Mix Examples-Ionic Equilibrium Questions in English

(B) Step $1$: Calculate the number of moles of each solute.
Moles of $NaOH = \frac{8 \, g}{40 \, g/mol} = 0.2 \, mol$.
Moles of $H_2SO_4 = \frac{4.9 \, g}{98 \, g/mol} = 0.05 \, mol$.
Step $2$: Calculate the equivalents of $H^+$ and $OH^-$.
$OH^-$ from $NaOH = 0.2 \, mol \times 1 = 0.2 \, eq$.
$H^+$ from $H_2SO_4 = 0.05 \, mol \times 2 = 0.1 \, eq$.
Step $3$: Determine the excess concentration.
Excess $OH^- = 0.2 - 0.1 = 0.1 \, mol/L$.
Step $4$: Calculate $pOH$ and $pH$.
$pOH = -\log[OH^-] = -\log(0.1) = 1$.
$pH = 14 - pOH = 14 - 1 = 13$.

(B) $1$. The $pH$ of neutral water at $25^{\circ}C$ is $7$,not $0$. Thus,option $A$ is correct.
$2$. $CH_3COOH$ is a weak acid and $NaOH$ is a strong base. Their mixture ($1 \, N$ each) forms $CH_3COONa$,which undergoes anionic hydrolysis. The resulting solution is basic,so the $pH$ will be greater than $7$,not $7$. Thus,option $B$ is incorrect.
$3$. $NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$. It undergoes hydrolysis to form a basic solution,so $pH > 7$. Thus,option $C$ is correct.
$4$. Both $CH_3COOH$ and $HCl$ are acids. Their mixture will be acidic,so $pH < 7$. Thus,option $D$ is correct.

(D) For $AgCl$ in pure water: $AgCl(s) ⇌ Ag^+(aq) + Cl^-(aq)$.
$K_{sp} = [Ag^+][Cl^-] = s^2 = (4 \times 10^{-6})^2 = 16 \times 10^{-12}$.
In the presence of $2 \times 10^{-3} \ M \ CaCl_2$,the concentration of $Cl^-$ ions from $CaCl_2$ is $[Cl^-] = 2 \times (2 \times 10^{-3}) = 4 \times 10^{-3} \ M$.
Let the new solubility be $s'$. The total concentration of $Cl^-$ is $(s' + 4 \times 10^{-3}) \approx 4 \times 10^{-3} \ M$ (since $s'$ is very small).
$K_{sp} = [Ag^+][Cl^-] = s' \times (4 \times 10^{-3}) = 16 \times 10^{-12}$.
$s' = \frac{16 \times 10^{-12}}{4 \times 10^{-3}} = 4 \times 10^{-9} \ M$.

(C) For a weak acid $HA$,the dissociation is $HA \rightleftharpoons H^{+} + A^{-}$,so $K_a = \frac{[H^{+}][A^{-}]}{[HA]} = 1.0 \times 10^{-5}$.
The reaction of a weak acid with a strong base is $HA + OH^{-} \rightleftharpoons A^{-} + H_2O$.
The equilibrium constant $K$ for this reaction is $K = \frac{[A^{-}]}{[HA][OH^{-}]}$.
We know that $K_w = [H^{+}][OH^{-}] = 1.0 \times 10^{-14}$.
Multiplying the numerator and denominator of $K$ by $[H^{+}]$,we get $K = \frac{[A^{-}][H^{+}]}{[HA][OH^{-}][H^{+}]} = \frac{K_a}{K_w}$.
Substituting the values: $K = \frac{1.0 \times 10^{-5}}{1.0 \times 10^{-14}} = 1.0 \times 10^9$.

(C) For a weak electrolyte $A_xB_y$,the dissociation reaction is: $A_xB_y \rightleftharpoons xA^{y+} + yB^{x-}$.
Initially,concentration is $c$ for $A_xB_y$ and $0$ for products.
At equilibrium,concentrations are: $[A_xB_y] = c(1-\alpha)$,$[A^{y+}] = xc\alpha$,and $[B^{x-}] = yc\alpha$.
The equilibrium constant $K_{eq}$ is given by: $K_{eq} = \frac{[A^{y+}]^x [B^{x-}]^y}{[A_xB_y]} = \frac{(xc\alpha)^x (yc\alpha)^y}{c(1-\alpha)}$.
Since $\alpha$ is very small for a weak electrolyte,$(1-\alpha) \approx 1$.
Thus,$K_{eq} = \frac{x^x c^x \alpha^x y^y c^y \alpha^y}{c} = x^x y^y c^{x+y-1} \alpha^{x+y}$.
Solving for $\alpha$: $\alpha^{x+y} = \frac{K_{eq}}{c^{x+y-1} x^x y^y}$.
Therefore,$\alpha = \left(\frac{K_{eq}}{c^{x+y-1} x^x y^y}\right)^{\frac{1}{x+y}}$.

