \ M$ and $\alpha = 2 \% = 0.02$.
Substituting the values: $K_a = 0.1 \times (0.02)^2$.
$K_a = 0.1 \times 0.0004 = 4.0 \times 10^{-5}$.

(D) For a weak base,the dissociation constant $K_b$ is given by the formula $K_b = \alpha^2 C$,where $\alpha$ is the degree of dissociation and $C$ is the molar concentration.
Given $\alpha = 1.2 \% = 0.012$ and $C = 0.2 \ M$.
Substituting the values: $K_b = (0.012)^2 \times 0.2$.
$K_b = 0.000144 \times 0.2 = 2.88 \times 10^{-5}$.

(D) For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given: $C = 0.02 \ M$ and percent dissociation = $3 \%$.
Therefore,$\alpha = 3 / 100 = 0.03$.
Substituting the values into the formula:
$K_a = 0.02 \times (0.03)^2$
$K_a = 0.02 \times 0.0009$
$K_a = 1.8 \times 10^{-5}$

(D) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given,$C = 0.01 \ M = 10^{-2} \ M$.
The degree of dissociation $\alpha = \frac{0.001}{100} = 10^{-5}$.
Substituting the values into the formula:
$K_a = (10^{-2}) \times (10^{-5})^2$
$K_a = 10^{-2} \times 10^{-10}$
$K_a = 10^{-12}$.

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is $HA \rightleftharpoons H^+ + A^-$.
Let $C$ be the initial molar concentration and $x$ be the degree of dissociation.
$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(Cx)(Cx)}{C(1-x)} = \frac{Cx^2}{1-x}$.
Since the acid is weak,$x \ll 1$,so $1-x \approx 1$,and $K_a \approx Cx^2$.
Given $K_a = 5 \times 10^{-8}$ and $x = 0.5 \% = 0.005 = 5 \times 10^{-3}$.
$C = \frac{K_a}{x^2} = \frac{5 \times 10^{-8}}{(5 \times 10^{-3})^2}$.
$C = \frac{5 \times 10^{-8}}{25 \times 10^{-6}} = \frac{5}{25} \times 10^{-2} = 0.2 \times 10^{-2} = 0.002 \ M$.

(B) Given: $pH = 2.34$,$c = 0.1 \ M$.
Step $1$: Calculate the concentration of hydrogen ions $[H^+]$.
$[H^+] = 10^{-pH} = 10^{-2.34} = 4.571 \times 10^{-3} \ M$.
Step $2$: Use the relation for a monobasic acid,$[H^+] = c \alpha$.
$\alpha = \frac{[H^+]}{c} = \frac{4.571 \times 10^{-3}}{0.1} = 4.571 \times 10^{-2}$.
Thus,the degree of dissociation is $4.57 \times 10^{-2}$.

(B) For a weak acid,the dissociation constant $K_a$ is related to the degree of dissociation $\alpha$ and concentration $C$ by the formula: $K_a = C\alpha^2$ (assuming $\alpha \ll 1$).
Given $K_a = 1.32 \times 10^{-5}$ and $C = 0.05 \ M$.
$\alpha = \sqrt{\frac{K_a}{C}}$
$\alpha = \sqrt{\frac{1.32 \times 10^{-5}}{0.05}}$
$\alpha = \sqrt{\frac{1.32 \times 10^{-5}}{5 \times 10^{-2}}} = \sqrt{0.264 \times 10^{-3}} = \sqrt{2.64 \times 10^{-4}}$
$\alpha = 1.624 \times 10^{-2} \approx 1.61 \times 10^{-2}$.

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is:
$HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}$
At equilibrium,the concentration of $H^+$ and $A^-$ is $C\alpha$ and the concentration of $HA$ is $C(1-\alpha)$,where $C$ is the initial concentration and $\alpha$ is the degree of dissociation.
The dissociation constant $K_a$ is given by:
$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{C\alpha^2}{1-\alpha}$
Since the acid is weak,$\alpha \ll 1$,so $1-\alpha \approx 1$.
Thus,$K_a = C\alpha^2$.
Rearranging to solve for $C$:
$C = \frac{K_a}{\alpha^2} = \frac{2.7 \times 10^{-5}}{(3 \times 10^{-2})^2} = \frac{2.7 \times 10^{-5}}{9 \times 10^{-4}} = 0.3 \times 10^{-1} = 0.03 \ M$.

