Mix Examples-Ionic Equilibrium Questions in English

(D) The degree of ionization $(\alpha)$ depends on factors such as temperature,the nature of the solute,the nature of the solvent,and the concentration of the solution.
$Current$ is the flow of electric charge and does not influence the intrinsic equilibrium state or the degree of ionization of a substance in a solution.
Therefore,the correct option is $(D)$.

(B) solution of $NaHCO_3$ in water is alkaline in nature due to the hydrolysis of the $HCO_3^-$ ion.
The hydrolysis reaction is: $HCO_3^- + H_2O \rightleftharpoons H_2CO_3 + OH^-$.
Due to the presence of $OH^-$ ions,the solution is basic.
Methyl orange is a pH indicator that turns yellow in basic solutions (pH range $4.4 - 6.2$ is the transition range,but it appears yellow in alkaline media).

(A) The normality of $NaOH$ is $N_1 = 1 \times 1 = 1 \ N$.
The normality of $H_2SO_4$ is $N_2 = 2 \times 10 = 20 \ N$ (since $n$-factor is $2$).
Milliequivalents ($M$.eq.) of $NaOH = 1 \times 100 = 100$.
Milliequivalents ($M$.eq.) of $H_2SO_4 = 20 \times 10 = 200$.
Since $M$.eq. of acid $(200)$ > $M$.eq. of base $(100)$,the excess acid remains in the solution.
Therefore,the resulting mixture will be acidic.

(A) The stability of the complex $ZnX^{+}$ depends on the hard-soft acid-base $(HSAB)$ theory.
$Zn^{2+}$ is a borderline acid.
$F^{-}$ is a hard base,while $I^{-}$ is a soft base.
According to $HSAB$ theory,hard acids prefer to bind with hard bases.
Among the given options,$F^{-}$ is the hardest base,leading to the strongest electrostatic interaction with the $Zn^{2+}$ ion,resulting in the highest stability constant $(K_{eq})$ for the formation of $ZnF^{+}$.

(B) mixed salt is a salt that contains more than one type of cation or anion (excluding $H^+$ or $OH^-$ ions).
$NaKSO_4$ is a mixed salt because it contains two different cations,$Na^+$ and $K^+$,and one anion,$SO_4^{2-}$.
$NaHSO_4$ is an acid salt,$K_4[Fe(CN)_6]$ is a complex salt,and $Mg(OH)Cl$ is a basic salt.

(A) For a weak acid like $CH_3COOH$,the concentration of hydrogen ions $[H^+]$ is given by the product of the concentration of the acid $(c)$ and its degree of dissociation $(\alpha)$.
Given:
Concentration $c = 0.1 \ N = 0.1 \ M$ (since the valency factor of $CH_3COOH$ is $1$).
Degree of dissociation $\alpha = 30 \% = \frac{30}{100} = 0.3$.
Therefore,$[H^+] = c \times \alpha = 0.1 \times 0.3 = 0.03 \ M$.

(C) The reaction between a weak acid $(CH_3COOH)$ and a weak base $(NH_4OH)$ forms a salt $(CH_3COONH_4)$.
The $pH$ of a salt of a weak acid and a weak base is given by the formula:
$pH = \frac{1}{2} [pK_w + pK_a - pK_b]$
Given:
$pK_w = 14$
$pK_a = 5.0$
$pK_b = 5.0$
Substituting the values:
$pH = \frac{1}{2} [14 + 5.0 - 5.0]$
$pH = \frac{1}{2} [14] = 7$

(D) Step $1$: Calculate the milliequivalents of $HCl$ and $NaOH$.
$mEq \text{ of } HCl = 20 \ mL \times 0.5 \ N = 10 \ mEq$.
$mEq \text{ of } NaOH = 35 \ mL \times 0.1 \ N = 3.5 \ mEq$.
Step $2$: Determine the nature of the resulting solution.
Since $mEq \text{ of } HCl > mEq \text{ of } NaOH$,the solution is acidic.
$mEq \text{ of } H^+ \text{ remaining} = 10 - 3.5 = 6.5 \ mEq$.
Step $3$: Analyze the indicators.
Phenolphthalein is colorless in acidic medium and pink in basic medium. Methyl orange is red in acidic medium $(pH < 3.1)$ and yellow in basic medium $(pH > 4.4)$.
Calculating the final concentration of $H^+$:
$[H^+] = \frac{6.5 \ mEq}{55 \ mL} \approx 0.118 \ N$.
$pH = -\log(0.118) \approx 0.93$.
At $pH \approx 0.93$,methyl orange turns red. Therefore,the correct option is $D$.