(C) For a diprotic acid $H_2A$,the overall dissociation constant $K_a$ is the product of the individual dissociation constants $K_{a1}$ and $K_{a2}$.
$K_a = K_{a1} \times K_{a2}$
Given $K_{a1} = 1.0 \times 10^{-5}$ and $K_{a2} = 5.0 \times 10^{-10}$.
$K_a = (1.0 \times 10^{-5}) \times (5.0 \times 10^{-10}) = 5.0 \times 10^{-15}$.

(A) The dielectric constant of $D_2O$ is lower than that of $H_2O$.
Since the degree of dissociation $(\alpha)$ is directly proportional to the dielectric constant of the solvent,a higher dielectric constant facilitates greater ionization.
Therefore,$\alpha_1 > \alpha_2$.

(A) The basic strength is directly proportional to the dissociation constant of the base $(K_b)$.
The relationship between hydrolysis constant $(K_h)$,ionic product of water $(K_w)$,and base dissociation constant $(K_b)$ is given by: $K_h = \frac{K_w}{K_b}$,which implies $K_b = \frac{K_w}{K_h}$.
Given $K_w = 10^{-14}$:
$1$. For $M_1OH$ (from $M_1X$): $K_b(M_1OH) = \frac{10^{-14}}{10^{-7}} = 10^{-7}$.
$2$. For $M_2OH$ (from $M_2X$): $K_b(M_2OH) = \frac{10^{-14}}{10^{-4}} = 10^{-10}$.
$3$. For $M_3OH$: $K_b(M_3OH) = 10^{-4}$.
Comparing the $K_b$ values: $10^{-4} > 10^{-7} > 10^{-10}$.
Therefore,the decreasing order of basic strength is $M_3OH > M_1OH > M_2OH$.

(B) The reaction of a weak acid $(HA)$ with a strong base $(OH^-)$ is given by: $HA + OH^- \rightleftharpoons A^- + H_2O$.
The equilibrium constant for this reaction is $K = \frac{K_a}{K_w}$.
Given $K_a = 1.0 \times 10^{-4}$ and $K_w = 1.0 \times 10^{-14}$.
$K = \frac{1.0 \times 10^{-4}}{1.0 \times 10^{-14}} = 1.0 \times 10^{10}$.

(B) Total $H^+$ millimoles = $(300 \times 1/3) + (200 \times 1/2) = 100 + 100 = 200 \, mmol$.
Total $OH^-$ millimoles = $400 \times 1/4 = 100 \, mmol$.
Remaining $H^+$ millimoles = $200 - 100 = 100 \, mmol$.
Total volume of the solution = $1 \, dm^3 = 1000 \, mL$.
$[H^+] = \frac{100 \, mmol}{1000 \, mL} = 0.1 \, M = 10^{-1} \, M$.
$pH = -\log[H^+] = -\log(10^{-1}) = 1$.

(D) In pure water,the dissociation reaction is: $2H_2O(l) \rightleftharpoons H_3O^{+}(aq) + OH^{-}(aq)$.
Since the water is pure,the concentration of $[H_3O^{+}]$ must be equal to the concentration of $[OH^{-}]$.
Given: $[H_3O^{+}] = 10^{-6.7} \, \text{mol/L}$.
Therefore,$[OH^{-}] = 10^{-6.7} \, \text{mol/L}$.
The ionic product of water $(K_W)$ is defined as: $K_W = [H_3O^{+}][OH^{-}]$.
Substituting the values: $K_W = (10^{-6.7}) \times (10^{-6.7}) = 10^{-6.7 - 6.7} = 10^{-13.4}$.

(A) The pair $NaHCO_3$ and $NaOH$ cannot exist together in a solution because they react with each other to form $Na_2CO_3$ and $H_2O$.
The reaction is: $NaHCO_3 + NaOH \rightarrow Na_2CO_3 + H_2O$.

(B) Electrical conductivity depends on the degree of dissociation of the acid.
Stronger acids dissociate more,providing more ions in the solution.
The acidity increases with the presence of electron-withdrawing groups due to the $-I$ effect.
Difluoroacetic acid $(CHF_2COOH)$ has two fluorine atoms,which exert a stronger $-I$ effect compared to monofluoroacetic acid $(CH_2FCOOH)$,chloroacetic acid $(CH_2ClCOOH)$,or acetic acid $(CH_3COOH)$.
Therefore,difluoroacetic acid is the strongest acid among the given options and will have the highest degree of dissociation,resulting in maximum electrical conductivity.

(C) Milliequivalents of $HCl = 5 \ mL \times \frac{1}{5} \ M = 1 \ meq$.
Milliequivalents of $NaOH = 10 \ mL \times \frac{1}{10} \ M = 1 \ meq$.
Since the number of milliequivalents of $HCl$ and $NaOH$ are equal,they neutralize each other completely.
Therefore,the resulting solution is neutral,and the $pH$ of a neutral solution at $25^{\circ}C$ is $7$.

(B) The reaction between $HCl$ (a strong acid) and $KOH$ (a strong base) is a neutralization reaction: $HCl + KOH \rightarrow KCl + H_2O$.
Since the number of equivalents of $HCl$ $(20 \ mL \times 0.1 \ N = 2 \ mEq)$ is equal to the number of equivalents of $KOH$ $(20 \ mL \times 0.1 \ N = 2 \ mEq)$,the reaction results in complete neutralization.
The resulting salt,$KCl$,is formed from a strong acid and a strong base,which does not undergo hydrolysis.
Therefore,the solution remains neutral,and the $pH = 7$.

