Mix Examples-Ionic Equilibrium Questions in English

(C) Dimethylammonium acetate is a salt of a weak acid (acetic acid) and a weak base (dimethylamine).
The formula for the $pH$ of a salt of a weak acid and a weak base is given by:
$pH = 7 + \frac{1}{2} pK_{a} - \frac{1}{2} pK_{b}$
Given:
$pK_{a} = 4.77$
$pK_{b} = 3.27$
Substituting the values into the formula:
$pH = 7 + \frac{1}{2} (4.77) - \frac{1}{2} (3.27)$
$pH = 7 + \frac{1}{2} (4.77 - 3.27)$
$pH = 7 + \frac{1}{2} (1.50)$
$pH = 7 + 0.75$
$pH = 7.75$

(B) First,calculate the molar conductivity $(\Lambda_{m})$ using the formula: $\Lambda_{m} = \frac{1000 \times \kappa}{C}$
$\Lambda_{m} = \frac{1000 \times 2.0 \times 10^{-5}}{0.001} = 20 \, S \, cm^{2} \, mol^{-1}$
Next,calculate the degree of dissociation $(\alpha)$:
$\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{20}{190} = \frac{2}{19}$
For a weak acid $HA$,the ionization constant $K_{a}$ is given by:
$K_{a} = \frac{C \alpha^{2}}{1 - \alpha}$
Since $\alpha$ is very small,$1 - \alpha \approx 1$,but we can calculate it precisely:
$K_{a} = \frac{0.001 \times (2/19)^{2}}{1 - (2/19)} = \frac{0.001 \times (4/361)}{17/19} = \frac{0.001 \times 4}{361} \times \frac{19}{17} = \frac{0.004}{19 \times 17} = \frac{0.004}{323} \approx 1.238 \times 10^{-5} = 12.38 \times 10^{-6}$
Rounding to the nearest integer,we get $12 \times 10^{-6}$.

(C) The dissociation of the salt is given by: $A_{2}X_{3(s)} \rightleftharpoons 2A^{3+}_{(aq)} + 3X^{2-}_{(aq)}$.
Let the solubility be $s \ mol \ L^{-1}$. Then $[A^{3+}] = 2s$ and $[X^{2-}] = 3s$.
The solubility product $K_{sp} = (2s)^2(3s)^3 = 108s^5 = 1.1 \times 10^{-23}$.
$s^5 = \frac{1.1 \times 10^{-23}}{108} \approx 1.018 \times 10^{-25}$.
$s \approx 10^{-5} \ mol \ L^{-1} = 10^{-5} \ mol \ (10^{-3} \ m^3)^{-1} = 0.01 \ mol \ m^{-3}$.
For a sparingly soluble salt,the molar conductivity $\Lambda_m \approx \Lambda_m^{\infty} = \frac{\kappa}{C}$,where $\kappa$ is the specific conductance and $C$ is the concentration in $mol \ m^{-3}$.
$\Lambda_m^{\infty} = \frac{3 \times 10^{-5} \ S \ m^{-1}}{0.01 \ mol \ m^{-3}} = 3 \times 10^{-3} \ S \ m^2 \ mol^{-1}$.
Comparing this with $x \times 10^{-3} \ S \ m^2 \ mol^{-1}$,we get $x = 3$.

(A) The titration of a weak base $(NH_4OH)$ with a strong acid $(HCl)$ involves the addition of acid to the base.
Initially,the $pH$ is high (basic).
As the strong acid is added,the $pH$ decreases gradually.
Near the equivalence point,there is a sharp drop in $pH$.
Since the resulting salt $(NH_4Cl)$ is acidic due to the hydrolysis of the $NH_4^+$ ion,the equivalence point $pH$ is less than $7$.
Therefore,the correct plot shows a downward curve starting from a high $pH$ value.

(C) The $pH$ at the equivalence point for the titration of a weak acid with a weak base is given by the formula:
$pH = \frac{1}{2}(pK_{w} + pK_{a} - pK_{b})$
From this expression,it is clear that the $pH$ is independent of the concentration of the acid and the base (statement $ii$ is correct,$i$ is incorrect).
The expression also shows that the $pH$ depends on the $pK_{a}$ of the acid and the $pK_{b}$ of the base (statement $iii$ is correct,$iv$ is incorrect).
Therefore,the correct statements are $(ii)$ and $(iii)$.

(B) $H_2SO_4$ is a diprotic acid,so $10 \ mL$ of $0.1 \ M$ $H_2SO_4$ provides $2 \times 10 \times 0.1 = 2 \ mmol$ of $H^{+}$ ions.
$KOH$ is a monoprotic base,so $10 \ mL$ of $0.1 \ M$ $KOH$ provides $10 \times 0.1 = 1 \ mmol$ of $OH^{-}$ ions.
After neutralization,the remaining $H^{+}$ ions $= 2 - 1 = 1 \ mmol$.
Total volume of the mixture $= 10 \ mL + 10 \ mL = 20 \ mL$.
Therefore,$[H^{+}] = \frac{1 \ mmol}{20 \ mL} = 0.05 \ M$.