(C) $1$. Calculate the moles of $Na_2CO_3 \cdot H_2O$: The molar mass is $124 \ g/mol$. Moles $= 0.62 \ g / 124 \ g/mol = 0.005 \ mol$.
$2$. Calculate the equivalents of $(NH_4)_2SO_4$: Equivalents $= N \times V(L) = 0.1 \ N \times 0.1 \ L = 0.01 \ eq$.
$3$. The reaction is: $Na_2CO_3 + (NH_4)_2SO_4 \rightarrow (NH_4)_2CO_3 + Na_2SO_4$.
$4$. $1 \ mol$ of $Na_2CO_3$ reacts with $1 \ mol$ of $(NH_4)_2SO_4$. Since $Na_2CO_3$ is a salt of a strong base $(NaOH)$ and a weak acid $(H_2CO_3)$,it provides $2$ equivalents per mole. Thus,$0.005 \ mol$ of $Na_2CO_3$ provides $0.01 \ eq$.
$5$. Since $0.01 \ eq$ of $Na_2CO_3$ reacts with $0.01 \ eq$ of $(NH_4)_2SO_4$,the resulting solution contains $(NH_4)_2CO_3$,which is a salt of a weak acid and a weak base. The pH of such a salt is given by $pH = 7 + 0.5(pK_a - pK_b)$. For $(NH_4)_2CO_3$,$pK_a$ of $HCO_3^-$ is $\approx 10.3$ and $pK_b$ of $NH_4OH$ is $\approx 4.75$,resulting in a slightly basic solution.

(B) Step $1$: Calculate the millimoles of $NaOH$.
$n(NaOH) = M \times V(mL) = \frac{1}{10} \times 100 = 10 \, mmol$.
Step $2$: Calculate the millimoles of $HCl$.
$n(HCl) = M \times V(mL) = \frac{1}{5} \times 50 = 10 \, mmol$.
Step $3$: Since $NaOH$ and $HCl$ react in a $1:1$ molar ratio $(NaOH + HCl \rightarrow NaCl + H_2O)$,the $10 \, mmol$ of $NaOH$ will exactly neutralize $10 \, mmol$ of $HCl$.
Step $4$: As both reactants are completely consumed,the resulting solution contains only $NaCl$ (a salt of a strong acid and a strong base),which is neutral.
Therefore,the $pH$ of the resulting solution is $7$.

(C) For $NH_4OH$,the concentration $C = 0.1 \ M$ (decinormal).
Given the degree of ionization $\alpha = 20\% = 0.2$.
The concentration of hydroxide ions is $[OH^-] = C \times \alpha = 0.1 \times 0.2 = 0.02 \ M = 2 \times 10^{-2} \ M$.
Calculate $pOH$ as $pOH = -\log[OH^-] = -\log(2 \times 10^{-2}) = 2 - \log 2 = 2 - 0.301 = 1.699 \approx 1.7$.
Finally,calculate $pH$ using the relation $pH + pOH = 14$.
$pH = 14 - 1.7 = 12.3$.