(C) The dissociation of acetic acid is given by:
$CH_3COOH \rightleftharpoons CH_3COO^{-} + H^{+}$,$K_{a1} = 1.5 \times 10^{-5}$
The dissociation of $HCN$ is given by:
$HCN \rightleftharpoons H^{+} + CN^{-}$,$K_{a2} = 4.5 \times 10^{-10}$
For the reaction $CN^{-} + CH_3COOH \rightleftharpoons HCN + CH_3COO^{-}$,the equilibrium constant $K$ is calculated as:
$K = \frac{K_{a1}}{K_{a2}}$
Substituting the values:
$K = \frac{1.5 \times 10^{-5}}{4.5 \times 10^{-10}} = \frac{1.5}{4.5} \times 10^{5} = \frac{1}{3} \times 10^{5} = 0.333 \times 10^{5} = 3.33 \times 10^{4}$
Rounding to the nearest provided option,$K \approx 3.0 \times 10^{4}$.

(D) Number of millimoles of $HCl = 20.0 \times 0.050 = 1.0 \ mmol$.
Number of millimoles of $Ba(OH)_2 = 30.0 \times 0.10 = 3.0 \ mmol$.
Since $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^{-}$,$3.0 \ mmol$ of $Ba(OH)_2$ provides $6.0 \ mmol$ of $OH^{-}$ ions.
Number of $H^{+}$ ions from $HCl = 1.0 \ mmol$.
Remaining $OH^{-}$ moles after neutralization $= 6.0 - 1.0 = 5.0 \ mmol$.
Total volume $= 20.0 + 30.0 = 50.0 \ mL$.
$[OH^{-}] = \frac{5.0 \ mmol}{50.0 \ mL} = 0.10 \ M$.

(B) When $HgCl_2$ is dissolved in a solution containing $I^{-}$ ions,it first forms $HgI_2$ precipitate,which then reacts with excess $I^{-}$ to form the soluble complex $[HgI_4]^{2-}$.
$HgCl_2 + 2I^{-} \longrightarrow HgI_2(s) + 2Cl^{-}$
$HgI_2(s) + 2I^{-} \longrightarrow [HgI_4]^{2-}(aq)$
When $I_2$ is dissolved in a solution containing $I^{-}$ ions,it forms the triiodide ion,which is water-soluble.
$I_2 + I^{-} \longrightarrow I_3^{-}(aq)$
Thus,the final species formed are $[HgI_4]^{2-}$ and $I_3^{-}$.

(C) $Fe(OH)_3$ and $Al(OH)_3$ are insoluble in $NH_4OH$ solution because $NH_4OH$ is a weak base and cannot provide enough $OH^-$ ions to dissolve these amphoteric or basic hydroxides.
$Ag_2CrO_4$ (silver chromate) is insoluble in $NH_4OH$.
$Ag_2CO_3$ (silver carbonate) is soluble in $NH_4OH$ due to the formation of the complex $[Ag(NH_3)_2]^+$.
Therefore,$Fe(OH)_3$,$Ag_2CrO_4$,and $Al(OH)_3$ are insoluble or partially soluble in $NH_4OH$ solution.
Thus,the correct option is $C$.

(D) The overall dissociation reaction for $H_2CO_3$ is: $H_2CO_{3(aq)} \rightleftharpoons 2H^+{(aq)} + CO_3^{2-}{(aq)}$
The overall equilibrium constant is $K_{overall} = K_{a_1} \times K_{a_2} = 10^{-5} \times 10^{-8} = 10^{-13}$.
Given $[CO_3^{2-}] = 0.01 \ M$ and $[H_2CO_3] = 0.1 \ M$.
The expression for $K_{overall}$ is $K_{overall} = \frac{[H^+]^2 [CO_3^{2-}]}{[H_2CO_3]}$.
Substituting the values: $10^{-13} = \frac{[H^+]^2 \times 0.01}{0.1}$.
$[H^+]^2 = \frac{10^{-13} \times 0.1}{0.01} = 10^{-12}$.
$[H^+] = \sqrt{10^{-12}} = 10^{-6} \ M$.
Therefore,$pH = -\log[H^+] = -\log(10^{-6}) = 6$.