(C) The concentration of $Ag^{+}$ ions from $AgNO_3$ is $[Ag^{+}] = 10^{-5} \ M$.
The concentration of $NO_3^{-}$ ions is $[NO_3^{-}] = 10^{-5} \ M$.
Due to the common ion effect,the concentration of $Br^{-}$ ions is determined by the solubility product constant: $[Br^{-}] = \frac{K_{sp}}{[Ag^{+}]} = \frac{4.9 \times 10^{-13}}{10^{-5}} = 4.9 \times 10^{-8} \ M$.
The conductivity $\kappa$ is given by $\kappa = \sum C_i \lambda_i$.
For $Ag^{+}$: $\kappa_{Ag^{+}} = 10^{-5} \times 6 \times 10^{-3} = 6 \times 10^{-8} \ S \ m^{-1}$.
For $Br^{-}$: $\kappa_{Br^{-}} = 4.9 \times 10^{-8} \times 8 \times 10^{-3} = 39.2 \times 10^{-11} \ S \ m^{-1} = 0.392 \times 10^{-8} \ S \ m^{-1}$.
For $NO_3^{-}$: $\kappa_{NO_3^{-}} = 10^{-5} \times 7 \times 10^{-3} = 7 \times 10^{-8} \ S \ m^{-1}$.
Total conductivity $\kappa = (6 + 0.392 + 7) \times 10^{-8} \ S \ m^{-1} = 13.392 \times 10^{-8} \ S \ m^{-1}$.
Rounding to the nearest integer,we get $13 \times 10^{-8} \ S \ m^{-1}$.

(A) The dissociation reaction is: $A_2B_3 (aq.) \rightleftharpoons 2 A^{3+} (aq.) + 3 B^{2-} (aq.)$
At equilibrium,the concentrations are: $[A_2B_3] = c(1 - \alpha)$,$[A^{3+}] = 2c\alpha$,and $[B^{2-}] = 3c\alpha$.
The equilibrium constant expression is: $K_{eq} = \frac{[A^{3+}]^2 [B^{2-}]^3}{[A_2B_3]} = \frac{(2c\alpha)^2 (3c\alpha)^3}{c(1 - \alpha)}$.
For a weak electrolyte,$\alpha \ll 1$,so $(1 - \alpha) \approx 1$.
Thus,$K_{eq} = \frac{(4c^2\alpha^2)(27c^3\alpha^3)}{c} = 108c^4\alpha^5$.
Solving for $\alpha$: $\alpha^5 = \frac{K_{eq}}{108c^4} \implies \alpha = \left(\frac{K_{eq}}{108 c^4}\right)^{\frac{1}{5}}$.

(A) $ (A) $ The $pH$ of $1 \times 10^{-8} \ M \ HCl$ is not $8$ because the contribution of $H^+$ ions from water cannot be neglected. The $pH$ is approximately $6.98$.
$ (B) $ The conjugate base of $H_2PO_4^{-}$ is obtained by removing one $H^+$ ion,which gives $HPO_4^{2-}$. This is correct.
$ (C) $ The auto-ionization of water is an endothermic process,so $K_w$ increases with an increase in temperature. This is correct.
$ (D) $ According to the Henderson-Hasselbalch equation,$pH = pK_a + \log \frac{[Salt]}{[Acid]}$. At the half-neutralization point,$[Salt] = [Acid]$,so $pH = pK_a$. This is correct.

(A) Ammonium carbonate $(NH_4)_2CO_3$ is a salt of a weak base $(NH_4OH)$ and a weak acid $(H_2CO_3)$.
The $pH$ of a salt solution formed from a weak acid and a weak base is given by the formula: $pH = 7 + \frac{1}{2}(pK_a - pK_b)$.
For $(NH_4)_2CO_3$,the $pK_a$ of $H_2CO_3$ is approximately $6.35$ and the $pK_b$ of $NH_4OH$ is approximately $4.75$. Since $pK_b < pK_a$,the solution is slightly basic.
Thus,both Statement $I$ and Statement $II$ are correct.

(A) When $CO_2$ dissolves in water,it forms carbonic acid $(H_2CO_3)$.
$H_2CO_3$ is a weak diprotic acid that undergoes partial dissociation in two steps:
$CO_2 + H_2O \rightleftharpoons H_2CO_3$
$H_2CO_3 \rightleftharpoons H^+ + HCO_3^-$
$HCO_3^- \rightleftharpoons H^+ + CO_3^{2-}$
Therefore,the species present in the equilibrium mixture are $CO_2$,$H_2CO_3$,$HCO_3^-$,and $CO_3^{2-}$.

(C) The reaction is $BOH + HCl \longrightarrow BCl + H_2O$.
At the equivalence point,the salt $BCl$ is formed.
The moles of $BOH = 2.5 \ mL \times 0.4 \ M = 1.0 \ mmol$.
The volume of $HCl$ required $= \frac{1.0 \ mmol}{2/15 \ M} = 7.5 \ mL$.
Total volume at equivalence point $= 2.5 \ mL + 7.5 \ mL = 10 \ mL$.
Concentration of salt $C = \frac{1.0 \ mmol}{10 \ mL} = 0.1 \ M$.
For the salt of a weak base and strong acid,$[H^{+}] = \sqrt{\frac{K_W \times C}{K_b}}$.
$[H^{+}] = \sqrt{\frac{10^{-14} \times 0.1}{10^{-12}}} = \sqrt{10^{-3}} = \sqrt{10 \times 10^{-4}} \approx 3.16 \times 10^{-2} \ M$.