(C) $1$. Calculate the milliequivalents of $NaOH$: $n_{NaOH} = 10 \ mL \times 0.1 \ N = 1 \ meq$.
$2$. Calculate the milliequivalents of $H_2SO_4$: $n_{H_2SO_4} = 10 \ mL \times 0.05 \ N = 0.5 \ meq$.
$3$. Since $NaOH$ is a strong base and $H_2SO_4$ is a strong acid,they neutralize each other.
$4$. Remaining $NaOH = 1 \ meq - 0.5 \ meq = 0.5 \ meq$.
$5$. Total volume of the solution = $10 \ mL + 10 \ mL = 20 \ mL$.
$6$. Concentration of $[OH^-] = \frac{0.5 \ meq}{20 \ mL} = 0.025 \ N = 2.5 \times 10^{-2} \ M$.
$7$. $pOH = -\log(2.5 \times 10^{-2}) = 2 - \log(2.5) = 2 - 0.3979 = 1.6021$.
$8$. $pH = 14 - pOH = 14 - 1.6021 = 12.3979$.
$9$. Since $pH > 7$,the correct option is $C$.

(B, C) Statement $A$ is incorrect because for a very dilute acid solution, the contribution of $H^+$ ions from water cannot be neglected. The $pH$ of $1.0 \times 10^{-8} \ M$ $HCl$ is approximately $6.98$, not $8$.
Statement $B$ is correct because the conjugate base is formed by the removal of one proton $(H^+)$ from the acid: $H_2PO_4^- \rightarrow H^+ + HPO_4^{2-}$.
Statement $C$ is correct because the autoprotolysis of water is an endothermic process $(\Delta H > 0)$, so the equilibrium constant $K_w$ increases with an increase in temperature.
Statement $D$ is incorrect because at the half-neutralization point of a weak acid, the concentration of the acid equals the concentration of its conjugate base $([HA] = [A^-])$. According to the Henderson-Hasselbalch equation, $pH = pK_a + \log(\frac{[A^-]}{[HA]})$, which simplifies to $pH = pK_a$.

(D) For $(a)$,$HCl$ is a strong acid. $[H^+] = 0.01 \, M = 10^{-2} \, M$. Thus,$pH = -\log(10^{-2}) = 2$.
For $(b)$,$H_2SO_4$ is a strong acid. $[H^+] = 2 \times 0.01 \, M = 0.02 \, M$. Thus,$pH = -\log(0.02) \approx 1.7$.
For $(c)$,$moles \, of \, H_2SO_4 = 0.02 \, M \times 0.050 \, L = 0.001 \, mol$. $moles \, of \, H^+ = 2 \times 0.001 = 0.002 \, mol$.
$moles \, of \, NaOH = 0.02 \, M \times 0.050 \, L = 0.001 \, mol$. $moles \, of \, OH^- = 0.001 \, mol$.
$Net \, moles \, of \, H^+ = 0.002 - 0.001 = 0.001 \, mol$.
$Total \, volume = 50 \, mL + 50 \, mL = 100 \, mL = 0.1 \, L$.
$[H^+] = \frac{0.001 \, mol}{0.1 \, L} = 0.01 \, M = 10^{-2} \, M$.
Thus,$pH = -\log(10^{-2}) = 2$.
Since $(a)$ and $(c)$ both result in a $pH$ of $2$,the correct option is $(d)$.

(C) Statement $(C)$ is incorrect.
For a very dilute acid solution like $1 \times 10^{-8} \ M \ HCl$,the contribution of $H^+$ ions from water cannot be neglected.
The total $[H^+] = 10^{-8} + 10^{-7} \approx 1.1 \times 10^{-7} \ M$.
Therefore,$pH = -\log(1.1 \times 10^{-7}) \approx 6.96$,which is slightly acidic,not $8$ (which is basic).

(A) $HCl$ is a strong acid,so its $pH$ is the lowest $(pH \approx 1)$.
$NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,so it undergoes cationic hydrolysis,resulting in an acidic solution $(pH < 7)$.
$NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,so it is neutral $(pH = 7)$.
$NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,so it undergoes anionic hydrolysis,resulting in a basic solution $(pH > 7)$.
Therefore,the order of increasing $pH$ is: $HCl < NH_4Cl < NaCl < NaCN$.