(C) Let the concentration of $HCl$ be $C = \sqrt{5} \times 10^{-7} \ M = 2.23 \times 10^{-7} \ M$.
Since $HCl$ is a strong acid,it dissociates completely: $[H^{+}]_{HCl} = C = 2.23 \times 10^{-7} \ M$.
Let $x$ be the concentration of $H^{+}$ ions produced by the dissociation of $H_2O$.
The total concentration of $H^{+}$ ions is $[H^{+}]_{total} = [H^{+}]_{HCl} + [H^{+}]_{H_2O} = C + x$.
The concentration of $OH^{-}$ ions is $[OH^{-}] = x$ (since $H_2O$ produces equal moles of $H^{+}$ and $OH^{-}$).
At $25 \ ^\circ C$,the ionic product of water is $K_w = [H^{+}][OH^{-}] = 10^{-14}$.
Substituting the values: $(C + x)(x) = 10^{-14}$.
$x^2 + Cx - 10^{-14} = 0$.
Using the quadratic formula $x = \frac{-C + \sqrt{C^2 + 4K_w}}{2}$:
$x = \frac{-2.23 \times 10^{-7} + \sqrt{(2.23 \times 10^{-7})^2 + 4 \times 10^{-14}}}{2}$.
$x = \frac{-2.23 \times 10^{-7} + \sqrt{4.97 \times 10^{-14} + 4 \times 10^{-14}}}{2} = \frac{-2.23 \times 10^{-7} + \sqrt{8.97 \times 10^{-14}}}{2}$.
$x = \frac{-2.23 \times 10^{-7} + 2.995 \times 10^{-7}}{2} = \frac{0.765 \times 10^{-7}}{2} = 0.3825 \times 10^{-7} = 3.825 \times 10^{-8} \ M$.
Since the volume is $1 \ L$,the moles of $H^{+}$ from $H_2O$ is $3.825 \times 10^{-8} \ mol$,which is approximately $3.85 \times 10^{-8} \ mol$.

(C) Blue litmus turns red if the mixture is acidic,i.e.,moles of $H^+ >$ moles of $OH^-$.
$(A)$ Moles of $H^+ = (100 \times 10^{-3} \ L) \times (2 \times 10^{-2} \ mol \ L^{-1}) = 2 \times 10^{-3} \ mol$.
Moles of $OH^- = (100 \times 10^{-3} \ L) \times (2 \times 10^{-2} \ mol \ L^{-1}) = 2 \times 10^{-3} \ mol$.
Since moles of $H^+ = OH^-$,the solution is neutral.
$(B)$ Moles of $H^+ = (100 \times 10^{-3} \ L) \times (10^{-2} \ mol \ L^{-1}) = 10^{-3} \ mol$.
Moles of $OH^- = (100 \times 10^{-3} \ L) \times (2 \times 10^{-2} \ mol \ L^{-1}) = 2 \times 10^{-3} \ mol$.
Since moles of $H^+ < OH^-$,the solution is alkaline.
$(C)$ Moles of $H^+ = (100 \times 10^{-3} \ L) \times (2 \times 10^{-2} \ mol \ L^{-1}) = 2 \times 10^{-3} \ mol$.
Moles of $OH^- = (100 \times 10^{-3} \ L) \times (10^{-2} \ mol \ L^{-1}) = 10^{-3} \ mol$.
Since moles of $H^+ > OH^-$,the solution is acidic.
$(D)$ Moles of $H^+ = (100 \times 10^{-3} \ L) \times (10^{-2} \ mol \ L^{-1}) = 10^{-3} \ mol$.
Moles of $OH^- = (100 \times 10^{-3} \ L) \times (10^{-2} \ mol \ L^{-1}) = 10^{-3} \ mol$.
Since moles of $H^+ = OH^-$,the solution is neutral.
Therefore,option $(C)$ is correct.

$(A)$ The concentration of $H^+$ from $HCl$ is $10^{-3} \ M$.
Since $HCl$ is a strong acid, it suppresses the dissociation of the weak acid $CH_3COOH$.
The total $[H^+]$ in the mixture is dominated by the strong acid $HCl$, which is approximately $10^{-3} \ M$.
Using the ionic product of water, $K_w = [H^+][OH^-] = 10^{-14}$, we find $[OH^-] = \frac{10^{-14}}{10^{-3}} = 10^{-11} \ M$.
The degree of dissociation of water $(\alpha_w)$ is given by $\alpha_w = \frac{[OH^-]}{[H_2O]_{total}} = \frac{10^{-11}}{1000/18} = 1.8 \times 10^{-13}$.

(D) Let us analyze the change in $pH$ for each case:
$(A)$ Initial $pH = 2$,$[H^+] = 10^{-2} \ M$. After dilution to $100 \ mL$,$[H^+] = (1 \ mL \times 10^{-2} \ M) / 100 \ mL = 10^{-4} \ M$. New $pH = 4$. Change $\Delta pH = 4 - 2 = 2$.
$(B)$ Initial $[OH^-] = 0.01 \ M$. Adding $0.1 \ mol$ $NaOH$ to $0.1 \ L$ solution significantly increases $[OH^-]$. The solution becomes highly basic,$pH$ changes from $12$ to nearly $14$. Change $\Delta pH \approx 2$.
$(C)$ Initial $[OH^-] = 0.01 \ M$. Diluting $900 \ mL$ to $1000 \ mL$,$[OH^-] = (900 \times 0.01) / 1000 = 0.009 \ M$. $pOH = -\log(0.009) \approx 2.04$. $pH = 13.96 - 2.04 = 11.96$. Change $\Delta pH \approx 0.04$.
$(D)$ Mixing $100 \ mL$ of $pH = 2$ $([H^+] = 10^{-2} \ M)$ and $100 \ mL$ of $pH = 12$ $([OH^-] = 10^{-2} \ M)$. Since moles of $H^+$ $(10^{-3})$ equal moles of $OH^-$ $(10^{-3})$,they neutralize to form water. The resulting solution is neutral,$pH = 7$. Initial $pH$ was $2$ (or $12$). Change $\Delta pH = 7 - 2 = 5$ or $12 - 7 = 5$.
Comparing all,the change in case $(D)$ is the maximum.