(B) At $pH = 7$,the salt $MX$ dissociates as $MX_{(s)} \rightleftharpoons M^+_{(aq)} + X^-_{(aq)}$. The solubility $S_1 = \sqrt{K_{sp}} = 10^{-4} \ mol \ L^{-1}$,so $K_{sp} = 10^{-8}$.
At $pH = 2$,$[H^+] = 10^{-2} \ M$. The salt dissociates and $X^-$ reacts with $H^+$: $X^-_{(aq)} + H^+_{(aq)} \rightleftharpoons HX_{(aq)}$.
The equilibrium constant for this reaction is $K = \frac{1}{K_a}$.
The total solubility $S = [M^+] = [X^-] + [HX] = 10^{-3} \ M$.
From $K_{sp} = [M^+][X^-]$,we have $[X^-] = \frac{K_{sp}}{[M^+]} = \frac{10^{-8}}{10^{-3}} = 10^{-5} \ M$.
Thus,$[HX] = S - [X^-] = 10^{-3} - 10^{-5} \approx 10^{-3} \ M$.
Using the equilibrium expression for $HX$ formation: $K_a = \frac{[H^+][X^-]}{[HX]} = \frac{10^{-2} \times 10^{-5}}{10^{-3}} = 10^{-4}$.
Therefore,$pK_a = -\log(10^{-4}) = 4$.

(A) $P$: This is an acidic buffer solution $(CH_3COOH + CH_3COONa)$. The $[H^{+}]$ of a buffer solution remains constant upon dilution. Thus,$P \rightarrow 1$.
$Q$: This is a salt of a weak acid and a strong base $(CH_3COONa)$. The $[H^{+}]$ is given by $[H^{+}] = \sqrt{K_w K_a / C}$. Since $[H^{+}] \propto 1/\sqrt{C}$,and the concentration $C$ decreases by a factor of $2$ (from $40 \ mL$ to $80 \ mL$),$[H^{+}]$ increases by $\sqrt{2}$. Thus,$Q \rightarrow 4$.
$R$: This is a salt of a strong acid and a weak base $(NH_4Cl)$. The $[H^{+}]$ is given by $[H^{+}] = \sqrt{K_h C} = \sqrt{(K_w/K_b) C}$. Since $[H^{+}] \propto \sqrt{C}$,and the concentration $C$ decreases by a factor of $2$ (from $40 \ mL$ to $80 \ mL$),$[H^{+}]$ decreases by $1/\sqrt{2}$. Thus,$R \rightarrow 3$.
$S$: This is a saturated solution of a sparingly soluble base in equilibrium with its solid. The $[OH^{-}]$ remains constant due to the common ion effect of the solid. Since $[H^{+}][OH^{-}] = K_w$,$[H^{+}]$ remains constant. Thus,$S \rightarrow 1$.
Therefore,the correct matching is $P$ $\rightarrow 1, Q$ $\rightarrow 4, R$ $\rightarrow 3, S$ $\rightarrow 1$.

(C) From the titration curve,the equivalence point is reached at $V = 6 \ mL$ of $HA$ added.
At the half-equivalence point,$V = 3 \ mL$,the concentration of the weak base $[B]$ is equal to the concentration of its conjugate acid $[BH^{+}]$.
According to the Henderson-Hasselbalch equation for a weak base:
$pOH = pK_{b} + \log\left(\frac{[BH^{+}]}{[B]}\right)$
At the half-equivalence point,$[BH^{+}] = [B]$,so $\log(1) = 0$.
Therefore,$pOH = pK_{b}$.
From the graph,at $V = 3 \ mL$,the $pH$ is approximately $11.15$.
Since $pH + pOH = 14$ at $25^{\circ}C$,we have $pOH = 14 - 11.15 = 2.85$.
Thus,$pK_{b} = 2.85$.

(A) The rate of acid-catalyzed hydrolysis of an ester is directly proportional to the concentration of $H^+$ ions,i.e.,$Rate \propto [H^+]$.
For a strong acid $(HX, 1 \ M)$,$[H^+]_{HX} = 1 \ M$.
Given that the rate with $HA$ is $1/100$th of the rate with $HX$,we have:
$\frac{Rate_{HA}}{Rate_{HX}} = \frac{[H^+]_{HA}}{[H^+]_{HX}} = \frac{1}{100}$
Since $[H^+]_{HX} = 1 \ M$,then $[H^+]_{HA} = 0.01 \ M$.
For the weak acid dissociation: $HA \rightleftharpoons H^+ + A^-$.
At equilibrium,$[H^+] = [A^-] = 0.01 \ M$ and $[HA] = 1 - 0.01 \approx 1 \ M$.
The acid dissociation constant $K_a$ is given by:
$K_a = \frac{[H^+][A^-]}{[HA]} = \frac{(0.01)(0.01)}{1} = 10^{-4}$.

(A) $1.$ $HCl + NaOH \longrightarrow NaCl + H_2O$
$n = 100 \times 1 = 100 \ mmol = 0.1 \ mol$
Energy evolved due to neutralization of $HCl$ and $NaOH = 0.1 \times 57 = 5.7 \ kJ = 5700 \ J$
Energy used to increase temperature of solution $= 200 \times 4.2 \times 5.7 = 4788 \ J$
Energy used to increase temperature of calorimeter $= 5700 - 4788 = 912 \ J$
Calorimeter constant $(C_{cal}) = \frac{912 \ J}{5.7^{\circ}C} = 160 \ J/^{\circ}C$
Energy evolved by neutralization of $CH_3COOH$ and $NaOH$ in Expt. $2$ $= (200 \times 4.2 \times 5.6) + (160 \times 5.6) = 4704 + 896 = 5600 \ J$
Energy used in dissociation of $0.1 \ mol \ CH_3COOH = 5700 - 5600 = 100 \ J = 0.1 \ kJ$
Enthalpy of dissociation $= \frac{0.1 \ kJ}{0.1 \ mol} = 1.0 \ kJ \ mol^{-1}$
$2.$ After mixing,$[CH_3COOH] = \frac{100 \times 2 - 100 \times 1}{200} = 0.5 \ M$ and $[CH_3COONa] = \frac{100 \times 1}{200} = 0.5 \ M$
$pK_a = -\log(2.0 \times 10^{-5}) = 5 - \log 2 = 5 - 0.3 = 4.7$
Using Henderson-Hasselbalch equation: $pH = pK_a + \log \frac{[salt]}{[acid]} = 4.7 + \log \frac{0.5}{0.5} = 4.7$