(B) The reaction of a weak acid $(HA)$ with a strong base $(OH^-)$ is: $HA + OH^- \rightleftharpoons A^- + H_2O$.
This reaction is the reverse of the hydrolysis of the conjugate base $(A^-)$.
The equilibrium constant for this reaction is $K = \frac{1}{K_h}$,where $K_h$ is the hydrolysis constant of $A^-$.
We know that $K_h = \frac{K_w}{K_a}$.
Therefore,$K = \frac{K_a}{K_w}$.
Given $K_a = 1 \times 10^{-4}$ and $K_w = 1 \times 10^{-14}$.
$K = \frac{10^{-4}}{10^{-14}} = 10^{10}$.

(A) The density of water is $1 \ g/mL$,so $1 \ litre$ of water weighs $1000 \ g$.
The number of moles of water in $1 \ litre$ is $n = \frac{1000 \ g}{18 \ g/mol} = 55.56 \ mol$.
The concentration of $H^+$ ions is $[H^+] = 10^{-7} \ mol/L$.
The degree of ionization $(\alpha)$ is defined as the ratio of the number of moles ionized to the total number of moles.
$\alpha = \frac{[H^+]}{[H_2O]} = \frac{10^{-7}}{55.56} = 1.8 \times 10^{-9}$.
To express this as a percentage: $\alpha \% = (1.8 \times 10^{-9}) \times 100 = 1.8 \times 10^{-7} \%$.

(D) The degree of dissociation $\alpha = \frac{1.9 \times 10^{-7}}{100} = 1.9 \times 10^{-9}$.
The molar concentration of water $C = \frac{1000 \, g/L}{18 \, g/mol} = 55.55 \, M$.
The ionisation constant $K_i$ is given by $K_i = C \alpha^2$.
$K_i = 55.55 \times (1.9 \times 10^{-9})^2$.
$K_i = 55.55 \times 3.61 \times 10^{-18} \approx 2.0 \times 10^{-16}$.

(A) Given: Concentration of solution $= 0.1 \ M$.
Degree of ionisation $(\alpha) = 2\% = 0.02$.
Ionic product of water $(K_w) = 1 \times 10^{-14}$.
Concentration of $[H^{+}] = C \times \alpha = 0.1 \times 0.02 = 2 \times 10^{-3} \ M$.
Concentration of $[OH^{-}] = \frac{K_w}{[H^{+}]} = \frac{1 \times 10^{-14}}{2 \times 10^{-3}} = 0.5 \times 10^{-11} = 5 \times 10^{-12} \ M$.

(A) The degree of ionization $\alpha$ is given as $1.34 \% = 0.0134$.
Given concentration $c = 0.1 \, M$.
The ionization constant $K_a$ for a weak acid is given by the formula $K_a = c \alpha^2$.
Substituting the values: $K_a = 0.1 \times (0.0134)^2$.
$K_a = 0.1 \times 0.00017956 = 1.7956 \times 10^{-5} \approx 1.8 \times 10^{-5}$.
Comparing this with the given options,the closest value is $1.82 \times 10^{-5}$.

(C) Initial concentration of $[H^{+}] = 1 \times 10^{-4} \ M$.
When the solution is diluted with an equal volume of water,the total volume becomes double,so the concentration of $[H^{+}]$ is halved:
$[H^{+}]_{new} = \frac{1 \times 10^{-4}}{2} = 0.5 \times 10^{-4} \ M$.
Using the ionic product of water,$K_w = [H^{+}][OH^{-}] = 1 \times 10^{-14}$.
$[OH^{-}] = \frac{K_w}{[H^{+}]_{new}} = \frac{1 \times 10^{-14}}{0.5 \times 10^{-4}} = 2 \times 10^{-10} \ mol \ dm^{-3}$.

(B) The dissociation of $Ba(OH)_2$ is given by: $Ba(OH)_2 \rightarrow Ba^{2+} + 2OH^{-}$.
Since $1 \ mol$ of $Ba(OH)_2$ produces $2 \ mol$ of $OH^{-}$ ions,the concentration of $OH^{-}$ ions is $[OH^{-}] = 2 \times [Ba(OH)_2] = 2 \times 0.005 \ M = 0.01 \ M$.
The number of moles of $OH^{-}$ ions is calculated as: $\text{Moles} = \text{Molarity} \times \text{Volume (in L)}$.
$\text{Moles of } OH^{-} = 0.01 \ M \times 0.3 \ L = 0.0030 \ mol$.