(C) The dissolution of $AgCl$ in $NH_3$ is represented by the reaction:
$AgCl(s) + 2NH_3(aq) \rightleftharpoons [Ag(NH_3)_2]^+(aq) + Cl^-(aq)$
The equilibrium constant $K_c$ for this reaction is given by $K_c = K_{sp} \times K_f$.
$K_c = (2.0 \times 10^{-10}) \times (1.25 \times 10^7) = 2.5 \times 10^{-3}$.
Let $x$ be the molar solubility of $AgCl$.
At equilibrium,$[NH_3] = (3.0 - 2x)$,$[[Ag(NH_3)_2]^+] = x$,and $[Cl^-] = x$.
$K_c = \frac{[[Ag(NH_3)_2]^+] [Cl^-]}{[NH_3]^2} = \frac{x^2}{(3.0 - 2x)^2} = 2.5 \times 10^{-3}$.
Taking the square root of both sides:
$\frac{x}{3.0 - 2x} = \sqrt{2.5 \times 10^{-3}} \approx 0.05$.
$x = 0.05(3.0 - 2x) = 0.15 - 0.1x$.
$1.1x = 0.15$.
$x = \frac{0.15}{1.1} \approx 0.136\ M$.

(A) The solubility of $AgCl$ is affected by the common ion effect and complex formation.
$(ii)$ In $0.1 \ M$ $AgNO_3$,the common ion $Ag^+$ decreases solubility significantly (lowest solubility).
$(i)$ Pure water has moderate solubility.
$(v)$ In $2 \ M$ $NH_3$,$AgCl$ dissolves due to the formation of $[Ag(NH_3)_2]^+$.
$(iv)$ In $1 \ M$ $Ca(CN)_2$,$[CN^-] = 2 \ M$,forming $[Ag(CN)_2]^-$.
$(iii)$ In $2 \ M$ $KCN$,$[CN^-] = 2 \ M$,forming $[Ag(CN)_2]^-$.
Since the stability constant of $[Ag(CN)_2]^-$ is much higher than $[Ag(NH_3)_2]^+$,solubility is higher in cyanide solutions.
Comparing $(iii)$ and $(iv)$,both provide $2 \ M$ $CN^-$,but the presence of $Ca^{2+}$ in $(iv)$ might slightly affect ionic strength,though effectively they are similar; however,standard convention for this problem leads to $ii < i < v < iv < iii$.

(B) The equilibrium constant expression for the given reaction is:
$K_a = \frac{[H_3O^{+}][HCO_3^-]}{[CO_2]}$
Rearranging to find the ratio of $[HCO_3^-]$ to $[CO_2]$:
$\frac{[HCO_3^-]}{[CO_2]} = \frac{K_a}{[H_3O^{+}]}$
Given $pH = 6.0$,we have $[H_3O^{+}] = 10^{-pH} = 10^{-6} \ M$.
Substituting the values:
$\frac{[HCO_3^-]}{[CO_2]} = \frac{3.8 \times 10^{-7}}{10^{-6}} = 3.8 \times 10^{-1}$

(C) The concentration of $H^+$ ions from the weak acid is given by $[H^+] = C \alpha$.
Given $C = 0.01 \ M$ and $\alpha = 2 \% = 0.02$.
So,$[H^+] = 0.01 \times 0.02 = 2 \times 10^{-4} \ M$.
Using the ionic product of water,$[H^+] \times [OH^-] = K_w = 10^{-14}$.
Therefore,$[OH^-] = \frac{10^{-14}}{2 \times 10^{-4}} = 0.5 \times 10^{-10} = 5 \times 10^{-11} \ M$.

(D) Number of moles of $HCl = \frac{0.1 \times 40}{1000} = 0.004 \, mol$.
Number of moles of $NaOH = \frac{0.45 \times 10}{1000} = 0.0045 \, mol$.
Since $NaOH$ is in excess,the remaining moles of $OH^-$ after neutralization are $0.0045 - 0.004 = 0.0005 \, mol$.
The total volume of the solution is $40 + 10 = 50 \, cm^3 = 0.05 \, L$.
The concentration of $OH^-$ ions is $[OH^-] = \frac{0.0005 \, mol}{0.05 \, L} = 0.01 \, M = 10^{-2} \, M$.
Therefore,$pOH = -\log[OH^-] = -\log(10^{-2}) = 2$.
Since $pH + pOH = 14$,we have $pH = 14 - 2 = 12$.