(B) . Incorrect. For a very dilute acid solution $(1 \times 10^{-8} \ M)$,the contribution of $H^{\oplus}$ from water cannot be ignored. The $pH$ will be slightly less than $7$,not $8$.
$B$. Correct. The conjugate base is formed by the removal of a proton $(H^{\oplus})$: $H_2PO_4^{\ominus} \rightarrow HPO_4^{2-} + H^{\oplus}$.
$C$. Correct. The auto-ionization of water is an endothermic process; thus,$K_w$ increases as temperature increases.
$D$. Incorrect. At the half-neutralization point of a weak acid titrated with a strong base,the Henderson-Hasselbalch equation gives $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$. Since $[Salt] = [Acid]$,$pH = pK_a$.

(C) Isohydric solutions are those solutions which have the same concentration of hydrogen ions $([H^{+}])$.
For a weak acid $HA$,the concentration of $H^{+}$ is given by $[H^{+}] = \sqrt{K_a \cdot C}$.
For two isohydric solutions,$[H^{+}]_1 = [H^{+}]_2$.
Therefore,$\sqrt{K_{a_1} C_1} = \sqrt{K_{a_2} C_2}$,which implies $K_{a_1} C_1 = K_{a_2} C_2$.
Since $C = \frac{1}{V}$ (where $V$ is dilution),we can write $K_{a_1} \cdot \frac{1}{V_1} = K_{a_2} \cdot \frac{1}{V_2}$,which simplifies to $K_{a_1} V_2 = K_{a_2} V_1$ or $\frac{K_{a_1}}{V_1} = \frac{K_{a_2}}{V_2}$.
Option $C$ states $\frac{K_{a_1}}{V_2} = \frac{K_{a_2}}{V_1}$,which is mathematically incorrect.

(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: $C = 0.1 \ M$ and $\alpha = 1.5 \% = \frac{1.5}{100} = 0.015$.
Substituting the values:
$K_a = 0.1 \times (0.015)^2$
$K_a = 0.1 \times 0.000225$
$K_a = 2.25 \times 10^{-5}$

(D) According to Kohlrausch's law,the molar conductivity at infinite dilution for $NH_4OH$ is given by: $\Lambda^0_{m}(NH_4OH) = \Lambda^0_{m}(NH_4Cl) + \Lambda^0_{m}(NaOH) - \Lambda^0_{m}(NaCl)$.
Substituting the given values: $\Lambda^0_{m}(NH_4OH) = 130 + 213 - 109 = 234 \ S \ cm^2 \ mol^{-1}$.
The degree of dissociation $(\alpha)$ is calculated as: $\alpha = \frac{\Lambda^m_c}{\Lambda^0_m} = \frac{9.0}{234}$.
The percent dissociation is: $\alpha \times 100 = \frac{9.0 \times 100}{234} = \frac{900}{234} = \frac{100}{26} \%$.

(C) The degree of dissociation $(\alpha)$ is given by the formula: $\alpha = \frac{\wedge_m^c}{\wedge_m^{\circ}}$.
First,calculate the molar conductivity at infinite dilution $(\wedge_m^{\circ})$ for $CH_3COOH$ using Kohlrausch's law:
$\wedge_m^{\circ}(CH_3COOH) = \wedge^{\circ}(CH_3COO^{-}) + \wedge^{\circ}(H^{+}) = 50 + 350 = 400 \ S \ cm^2 \ mol^{-1}$.
Given the molar conductivity at concentration $c$ $(\wedge_m^c)$ is $20 \ S \ cm^2 \ mol^{-1}$.
Now,calculate $\alpha$:
$\alpha = \frac{20}{400} = \frac{1}{20} = 0.05 = 5 \times 10^{-2}$.

(D) The concentration of $H^{+}$ ions is given by $[H^{+}] = C \cdot \alpha = 0.02 \times 0.04 = 8 \times 10^{-4} \ M$.
Using the ionic product of water,$[H^{+}][OH^{-}] = 10^{-14}$.
Therefore,$[OH^{-}] = \frac{10^{-14}}{8 \times 10^{-4}} = 1.25 \times 10^{-11} \ M$.

(C) Given: $[H^{+}] = 0.2 \text{ M}$,Volume $V = 10 \text{ cm}^3 = 10^{-2} \text{ L}$.
At $298 \text{ K}$,the ionic product of water is $K_w = [H^{+}][OH^{-}] = 10^{-14}$.
$[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{10^{-14}}{0.2} = 5 \times 10^{-14} \text{ mol/L}$.
Number of $OH^{-}$ ions $= [OH^{-}] \times V \times N_A$.
$= (5 \times 10^{-14} \text{ mol/L}) \times (10^{-2} \text{ L}) \times (6.022 \times 10^{23} \text{ ions/mol}) \approx 3.011 \times 10^8$.
Note: The calculated value is $3.011 \times 10^8$. The provided option $(c)$ $3 \times 10^{12}$ is mathematically inconsistent with the given parameters.

(B) The degree of dissociation $(\alpha)$ is almost unity for strong electrolytes,which dissociate completely in aqueous solution.
$KCl$ (Potassium chloride) is a strong electrolyte.
$NH_4Cl$ (Ammonium chloride) is a salt of a weak base and a strong acid,which undergoes hydrolysis.
$CH_3COONa$ (Sodium acetate) is a salt of a weak acid and a strong base,which also undergoes hydrolysis.
Therefore,only $KCl$ has a degree of dissociation close to $1$.