(A) The concentration of $H^{+}$ ions is given by $[H^{+}] = C \times \alpha$.
Given $C = 0.1 \ N$ and $\alpha = 0.1$,we have $[H^{+}] = 0.1 \times 0.1 = 10^{-2} \ M$.
At room temperature,the ionic product of water is $K_w = [H^{+}][OH^{-}] = 10^{-14}$.
Therefore,$[OH^{-}] = \frac{10^{-14}}{[H^{+}]} = \frac{10^{-14}}{10^{-2}} = 10^{-12} \ M$.

(C) To determine the concentration of $[H_3O^{+}]$,we analyze the nature of the solutes in $0.01 \ M$ aqueous solutions:
$1$. $NaNO_2$ is a salt of a weak acid $(HNO_2)$ and a strong base $(NaOH)$,making it basic ($pH > 7$,low $[H_3O^{+}]$).
$2$. $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,making it neutral ($pH = 7$,$[H_3O^{+}] = 10^{-7} \ M$).
$3$. $H_2S$ is a weak acid,which partially dissociates to produce $[H_3O^{+}]$.
$4$. $H_2SO_4$ is a strong acid,which completely dissociates to produce a high concentration of $[H_3O^{+}]$.
Thus,the increasing order of $[H_3O^{+}]$ is: $NaNO_2 < NaCl < H_2S < H_2SO_4$.

(D) When $NaCl$ is dissolved in water,the sodium ion becomes hydrated.
When sodium chloride dissolves in water,the sodium and chloride ions and the polar water molecules are strongly attracted to one another by ion-dipole interactions.
The solvent molecules (water in this case) surround the ions,removing them from the crystal lattice and forming the solution.
As the dissolving process proceeds,the individual ions are removed from the solid surface,becoming completely separate,hydrated species in the solution.

(B) The reaction between potassium chromate $(K_2CrO_4)$ and dilute nitric acid $(HNO_3)$ is an acid-base equilibrium reaction where the chromate ion $(CrO_4^{2-})$ is converted into the dichromate ion $(Cr_2O_7^{2-})$.
The balanced chemical equation is:
$2K_2CrO_4 + 2HNO_3 \rightarrow K_2Cr_2O_7 + 2KNO_3 + H_2O$
In ionic form:
$2CrO_4^{2-} + 2H^{+} \rightarrow Cr_2O_7^{2-} + H_2O$
This is not a redox reaction; the oxidation state of chromium remains $+6$ in both $CrO_4^{2-}$ and $Cr_2O_7^{2-}$.

(C) $Al(OH)_3$ is an amphoteric hydroxide,which dissolves in excess $NaOH$ to form soluble sodium aluminate $(NaAlO_2)$.
$Fe(OH)_3$ is a basic hydroxide and does not react with $NaOH$.
Therefore,adding $NaOH$ solution allows for the separation of $Fe(OH)_3$ (as a precipitate) from $Al(OH)_3$ (in solution).

(D) The number of equivalents of acid is $N_1 V_1 = 0.25 \times 20 = 5 \, \text{meq}$.
The number of equivalents of base is $N_2 V_2 = 0.2 \times 30 = 6 \, \text{meq}$.
Since the number of equivalents of base is greater than the number of equivalents of acid,the resulting solution will be basic.
The excess equivalents of base = $6 - 5 = 1 \, \text{meq}$.
The total volume of the mixture = $20 + 30 = 50 \, mL$.
The normality of the resulting solution = $\frac{\text{Excess equivalents}}{\text{Total volume}} = \frac{1 \, \text{meq}}{50 \, mL} = 0.02 \, N$.
Therefore,the resulting solution is $0.02 \, N$ basic.

(D) In the titration of a strong acid $(HCl)$ with a strong base $(NaOH)$,the initial $pH$ is very low due to the presence of $HCl$.
As the base is added,the $pH$ increases slowly at first.
Near the equivalence point,there is a sharp and sudden increase in $pH$ because the concentration of $H^+$ ions decreases rapidly.
After the equivalence point,the $pH$ becomes almost constant as it is determined by the excess of $NaOH$ added.
Graph $D$ correctly represents this characteristic sigmoidal curve for a strong acid-strong base titration.