(B) For a monoacidic weak base $BOH$:
$BOH_{(aq)} \rightleftharpoons B_{(aq)}^{+} + OH_{(aq)}^{-}$
Given $pH = 10.9$,we calculate $pOH$:
$pOH = 14 - pH = 14 - 10.9 = 3.1$
The concentration of hydroxide ions is:
$[OH^{-}] = 10^{-pOH} = 10^{-3.1} = 10^{-4} \times 10^{0.9} \approx 7.94 \times 10^{-4} \ M$
For a weak base,$[OH^{-}] = C \times \alpha$,where $C = 0.02 \ M$:
$\alpha = \frac{[OH^{-}]}{C} = \frac{7.94 \times 10^{-4}}{0.02} = 3.97 \times 10^{-2}$
The percent dissociation is $\alpha \times 100$:
$\% \alpha = 3.97 \times 10^{-2} \times 100 = 3.97 \% \approx 3.95 \%$
Thus,the correct option is $B$.

(D) For the dissociation of acetic acid: $CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$
Given concentration $C = 0.01 \ M$ and dissociation constant $K_a = 1.34 \times 10^{-6}$.
The expression for dissociation constant is $K_a = \frac{C \alpha^2}{1-\alpha}$.
Since $K_a$ is very small,$1-\alpha \approx 1$,so $K_a \approx C \alpha^2$.
Thus,$\alpha = \sqrt{\frac{K_a}{C}} = \sqrt{\frac{1.34 \times 10^{-6}}{0.01}} = \sqrt{1.34 \times 10^{-4}} = 1.157 \times 10^{-2}$.
Percent dissociation $\% \alpha = \alpha \times 100 = 1.157 \times 10^{-2} \times 100 = 1.157 \% \approx 1.15 \%$.

(B) For a weak monobasic acid $HA$,the dissociation equilibrium is: $HA_{(aq)} \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}$
Given concentration $C = 0.04 \ M$ and degree of dissociation $\alpha = 3.0 \% = 0.03$.
The dissociation constant $K_a$ is given by the formula: $K_a = \frac{C \alpha^2}{1 - \alpha}$.
Since the acid is weak,$\alpha \ll 1$,so $1 - \alpha \approx 1$.
Thus,$K_a \approx C \alpha^2$.
Substituting the values: $K_a = 0.04 \times (0.03)^2$.
$K_a = 0.04 \times 0.0009 = 3.6 \times 10^{-5}$.

(B) The dissociation of acetic acid is given by: $CH_3COOH_{(aq)} \rightleftharpoons CH_3COO^-_{(aq)} + H^+_{(aq)}$
Given concentration $C = 0.1 \ M$ and degree of ionization $\alpha = 1.34 \ \% = 0.0134$.
The dissociation constant $K_a$ is given by the formula: $K_a = \frac{C\alpha^2}{1-\alpha}$.
Since $\alpha$ is very small,we can approximate $1-\alpha \approx 1$.
Thus,$K_a \approx C\alpha^2$.
Substituting the values: $K_a = 0.1 \times (0.0134)^2$.
$K_a = 0.1 \times 0.00017956 = 1.7956 \times 10^{-5} \approx 1.8 \times 10^{-5}$.

(A) Given: $K_{a} = 2.25 \times 10^{-6}$ and concentration $c = 0.01 \ M$.
For a weak monobasic acid,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_{a}}{c}}$.
Substituting the values: $\alpha = \sqrt{\frac{2.25 \times 10^{-6}}{0.01}} = \sqrt{2.25 \times 10^{-4}} = 0.015$.
Percent dissociation = $\alpha \times 100 = 0.015 \times 100 = 1.5 \%$.

(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{\text{Percentage dissociation}}{100} = \frac{0.04}{100} = 4 \times 10^{-4}$.
For a monobasic acid,the concentration of hydronium ions is given by $[H_3O^{+}] = c \times \alpha$.
Substituting the values: $[H_3O^{+}] = 0.05 \ M \times 4 \times 10^{-4} = 2.0 \times 10^{-5} \ M$.

(B) For a weak monobasic acid,the dissociation constant $K_a$ is given by $K_a = c \alpha^2$.
Thus,the degree of dissociation $\alpha = \sqrt{\frac{K_a}{c}}$.
Substituting the given values: $\alpha = \sqrt{\frac{1.8 \times 10^{-5}}{0.02}} = \sqrt{9 \times 10^{-4}} = 3 \times 10^{-2}$.
The concentration of hydronium ions is given by $[H_3O^{+}] = c \alpha$.
$[H_3O^{+}] = 0.02 \times 3 \times 10^{-2} = 6.0 \times 10^{-4} \ M$.