(D) $HCl$ is a strong acid $(SA)$,so its $pH$ is very low (approx $1$).
$NH_4NO_3$ is a salt of a strong acid and a weak base $(SAWB)$,which undergoes cationic hydrolysis,making the solution acidic $(pH < 7)$.
$NaCl$ is a salt of a strong acid and a strong base $(SASB)$,which does not undergo hydrolysis,making the solution neutral $(pH = 7)$.
$NaCN$ is a salt of a weak acid and a strong base $(WASB)$,which undergoes anionic hydrolysis,making the solution basic $(pH > 7)$.
Therefore,the increasing order of $pH$ is $HCl < NH_4NO_3 < NaCl < NaCN$.

(B) The degree of dissociation $(\alpha)$ is given as $4.2 \times 10^{-2}$.
To find the percent dissociation,multiply the degree of dissociation by $100$.
$\text{Percent dissociation} = \alpha \times 100$
$\text{Percent dissociation} = 4.2 \times 10^{-2} \times 100 = 4.2 \%$

(D) The reactions are: $(a)$ $NH_{3} + HCl \rightarrow NH_{4}Cl$ (Salt of strong acid and weak base,$pH < 7$). $(b)$ $NH_{3} + CH_{3}COOH \rightarrow CH_{3}COONH_{4}$ (Salt of weak acid and weak base,$pH \approx 7$). $(c)$ $2NH_{3} + H_{2}SO_{4} \rightarrow (NH_{4})_{2}SO_{4}$ (Salt of strong acid and weak base,$pH < 7$).
For $(a)$,$pH = 7 - \frac{1}{2}(pK_{b} + \log C) = 7 - \frac{1}{2}(4.74 + 0) = 4.63$.
For $(b)$,$pH = 7 + \frac{1}{2}(pK_{a} - pK_{b}) = 7 + \frac{1}{2}(4.74 - 4.74) = 7.0$.
For $(c)$,$H_{2}SO_{4}$ is a strong diprotic acid. $1 \ M \ H_{2}SO_{4}$ provides $2 \ M \ H^{+}$,which reacts with $1 \ M \ NH_{3}$ to leave $1 \ M \ H^{+}$ and $0.5 \ M \ (NH_{4})_{2}SO_{4}$. This results in a highly acidic solution $(pH < 1)$.
Thus,the order of $pH$ is $b (7.0) > a (4.63) > c (< 1)$,which corresponds to $b > a > c$.

(A) Calculate the milliequivalents $(meq)$ of $NaOH$ and $HCl$ present in the solution.
$meq \ of \ NaOH = N_1 \times V_1 = 0.4 \times 5 = 2.0 \ meq$.
$meq \ of \ HCl = N_2 \times V_2 = 0.1 \times 20 = 2.0 \ meq$.
Since the number of milliequivalents of the strong base $(NaOH)$ is equal to the number of milliequivalents of the strong acid $(HCl)$,the neutralization reaction is complete.
The resulting solution contains only the salt $(NaCl)$ and water,making the solution neutral.
Therefore,the $pH$ of the resulting solution is $7$.

(B) Total milliequivalents of $H^{+}$ ions from $HCl$ and $HNO_{3}$:
$= (30 \times \frac{1}{3}) + (20 \times \frac{1}{2}) = 10 + 10 = 20 \ mEq$.
Total milliequivalents of $OH^{-}$ ions from $NaOH$:
$= 40 \times \frac{1}{4} = 10 \ mEq$.
Since $H^{+}$ reacts with $OH^{-}$ to form water,the remaining milliequivalents of $H^{+}$:
$= 20 - 10 = 10 \ mEq$.
The final volume of the solution is $1 \ dm^{3} = 1000 \ mL$.
Concentration of $H^{+}$ ions $[H^{+}] = \frac{10 \ mEq}{1000 \ mL} = 10^{-2} \ M$.
$pH = -\log[H^{+}] = -\log(10^{-2}) = 2$.

(A) The dissociation reaction is: $HA + H_2O \rightleftharpoons H_3O^{+} + A^{-}$.
Initial concentration: $0.5 \ M$,$0$,$0$.
Equilibrium concentration: $0.5(1 - \alpha)$,$0.5\alpha$,$0.5\alpha$.
Given $\alpha = 1 \% = 0.01$.
$[H_3O^{+}] = [A^{-}] = 0.5 \times 0.01 = 0.005 \ M$.
$[HA] = 0.5(1 - 0.01) = 0.5 \times 0.99 = 0.495 \ M$.

(C) Option $A$ is correct because Bronsted-Lowry theory defines acids as proton donors,but $BCl_3$ is a Lewis acid that accepts electron pairs,not protons.
Option $B$ is incorrect because for $0.01 \ M \ NaOH$,$[OH^-] = 10^{-2} \ M$,so $pOH = 2$ and $pH = 14 - 2 = 12$.
Option $C$ is correct as the ionic product of water $(K_w)$ at $25^{\circ} C$ is $1.0 \times 10^{-14} \ mol^2 \ L^{-2}$.
Option $D$ is correct as the definition of $pH$ is $pH = -\log [H^+]$.
Note: In many contexts,multiple statements may be technically correct. However,$C$ and $D$ are fundamental definitions. Given the standard format,$C$ is a specific constant value.