(B) The number of equivalents of $HCl$ = $N_1 V_1 = 0.1 \times 100 = 10 \ mEq$.
The number of equivalents of $NaOH$ = $N_2 V_2 = 0.2 \times 100 = 20 \ mEq$.
Since $NaOH$ is in excess,the resulting solution will be basic.
Remaining equivalents of $NaOH$ = $20 - 10 = 10 \ mEq$.
Total volume of the mixture = $100 \ cm^3 + 100 \ cm^3 = 200 \ cm^3$.
Resulting normality $(N_{res})$ = $\frac{\text{Remaining equivalents}}{\text{Total volume}} = \frac{10}{200} = 0.05 \ N$.

(B) $NaOH$ is hygroscopic and deliquescent in nature. It absorbs moisture and $CO_2$ from the atmosphere. The absorption of water increases the total volume of the solution,which leads to a decrease in the molarity (strength) of the $NaOH$ solution.

(D) The correct option is $D$. The degree of ionization $(\alpha)$ of an electrolyte depends on several factors:
$1$. Nature of the electrolyte: Strong electrolytes ionize completely,while weak electrolytes ionize partially.
$2$. Nature of the solvent: Solvents with high dielectric constants facilitate greater ionization.
$3$. Temperature: Generally,an increase in temperature increases the degree of ionization.
$4$. Concentration: Dilution increases the degree of ionization.
Therefore,all the given factors influence the degree of ionization.

(C) The reaction of $ZnSO_4$ with $NaHCO_3$ results in the precipitation of normal zinc carbonate $(ZnCO_3)$.
The chemical equation is: $ZnSO_4 + 2NaHCO_3 \to ZnCO_3 + Na_2SO_4 + H_2O + CO_2$.
Using $Na_2CO_3$ would lead to the formation of basic zinc carbonate,while $NaHCO_3$ is used to obtain the normal carbonate.

(C) $CH_3COONa$ is a salt of a weak acid and a strong base,which undergoes anionic hydrolysis in water to form $CH_3COOH$ and $OH^-$.
$CH_3CONH_2$ (acetamide) undergoes hydrolysis in the presence of water (often catalyzed by acid or base) to form $CH_3COOH$ and $NH_3$.
Since both compounds undergo hydrolysis,the correct answer is $(c)$.

(A) The number of equivalents of acid $(N_1V_1)$ = $20 \, mL \times 0.25 \, N = 5 \, meq$.
The number of equivalents of base $(N_2V_2)$ = $30 \, mL \times 0.2 \, N = 6 \, meq$.
Since the number of equivalents of base is greater than the acid,the resulting solution will be basic.
Excess equivalents of base = $6 \, meq - 5 \, meq = 1 \, meq$.
Total volume of the mixture = $20 \, mL + 30 \, mL = 50 \, mL$.
Normality of the resulting solution = $\frac{\text{Excess equivalents}}{\text{Total volume}} = \frac{1 \, meq}{50 \, mL} = 0.02 \, N$.
Therefore,the solution is $0.02 \, N$ basic.

(A) In pure water,the dissociation equilibrium is $H_2O \rightleftharpoons H^{+} + OH^{-}$.
At $25 \, ^oC$,the concentration of $H^{+}$ ions is $[H^{+}] = 1 \times 10^{-7} \, M$ and the concentration of $OH^{-}$ ions is $[OH^{-}] = 1 \times 10^{-7} \, M$.
Therefore,the sum of the concentrations is $[H^{+}] + [OH^{-}] = 1 \times 10^{-7} + 1 \times 10^{-7} = 2 \times 10^{-7} \, M$.