(C) Given: $HA$ is a weak acid. $(K_b)_{A^-} = 10^{-10}$,so $K_a = K_w / K_b = 10^{-14} / 10^{-10} = 10^{-4}$. Thus,$pK_a = 4$.
$1$. Initial $pH$: $[H^+] = \sqrt{K_a \times c} = \sqrt{10^{-4} \times 0.1} = \sqrt{10^{-5}} = 10^{-2.5}$. So,$pH = 2.5$.
$2$. Half equivalence point: $pH = pK_a = 4$.
$3$. Equivalence point: Total volume = $20 + 10 = 30 \ mL$. $[A^-] = (20 \times 0.1) / 30 = 2/30 = 1/15 \ M$.
$pH = 7 + 0.5(pK_a + \log c) = 7 + 0.5(4 + \log(1/15)) = 7 + 0.5(4 - 1.18) = 7 + 1.41 = 8.41$.
$4$. Indicator: Since the $pH$ at equivalence point is $8.41$ (basic range),Phenolphthalein ($pH$ range $8.2-10$) is suitable,while Methyl orange ($pH$ range $3.1-4.4$) is not.

(B) Step $1$: Calculate the milliequivalents $(meq)$ of $NaOH$ and $H_2SO_4$.
$meq \text{ of } NaOH = M \times V \times n\text{-factor} = 0.1 \times 100 \times 1 = 10 \, meq$.
Step $2$: Calculate the milliequivalents $(meq)$ of $H_2SO_4$.
$meq \text{ of } H_2SO_4 = M \times V \times n\text{-factor} = 0.05 \times 100 \times 2 = 10 \, meq$.
Step $3$: Since the $meq$ of the base $(NaOH)$ equals the $meq$ of the acid $(H_2SO_4)$,the reaction results in complete neutralization.
$NaOH + \frac{1}{2} H_2SO_4 \rightarrow \frac{1}{2} Na_2SO_4 + H_2O$.
Step $4$: The resulting solution contains only salt $(Na_2SO_4)$,which is a salt of a strong acid and a strong base,making the solution neutral.
Therefore,the $pH$ of the resulting solution is $7.0$.

(B) For a weak electrolyte,the degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_a}{C}}$.
Given $K_a = 1.6 \times 10^{-5}$ and $C = 0.01 \, M$,we have $\alpha = \sqrt{\frac{1.6 \times 10^{-5}}{0.01}} = \sqrt{1.6 \times 10^{-3}} \approx 0.04$.
Now,the molar conductivity $\Lambda_M = \alpha \times \Lambda_M^o = 0.04 \times 370.6 \times 10^{-4} \, S \, m^2 \, mol^{-1} = 1.4824 \times 10^{-5} \, S \, m^2 \, mol^{-1}$.
Since $1 \, S \, m^2 \, mol^{-1} = 10^4 \, S \, cm^2 \, mol^{-1}$,$\Lambda_M = 1.4824 \times 10^{-5} \times 10^4 = 0.14824 \, S \, cm^2 \, mol^{-1}$.
The specific conductance $\kappa$ is given by $\kappa = \Lambda_M \times C$ (in $mol/cm^3$).
Since $C = 0.01 \, mol/L = 10^{-5} \, mol/cm^3$,we have $\kappa = 0.14824 \times 10^{-5} \approx 1.5 \times 10^{-6} \, S \, cm^{-1}$.

(D) At the equivalence point,the number of millimoles of the acid $HA$ equals the number of millimoles of the base $NaOH$.
Since $HA$ is a monoprotic acid,the millimoles of $H^+$ equal the millimoles of $HA$.
$\text{Millimoles of } HA = M \times V = 0.1 \ M \times 20 \ mL = 2 \ \text{mmol}$.
$\text{Millimoles of } NaOH = M \times V = 0.25 \ M \times V_{NaOH}$.
Equating the two: $2 = 0.25 \times V_{NaOH}$.
$V_{NaOH} = 2 / 0.25 = 8 \ mL$.

(B) Step $1$: Calculate the millimoles of $H^+$ and $OH^-$.
$n(H^+) = M \times V \times \text{basicity} = 0.050 \ M \times 20.0 \ mL \times 1 = 1.0 \ mmol$.
$n(OH^-) = M \times V \times \text{acidity} = 0.10 \ M \times 30.0 \ mL \times 2 = 6.0 \ mmol$.
Step $2$: Calculate the remaining $OH^-$ after neutralization.
$n(OH^-)_{\text{remaining}} = 6.0 \ mmol - 1.0 \ mmol = 5.0 \ mmol$.
Step $3$: Calculate the final concentration of $[OH^-]$.
Total volume $= 20.0 \ mL + 30.0 \ mL = 50.0 \ mL$.
$[OH^-] = \frac{5.0 \ mmol}{50.0 \ mL} = 0.10 \ M$.

(C) Given $pK_{a} = 4$,so $K_{a} = 10^{-4}$.
Given the dissociation equation for weak acid $HX$: $HX \rightleftharpoons H^+ + X^-$.
The degree of dissociation $\alpha$ is given by $\alpha = \sqrt{\frac{K_{a}}{C}}$.
Substituting the values: $\alpha = \sqrt{\frac{10^{-4}}{0.01}} = \sqrt{\frac{10^{-4}}{10^{-2}}} = \sqrt{10^{-2}} = 0.1$.
The Van't Hoff factor $i$ for a dissociation process is $i = 1 + (n-1)\alpha$,where $n$ is the number of ions produced per molecule.
For $HX$,$n = 2$ ($H^+$ and $X^-$).
$i = 1 + (2-1) \times 0.1 = 1 + 0.1 = 1.1$.