(B) For a weak acid,the degree of ionization $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$,where $C$ is the molar concentration.
Given $K_a = 1.8 \times 10^{-5}$.
For solution "$A$": $\alpha_A = \frac{4.242}{100} = 0.04242$. Thus,$0.04242 = \sqrt{\frac{1.8 \times 10^{-5}}{x}}$. Squaring both sides: $0.0018 = \frac{1.8 \times 10^{-5}}{x}$,which gives $x = \frac{1.8 \times 10^{-5}}{1.8 \times 10^{-3}} = 0.01 \ M$.
For solution "$B$": $\alpha_B = \frac{3}{100} = 0.03$. Thus,$0.03 = \sqrt{\frac{1.8 \times 10^{-5}}{y}}$. Squaring both sides: $0.0009 = \frac{1.8 \times 10^{-5}}{y}$,which gives $y = \frac{1.8 \times 10^{-5}}{9 \times 10^{-4}} = 0.02 \ M$.
When $1 \ L$ of solution "$A$" $(0.01 \ M)$ is mixed with $1 \ L$ of solution "$B$" $(0.02 \ M)$,the total volume becomes $2 \ L$.
The total moles of acetic acid = $(1 \ L \times 0.01 \ M) + (1 \ L \times 0.02 \ M) = 0.03 \ \text{moles}$.
The resultant concentration = $\frac{0.03 \ \text{moles}}{2 \ L} = 0.015 \ M$.

(A) The ionic product of water $(K_w)$ at $100^{\circ} C$ is approximately $51.3 \times 10^{-14}$,which is greater than $10^{-14}$. Thus,statement $(A)$ is incorrect.
$(B)$ The dissociation of water is an endothermic process. As temperature increases,$K_w$ increases,which leads to a decrease in the $pH$ of neutral water $(pH = -\log[H^+])$. Thus,statement $(B)$ is correct.
$(C)$ $NaH_2PO_4$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_3PO_4)$. However,the anion $H_2PO_4^-$ acts as an acid $(K_a > K_b)$,making the solution acidic,not basic. Thus,statement $(C)$ is incorrect.
$(D)$ $NH_3$ can accept a proton to form $NH_4^+$ (acting as a Bronsted base) and can donate a proton to form $NH_2^-$ (acting as a Bronsted acid). Thus,statement $(D)$ is correct.
Note: Based on the analysis,only $(B)$ and $(D)$ are correct. However,if the question implies $Na_2HPO_4$ in option $(C)$,it would be basic. Given the options provided,there is a discrepancy in the standard interpretation.

(A) Moles of $HCl = 0.4 \ M \times 0.1 \ L = 0.04 \ \text{moles}$.
Moles of $NaOH = 0.5 \ M \times 0.1 \ L = 0.05 \ \text{moles}$.
$HCl$ and $NaOH$ react in a $1:1$ ratio according to the equation: $HCl + NaOH \rightarrow NaCl + H_2O$.
Since $0.05 \ \text{moles}$ of $NaOH$ react with $0.04 \ \text{moles}$ of $HCl$,$NaOH$ is in excess.
Remaining moles of $NaOH = 0.05 - 0.04 = 0.01 \ \text{moles}$.
Total volume of the solution = $100 \ mL + 100 \ mL + 800 \ mL = 1000 \ mL = 1 \ L$.
Concentration of $[OH^-] = \frac{0.01 \ \text{moles}}{1 \ L} = 0.01 \ M = 10^{-2} \ M$.
$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
Since $pH + pOH = 14$ at $27^{\circ}C$,$pH = 14 - 2 = 12$.

(A) The $pH$ of an acidic solution decreases as the concentration of $H^+$ ions increases. The initial concentration of $HCl$ is $0.01 \ M$ in $20 \ mL$. We calculate the final concentration $(M_{final})$ for each option:
$M_{final} = \frac{M_1 V_1 + M_2 V_2}{V_1 + V_2}$
For option $A$: $M_A = \frac{0.01 \times 20 + 0.02 \times 20}{20 + 20} = \frac{0.2 + 0.4}{40} = 0.015 \ M$. Since $0.015 \ M > 0.01 \ M$,the $pH$ will decrease.
For option $B$: $M_B = \frac{0.01 \times 20 + 0.005 \times 20}{40} = 0.0075 \ M$. Since $0.0075 \ M < 0.01 \ M$,the $pH$ will increase.
For option $C$: $M_C = \frac{0.01 \times 20 + 0.01 \times 20}{40} = 0.01 \ M$. The $pH$ remains unchanged.
For option $D$: $M_D = \frac{0.01 \times 20 + 0.005 \times 40}{20 + 40} = \frac{0.2 + 0.2}{60} = 0.0066 \ M$. Since $0.0066 \ M < 0.01 \ M$,the $pH$ will increase.
Therefore,only option $A$ results in a higher concentration of $H^+$,leading to a decrease in $pH$.

(B) For $HCl$ solution: $pH = 2$,so $[H^+] = 10^{-2} \ M = 0.01 \ M$.
Number of moles of $H^+ = M \times V(L) = 0.01 \times 0.2 = 0.002 \ mol$.
For $NaOH$ solution: $pH = 12$,so $pOH = 14 - 12 = 2$.
$[OH^-] = 10^{-2} \ M = 0.01 \ M$.
Number of moles of $OH^- = M \times V(L) = 0.01 \times 0.3 = 0.003 \ mol$.
Since $n_{OH^-} > n_{H^+}$,the resulting solution is basic.
Remaining moles of $OH^- = 0.003 - 0.002 = 0.001 \ mol$.
Total volume = $200 \ mL + 300 \ mL = 500 \ mL = 0.5 \ L$.
$[OH^-]_{final} = \frac{0.001 \ mol}{0.5 \ L} = 0.002 \ M = 2 \times 10^{-3} \ M$.
$pOH = -\log(2 \times 10^{-3}) = 3 - \log 2 = 3 - 0.3 = 2.7$.
$pH = 14 - pOH = 14 - 2.7 = 11.3$.