(A) For a weak acid $HA$,the degree of dissociation $\alpha$ is very small,so $(1 - \alpha) \cong 1$.
The formula for the degree of dissociation is $\alpha = \sqrt{\frac{K_{a}}{c}}$.
Given $K_{a} = 1.0 \times 10^{-5}$ and $c = 0.1 \ M$.
Substituting the values: $\alpha = \sqrt{\frac{1.0 \times 10^{-5}}{0.1}} = \sqrt{1.0 \times 10^{-4}} = 10^{-2}$.

(B) For a weak monobasic acid,the relationship between dissociation constant $(K_{a})$,degree of dissociation $(\alpha)$,and concentration $(c)$ is given by:
$K_{a} = \alpha^2 c$
Substituting the given values:
$c = \frac{K_{a}}{\alpha^2} = \frac{5.0 \times 10^{-9}}{(5.0 \times 10^{-4})^2}$
$c = \frac{5.0 \times 10^{-9}}{25 \times 10^{-8}} = 0.2 \times 10^{-1} = 0.02 \ M$

(D) For a weak monobasic acid $HA$,the dissociation is given by: $HA \rightleftharpoons H^+_{(aq)} + A^-_{(aq)}$
Given degree of dissociation $\alpha = 0.05 \% = 0.05 \times 10^{-2} = 5 \times 10^{-4}$.
Concentration $C = 0.02 \ M = 2 \times 10^{-2} \ M$.
For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 C$ (since $\alpha \ll 1$).
Substituting the values: $K_a = (5 \times 10^{-4})^2 \times (2 \times 10^{-2})$.
$K_a = (25 \times 10^{-8}) \times (2 \times 10^{-2}) = 50 \times 10^{-10} = 5.0 \times 10^{-9}$.

(C) The degree of dissociation $\alpha$ is given by $\alpha = \frac{\text{Percent dissociation}}{100} = \frac{5}{100} = 0.05 = 5 \times 10^{-2}$.
Given concentration $c = 0.02 \ M = 2 \times 10^{-2} \ M$.
For a weak monobasic acid,the dissociation constant $K_a$ is given by the formula $K_a = \alpha^2 c$ (since $\alpha$ is very small,$1 - \alpha \approx 1$).
Substituting the values: $K_a = (5 \times 10^{-2})^2 \times (2 \times 10^{-2})$.
$K_a = (25 \times 10^{-4}) \times (2 \times 10^{-2}) = 50 \times 10^{-6} = 5 \times 10^{-5}$.

(A) For a monobasic acid $HA$,the dissociation is given by: $HA \rightleftharpoons H_{(aq)}^{+} + A_{(aq)}^{-}$
Concentration of $H^{+}$ ions is calculated as $[H^{+}] = \alpha \times C$,where $\alpha$ is the degree of dissociation and $C$ is the molar concentration.
Given $\alpha = 2 \% = 0.02$ and $C = 0.02 \ M$.
$[H^{+}] = 0.02 \times 0.02 = 0.0004 \ M = 4 \times 10^{-4} \ M$.
$pH = -\log[H^{+}] = -\log(4 \times 10^{-4}) = 4 - \log(4) = 4 - 0.602 = 3.398 \approx 3.4$.

(B) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given: $C = 0.01 \ M$,$\alpha = 1.2 \% = 0.012$.
Substituting the values: $K_a = 0.01 \times (0.012)^2$.
$K_a = 0.01 \times 0.000144$.
$K_a = 1.44 \times 10^{-6}$.

(D) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given: Concentration $C = 0.1 \ M = 10^{-1} \ M$.
Degree of dissociation $\alpha = 0.01 \% = \frac{0.01}{100} = 10^{-4}$.
Substituting the values into the formula:
$K_a = (10^{-1}) \times (10^{-4})^2$
$K_a = 10^{-1} \times 10^{-8}$
$K_a = 10^{-9}$.

(B) For a weak acid,the dissociation constant $K_a$ is given by the formula $K_a = C \alpha^2$,where $C$ is the concentration and $\alpha$ is the degree of dissociation.
Given: $C = 0.002 \ M = 2 \times 10^{-3} \ M$.
Degree of dissociation $\alpha = 2 \% = 0.02 = 2 \times 10^{-2}$.
Substituting the values into the formula:
$K_a = (2 \times 10^{-3}) \times (2 \times 10^{-2})^2$
$K_a = (2 \times 10^{-3}) \times (4 \times 10^{-4})$
$K_a = 8 \times 10^{-7}$.