(C) When equal volumes are mixed,the concentration of each species is halved.
$[Ca^{2+}] = \frac{0.004}{2} = 0.002 \, M = 2 \times 10^{-3} \, M$
$[SO_4^{2-}] = [SO_4^{2-}]_{CaSO_4} + [SO_4^{2-}]_{H_2SO_4}$
$[SO_4^{2-}] = \frac{0.004}{2} + \frac{0.002}{2} = 0.002 + 0.001 = 0.003 \, M = 3 \times 10^{-3} \, M$
Ionic product $Q_{sp} = [Ca^{2+}] [SO_4^{2-}] = (2 \times 10^{-3}) \times (3 \times 10^{-3}) = 6 \times 10^{-6}$.

(C) For a weak acid $HCN$,the dissociation equilibrium is $HCN \rightleftharpoons H^+ + CN^-$.
The dissociation constant $K_a$ is given by $K_a = \frac{[H^+][CN^-]}{[HCN]}$.
For a weak acid,$[H^+] = [CN^-] = C\alpha$ and $[HCN] = C(1-\alpha) \approx C$.
Thus,$K_a = C\alpha^2$,which implies $\alpha = \sqrt{\frac{K_a}{C}} = \frac{[H^+]}{C}$.
From $K_w = [H^+][OH^-]$,we have $[H^+] = \frac{K_w}{[OH^-]}$.
Substituting this into the expression for $\alpha$,we get $\alpha = \frac{K_a}{[H^+]} = \frac{K_a \times [OH^-]}{K_w}$.
Therefore,both statements $(A)$ and $(B)$ are correct.

(A) For the dissociation of water: $H_2O(l) \rightleftharpoons H^+(aq) + OH^-(aq)$.
The equilibrium constant $K_c$ is given by $K_c = \frac{[H^+][OH^-]}{[H_2O]}$.
At $25 \, ^\circ C$,the ionic product of water $K_w = [H^+][OH^-] = 1 \times 10^{-14}$.
The concentration of pure water $[H_2O] = \frac{1000 \, g/L}{18 \, g/mol} = 55.55 \, M$.
Therefore,the dissociation constant $K_a = \frac{K_w}{[H_2O]} = \frac{1 \times 10^{-14}}{55.55} = (55.55 \times 10^{14})^{-1}$.

(A) The hydrolysis constant $K_h$ for a salt of a weak acid and strong base is given by $K_h = \frac{K_w}{K_a}$.
For $NaA$: $K_h(A^-) = 10^{-8} = \frac{10^{-14}}{K_a(HA)}$,so $K_a(HA) = 10^{-6}$.
For $NaB$: $K_h(B^-) = 10^{-10} = \frac{10^{-14}}{K_a(HB)}$,so $K_a(HB) = 10^{-4}$.
For $HC$: $K_a(HC) = 10^{-5}$.
Comparing the dissociation constants: $K_a(HB) = 10^{-4} > K_a(HC) = 10^{-5} > K_a(HA) = 10^{-6}$.
Since acidic strength is directly proportional to $K_a$,the order is $HB > HC > HA$.

(A) The dissociation reaction for carbonic acid is: $H_2CO_3 \rightleftharpoons 2H^{+} + CO_3^{2-}$.
Given concentration $C = 5 \times 10^{-3} \, M$ and degree of dissociation $\alpha = 10\% = 0.1$.
The concentration of $H^{+}$ ions is given by $[H^{+}] = 2 \times C \times \alpha$.
Substituting the values: $[H^{+}] = 2 \times (5 \times 10^{-3}) \times 0.1$.
$[H^{+}] = 10 \times 10^{-3} \times 0.1 = 10^{-3} \, M$.

(A) The number of $OH^{-}$ ions is calculated by multiplying the concentration of $OH^{-}$ ions by Avogadro's number $(N_A)$.
Number of $OH^{-}$ ions = $[OH^{-}] \times N_A$
$= 10^{-7} \times 6.023 \times 10^{23} = 6.023 \times 10^{16}$ ions.

(C) In pure water,the concentration of $H^+$ ions is equal to the concentration of $OH^-$ ions,i.e.,$[H^+] = [OH^-]$.
Given that at $90 \, ^\circ C$,$[H^+] = 10^{-6} \, M$.
Therefore,$[OH^-] = 10^{-6} \, M$.
Calculating the sum: $[H^+] + [OH^-] = 10^{-6} + 10^{-6} = 2 \times 10^{-6} \, M$.
Thus,the correct option is $C$.