(B) The dissociation of water is an endothermic process. Therefore,as temperature increases,the ionic product of water $(K_W)$ increases. Conversely,as temperature decreases,$K_W$ decreases. Thus,Statement $B$ is correct.
For a buffer solution,the $pH$ is given by the Henderson-Hasselbalch equation: $pH = pK_a + \log(\frac{[Salt]}{[Acid]})$. As temperature increases,the $pK_a$ value generally decreases,leading to an increase in the $pH$ of the buffer solution. Thus,Statement $A$ is also correct.

(C) The dissociation of acetic acid is represented as: $CH_3COOH \rightleftharpoons H^{+} + CH_3COO^{-}$
Using the expression for the dissociation constant $K_a = \frac{C\alpha^2}{1-\alpha}$,where $C = 0.5 \ M$ and $\alpha = 0.15$.
$K_a = \frac{0.5 \times (0.15)^2}{1 - 0.15} = \frac{0.5 \times 0.0225}{0.85} = \frac{0.01125}{0.85} \approx 0.0132$.
Now,calculate $pK_a$ using $pK_a = -\log K_a$.
$pK_a = -\log(0.0132) = -\log(1.32 \times 10^{-2})$.
$pK_a = -(\log 1.32 + \log 10^{-2}) = -(0.12 - 2) = 1.88$.

(A) Given: $\lambda^{\circ}(H^{+}) = 349.6 \ S \ cm^2 \ mol^{-1}$ and $\lambda^{\circ}(HCOO^{-}) = 54.6 \ S \ cm^2 \ mol^{-1}$.
Degree of dissociation $\alpha = 11.0 \% = 0.11$.
The molar conductivity at infinite dilution is calculated as: $\lambda_{m}^{\circ}(HCOOH) = \lambda^{\circ}(H^{+}) + \lambda^{\circ}(HCOO^{-}) = 349.6 + 54.6 = 404.2 \ S \ cm^2 \ mol^{-1}$.
The molar conductivity $\lambda_{m}$ at a given concentration is related to the degree of dissociation by the formula: $\lambda_{m} = \alpha \times \lambda_{m}^{\circ}$.
Substituting the values: $\lambda_{m} = 0.11 \times 404.2 = 44.46 \ S \ cm^2 \ mol^{-1}$.

(A) $AlCl_3$ is a salt of a weak base and a strong acid; it undergoes cationic hydrolysis:
$Al^{3+} + 3H_2O \rightarrow Al(OH)_3 + 3H^+$
$H^+$ ions are released into the solution,making the aqueous solution of $AlCl_3$ acidic $(A-II)$.
$CH_3COONa$ is a salt of a weak acid and a strong base; it undergoes anionic hydrolysis:
$CH_3COO^- + H_2O \rightarrow CH_3COOH + OH^-$
$OH^-$ ions are released,making the solution basic $(B-I)$.
$KCl$ is a strong electrolyte and an ionic compound,making its aqueous solution highly conductive $(C-III)$.
$Al_2O_3$ is an amphoteric oxide as it reacts with both acids and bases $(D-V)$.
Therefore,the correct match is $A-II, B-I, C-III, D-V$.

(C) The balanced chemical equation for the reaction between oxalic acid $(H_2C_2O_4)$ and acidified $KMnO_4$ is:
$2MnO_4^- + 5H_2C_2O_4 + 6H^+ \rightarrow 2Mn^{2+} + 10CO_2 + 8H_2O$
From the stoichiometry,$5$ moles of $H_2C_2O_4$ react with $2$ moles of $MnO_4^-$.
Using the relation $n_{H_2C_2O_4} / 5 = n_{MnO_4^-} / 2$:
$(40 \times 10^{-3} \times x) / 5 = (16 \times 10^{-3} \times 0.05) / 2$
$8 \times 10^{-3} \times x = 4 \times 10^{-4}$
$x = (4 \times 10^{-4}) / (8 \times 10^{-3}) = 0.05 \ M$
Oxalic acid is a diprotic acid: $H_2C_2O_4 \rightarrow 2H^+ + C_2O_4^{2-}$.
Assuming complete dissociation,$[H^+] = 2 \times [H_2C_2O_4] = 2 \times 0.05 = 0.1 \ M$.
$pH = -\log[H^+] = -\log(0.1) = 1$.

(A) $(i)$ In a saturated $(NH_4)_2SO_4$ solution,$NH_4^+$ ions are in equilibrium with $NH_3$ and $H^+$. When $CuSO_4$ is added,$Cu^{2+}$ reacts with $NH_3$ to form the stable complex $[Cu(NH_3)_4]^{2+}$.
$Cu^{2+}_{(aq)} + 4NH_{3(aq)} \rightarrow [Cu(NH_3)_4]^{2+}_{(aq)}$
This reaction consumes $NH_3$,shifting the equilibrium $NH_4^+ \rightleftharpoons NH_3 + H^+$ to the right,which increases the $H^+$ ion concentration and thus increases the acidity.
$(ii)$ In anhydrous $HF$,the auto-ionization equilibrium is $2HF \rightleftharpoons H_2F^+ + F^-$.
When $SbF_5$ is added,it acts as a strong Lewis acid and reacts with $F^-$ to form the stable $SbF_6^-$ ion $(SbF_5 + F^- \rightarrow SbF_6^-)$.
This removes $F^-$ from the solution,shifting the equilibrium to the right,which increases the concentration of $H_2F^+$ (the superacidic species),thereby increasing the acidity.