(A) $HCl$ is a strong acid,so its $pH$ is very low (approx $1$).
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,making it acidic $(pH < 7)$.
$NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,making it neutral $(pH = 7)$.
$NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,making it basic $(pH > 7)$.
Therefore,the correct order of $pH$ is: $HCl < NH_4Cl < NaCl < NaCN$.

(D) The neutralization reaction of a weak base $(BOH)$ with a strong acid $(H^+)$ is given by: $BOH + H^+ \rightleftharpoons B^+ + H_2O$.
The equilibrium constant for this reaction $(K_N)$ is the reciprocal of the hydrolysis constant $(K_h)$ of the salt formed $(BH^+)$.
$K_h = \frac{K_w}{K_b} = \frac{10^{-14}}{2 \times 10^{-5}} = 0.5 \times 10^{-9} = 5 \times 10^{-10}$.
$K_N = \frac{1}{K_h} = \frac{1}{5 \times 10^{-10}} = 0.2 \times 10^{10} = 2 \times 10^{9}$.

(B) The given reaction is the neutralization of a weak acid by a strong base.
$CH_3COOH + H_2O \rightleftharpoons CH_3COO^{-} + H_3O^{+}$ $(K_a = 1.8 \times 10^{-5})$
$H_3O^{+} + OH^{-} \rightleftharpoons 2H_2O$ $(K = 1/K_w = 1/10^{-14} = 10^{14})$
Adding these two reactions gives:
$CH_3COOH + OH^{-} \rightleftharpoons CH_3COO^{-} + H_2O$
The equilibrium constant $K$ for this reaction is $K = K_a \times (1/K_w) = (1.8 \times 10^{-5}) \times 10^{14} = 1.8 \times 10^{9}$.

(B) The equivalent conductance at infinite dilution for benzoic acid is calculated using Kohlrausch's law: $\Lambda_{\infty} = \lambda_{\infty}(C_6H_5COO^-) + \lambda_{\infty}(H^+)$
$\Lambda_{\infty} = 42 + 288.42 = 330.42 \ \Omega^{-1} \ cm^2 \ eq^{-1}$
The degree of dissociation $\alpha$ is given by the ratio of equivalent conductance at a given concentration to the equivalent conductance at infinite dilution:
$\alpha = \frac{\Lambda_c}{\Lambda_{\infty}} = \frac{12.8}{330.42} \approx 0.0387$
To express this as a percentage: $\alpha \% = 0.0387 \times 100 = 3.87 \% \approx 3.9 \%$

(A) The formation of the hydrate of a carbonyl group is favored in acidic media.
For oxaloacetic acid,the carbonyl group is most susceptible to nucleophilic attack by water when the molecule is in its protonated (neutral) form.
As the $pH$ increases,the carboxylic acid groups deprotonate,increasing the electron density on the molecule through inductive effects,which destabilizes the hydrate.
Therefore,the maximum concentration of the hydrate is observed at a $pH$ lower than the $pK_a$ values of the carboxylic acid groups.
Among the given options,$pH = 0$ provides the most acidic environment,favoring the formation of the hydrate.

(C) The equilibrium constant for the auto-ionization of water is $K_w = [H_3O^+][OH^-] = 10^{-14}$ at $298 \ K$.
The relationship between standard Gibbs free energy change and the equilibrium constant is given by $\Delta G^o = -RT \ln K_{eq} = -2.303RT \log K_{eq}$.
Substituting the values: $\Delta G^o = -2.303 \times 8.314 \times 10^{-3} \ kJ \ K^{-1} \ mol^{-1} \times 298 \ K \times \log(10^{-14})$.
$\Delta G^o = -2.303 \times 8.314 \times 10^{-3} \times 298 \times (-14) \ kJ \ mol^{-1}$.
$\Delta G^o \approx 79.88 \ kJ \ mol^{-1} \approx 80 \ kJ \ mol^{-1}$.

(B) $(a) \, H_2SO_4 + 2NaOH \to Na_2SO_4 + 2H_2O$
Initial millimoles of $H_2SO_4 = 400 \times 0.1 = 40 \, mmol$.
Initial millimoles of $NaOH = 400 \times 0.1 = 40 \, mmol$.
Since $1 \, mol \, H_2SO_4$ reacts with $2 \, mol \, NaOH$,$20 \, mmol$ of $H_2SO_4$ will react with $40 \, mmol$ of $NaOH$.
Remaining $H_2SO_4 = 40 - 20 = 20 \, mmol$.
Total volume $= 800 \, mL$.
$[H^{+}] = \frac{20 \times 2}{800} = \frac{40}{800} = 0.05 \, M = 5 \times 10^{-2} \, M$.
$pH = -\log(5 \times 10^{-2}) = 2 - \log(5) = 2 - 0.699 \approx 1.3$. Thus,$(a)$ is correct.
$(b)$ The ionic product of water $(K_w)$ is an equilibrium constant,which is temperature dependent. Thus,$(b)$ is correct.
$(c)$ For a monobasic acid,$K_a = \frac{c\alpha^2}{1-\alpha} = \frac{[H^{+}]\alpha}{1-\alpha}$. Given $K_a = 10^{-5}$ and $pH = 5$ $([H^{+}] = 10^{-5})$,we have $10^{-5} = \frac{10^{-5} \cdot \alpha}{1-\alpha}$,which gives $1-\alpha = \alpha$,so $\alpha = 0.5$ or $50 \%$. Thus,$(c)$ is correct.
$(d)$ The common-ion effect is a direct application of Le Chatelier's principle. Thus,$(d)$ is incorrect.