(D) Degree of dissociation,$\alpha = \frac{\Lambda_{m}^{c}}{\Lambda_{m}^{\circ}} = \frac{150}{500} = 0.3$
Given,$C = 0.007 \ M$
Hydrofluoric acid dissociates as follows:
$HF \rightleftharpoons H^{+} + F^{-}$
Initially: $C, 0, 0$
At equilibrium: $C(1 - \alpha), C\alpha, C\alpha$
Dissociation constant,$K_{a} = \frac{[H^{+}] [F^{-}]}{[HF]} = \frac{C \alpha \cdot C \alpha}{C(1 - \alpha)} = \frac{C \alpha^{2}}{(1 - \alpha)}$
On substituting the values,we get:
$K_{a} = \frac{0.007 \times (0.3)^{2}}{(1 - 0.3)} = \frac{0.007 \times 0.09}{0.7} = 0.01 \times 0.09 = 9 \times 10^{-4} \ M$

(B, C) $1$. The salt of a strong acid and a weak base undergoes cationic hydrolysis,resulting in an acidic solution with $pH < 7$. This is a correct statement.
$2$. For a salt of a weak acid and a weak base,the $pH$ is determined by the relation $pH = 7 + \frac{1}{2}(pK_{a} - pK_{b})$. If $K_{b} < K_{a}$,then $pK_{b} > pK_{a}$,which implies $pH < 7$ (acidic). Thus,the statement that it is basic if $K_{b} < K_{a}$ is incorrect.
$3$. For a very dilute solution of $HCl$ $(10^{-8} \ M)$,the contribution of $H^{+}$ ions from water cannot be neglected. The total $[H^{+}] = 10^{-8} + 10^{-7} \approx 1.1 \times 10^{-7} \ M$. Thus,$pH = -\log(1.1 \times 10^{-7}) \approx 6.96$,which is less than $7$. The statement that $pH$ is $8$ is incorrect.
$4$. The conjugate acid of $NH_{2}^{-}$ is formed by adding a proton $(H^{+})$,resulting in $NH_{3}$. This is a correct statement.
Therefore,statements $B$ and $C$ are incorrect.

(D) The $pH$ of an acidic buffer solution is given by the Henderson-Hasselbalch equation:
$pH = pK_{a} + \log \frac{[salt]}{[acid]}$
For $pH = pK_{a}$,we must have $\log \frac{[salt]}{[acid]} = 0$,which implies $[salt] = [acid]$.
$(1)$ $100 \ mL$ of $0.1 \ M \ CH_{3}COOH$ $(10 \ mmol)$ + $100 \ mL$ of $0.1 \ M \ CH_{3}COONa$ $(10 \ mmol)$.
Here,$[salt] = [acid] = 10 \ mmol$. Thus,$pH = pK_{a} + \log(1) = pK_{a}$.
$(2)$ $100 \ mL$ of $0.1 \ M \ CH_{3}COOH$ $(10 \ mmol)$ + $50 \ mL$ of $0.1 \ M \ NaOH$ $(5 \ mmol)$.
The reaction is: $CH_{3}COOH + NaOH \rightarrow CH_{3}COONa + H_{2}O$.
Initial: $10 \ mmol \ CH_{3}COOH, 5 \ mmol \ NaOH$.
Final: $5 \ mmol \ CH_{3}COOH, 5 \ mmol \ CH_{3}COONa$.
Since $[salt] = [acid] = 5 \ mmol$,$pH = pK_{a} + \log(1) = pK_{a}$.
$(3)$ $100 \ mL$ of $0.1 \ M \ CH_{3}COOH$ $(10 \ mmol)$ + $100 \ mL$ of $0.1 \ M \ NaOH$ $(10 \ mmol)$.
This results in complete neutralization to form $CH_{3}COONa$. No buffer is formed.
$(4)$ $CH_{3}COOH + NH_{3} \rightarrow CH_{3}COONH_{4}$.
This forms a salt of a weak acid and a weak base,not a buffer solution.
Therefore,both $(1)$ and $(2)$ satisfy the condition $pH = pK_{a}$.

(C) The self-ionization reaction of water is: $H_2O(\ell) \rightleftharpoons H^{+}_{(aq)} + OH^{-}_{(aq)}$ with $K_w = 10^{-14}$.
The standard Gibbs free energy change is given by: $\Delta G^{\circ} = -RT \ln K_w = -2.303 RT \log K_w$.
Given $R = 1.987 \times 10^{-3} \ kCal \ K^{-1} \ mol^{-1}$ and $T = 298 \ K$:
$\Delta G^{\circ} = -2.303 \times (1.987 \times 10^{-3}) \times 298 \times \log(10^{-14})$
$\Delta G^{\circ} = -2.303 \times 1.987 \times 10^{-3} \times 298 \times (-14)$
$\Delta G^{\circ} \approx 19.1 \ kCal \ mol^{-1}$.

(B) The reaction proceeds as follows:
$1$. $MgCO_3(s) + 2HCl(aq) \rightarrow MgCl_2(aq) + H_2O(l) + CO_2(g)$
$2$. The solution is neutralized with ammonia and buffered with $NH_4Cl / NH_4OH$.
$3$. Upon adding disodium hydrogen phosphate $(Na_2HPO_4)$ to the solution containing $Mg^{2+}$ ions in the presence of $NH_4^+$ ions,a white crystalline precipitate of magnesium ammonium phosphate is formed:
$Mg^{2+}(aq) + NH_4^+(aq) + HPO_4^{2-}(aq) \rightarrow Mg(NH_4)PO_4(s) \downarrow$
Thus,the formula of the white precipitate is $Mg(NH_4)PO_4$.