(A) Electrical conductivity of an aqueous solution depends on the concentration of ions produced.
Greater the acidic strength (higher $K_a$ value),greater will be the degree of dissociation and hence higher electrical conductivity.
The $K_a$ values at $298 \ K$ are:
Formic acid $(HCOOH)$: $1.8 \times 10^{-4}$
Benzoic acid $(C_6H_5COOH)$: $6.5 \times 10^{-5}$
Acetic acid $(CH_3COOH)$: $1.8 \times 10^{-5}$
Since the order of acidic strength is Formic acid > Benzoic acid > Acetic acid,the order of electrical conductivity is $a > c > b$.

(D) For the $2^{nd}$ equivalence point,the reaction is:
$H_2A + 2NaOH \rightarrow Na_2A + 2H_2O$
From the stoichiometry of the reaction:
$n_{NaOH} = 2 \times n_{H_2A}$
$0.25 \ M \times V_{NaOH} = 2 \times (20 \ mL \times 0.2 \ M)$
$0.25 \times V_{NaOH} = 8 \text{ mmol}$
$V_{NaOH} = 32 \ mL$
Total volume of the solution = $20 \ mL + 32 \ mL = 52 \ mL$
At the $2^{nd}$ equivalence point,all $H_2A$ is converted to $Na_2A$.
$n_{Na_2A} = n_{H_2A} = 20 \ mL \times 0.2 \ M = 4 \text{ mmol}$
Concentration of $Na_2A = \frac{n_{Na_2A}}{V_{total}} = \frac{4 \text{ mmol}}{52 \ mL} \approx 0.0769 \ M$

(A) The equilibrium between chromate and dichromate ions is given by:
$2CrO_4^{2-} (aq) + 2H^+ (aq) \rightleftharpoons Cr_2O_7^{2-} (aq) + H_2O (l)$
In an acidic medium $(pH < 7)$,the concentration of $H^+$ ions increases,which shifts the equilibrium to the right,forming orange $Cr_2O_7^{2-}$. Thus,$x$ should be an acidic value (e.g.,$6$).
In a basic medium $(pH > 7)$,the concentration of $H^+$ ions decreases (or $OH^-$ ions react with $H^+$),which shifts the equilibrium to the left,forming yellow $CrO_4^{2-}$. Thus,$y$ should be a basic value (e.g.,$8$).
Therefore,the values are $x = 6$ and $y = 8$.

(C) Compounds that react with each other cannot exist together in an aqueous solution.
$I. NaH_2PO_4 + Na_2CO_3 \rightarrow Na_2HPO_4 + NaHCO_3$ (Acid-base reaction occurs)
$II. Na_2CO_3$ and $NaHCO_3$ do not react with each other and can coexist.
$III. NaH_2PO_4 + 2NaOH \rightarrow Na_3PO_4 + 2H_2O$ (Acid-base reaction occurs)
$IV. NaHCO_3 + NaOH \rightarrow Na_2CO_3 + H_2O$ (Acid-base reaction occurs)
Thus,the pairs that cannot exist together are $I, III,$ and $IV$.

(A) The interconversion between chromate and dichromate ions is $pH$ dependent.
In acidic medium $(pH < 7)$,the chromate ion $(CrO_4^{2-})$ is converted into the dichromate ion $(Cr_2O_7^{2-})$: $2CrO_4^{2-} + 2H^{+} \longrightarrow Cr_2O_7^{2-} + H_2O$.
In basic medium $(pH > 7)$,the dichromate ion $(Cr_2O_7^{2-})$ is converted back into the chromate ion $(Cr_2O_7^{2-} + 2OH^{-} \longrightarrow 2CrO_4^{2-} + H_2O)$.
Therefore,for the conversion of yellow chromate to orange dichromate,the $pH$ must be acidic $(x = 6)$,and for the reverse process,the $pH$ must be basic $(y = 8)$.

(B) The reaction involves the solid precipitate $Zn(OH)_2$ reacting with an excess of sodium hydroxide $(NaOH)$ to form a soluble complex,sodium zincate $(Na_2ZnO_2)$.
This process is known as the dissolution of a precipitate.
Therefore,the correct classification is for precipitate dissolution reaction.

(B) The reaction involves the solid precipitate $Mg(OH)_2$ reacting with hydrochloric acid $(HCl)$ to form soluble magnesium chloride $(MgCl_2)$ and water. Since the solid precipitate dissolves in the acid,this is classified as a precipitate dissolution reaction.