(A) For a weak acid,the dissociation constant is given by $K_a = C \alpha^2$.
Initially,$C_1 = 0.1 \, M$ and $\alpha_1 = \frac{1}{100} = 0.01$.
$K_a = 0.1 \times (0.01)^2 = 0.1 \times 10^{-4} = 10^{-5}$.
Now,for the new concentration $C_2$ where $\alpha_2 = \frac{10}{100} = 0.1$:
$K_a = C_2 \times (\alpha_2)^2$
$10^{-5} = C_2 \times (0.1)^2$
$10^{-5} = C_2 \times 10^{-2}$
$C_2 = \frac{10^{-5}}{10^{-2}} = 10^{-3} \, M = 0.001 \, M$.

(B) The $pH$ at the equivalence point of an acid-base titration depends on the nature of the salt formed.
$1$. For a strong acid and strong base,the salt formed is neutral $(pH \approx 7)$.
$2$. For a strong acid and weak base,the salt undergoes cationic hydrolysis,resulting in an acidic solution $(pH < 7)$.
$3$. For a weak acid and strong base,the salt undergoes anionic hydrolysis,resulting in a basic solution $(pH > 7)$.
In the titration of acetic acid ($CH_3COOH$,a weak acid) against sodium hydroxide ($NaOH$,a strong base),the salt formed is sodium acetate $(CH_3COONa)$.
$CH_3COO^- + H_2O \rightleftharpoons CH_3COOH + OH^-$
The production of $OH^-$ ions makes the solution basic,hence the $pH$ at the equivalence point is greater than $8$.

(A) Given: Concentration $C = 0.1 \, M$,Degree of ionization $\alpha = 2\% = 0.02$.
For a weak acid,$[H^{+}] = C \times \alpha$.
$[H^{+}] = 0.1 \times 0.02 = 0.002 \, M = 2 \times 10^{-3} \, M$.
We know that $[H^{+}] \times [OH]^{-} = K_w = 1 \times 10^{-14}$.
$[OH]^{-} = \frac{1 \times 10^{-14}}{2 \times 10^{-3}} = 0.5 \times 10^{-11} = 5 \times 10^{-12} \, M$.
Therefore,$[H^{+}] = 2 \times 10^{-3} \, M$ and $[OH]^{-} = 5 \times 10^{-12} \, M$.

(B) The concentration of $H^{+}$ ions in pure water is $10^{-7} \ M$.
Number of $H^{+}$ ions in $1 \ L = 10^{-7} \times 6.022 \times 10^{23}$.
The concentration of $H_2O$ in $1 \ L$ is $\frac{1000 \ g}{18 \ g/mol} \approx 55.56 \ M$.
Number of $H_2O$ molecules in $1 \ L = 55.56 \times 6.022 \times 10^{23}$.
Ratio = $\frac{10^{-7} \times 6.022 \times 10^{23}}{55.56 \times 6.022 \times 10^{23}} = \frac{1}{55.56 \times 10^7} \approx 1 : 55.6 \times 10^7$.
Given the options provided,the closest value is $1 : 55.4 \times 10^7$.

(B) To determine the $pH$ order,we analyze the nature of the salts:
$1$. $HCl$ is a strong acid,so its $0.1 \ M$ solution has the lowest $pH$ (approx $1$).
$2$. $NH_4Cl$ is a salt of a strong acid $(HCl)$ and a weak base $(NH_4OH)$,making it acidic $(pH < 7)$.
$3$. $NaCl$ is a salt of a strong acid $(HCl)$ and a strong base $(NaOH)$,making it neutral $(pH = 7)$.
$4$. $NaCN$ is a salt of a weak acid $(HCN)$ and a strong base $(NaOH)$,making it basic $(pH > 7)$.
Thus,the increasing order of $pH$ is: $HCl < NH_4Cl < NaCl < NaCN$.

(D) The reaction is $2NaOH + H_2SO_4 \rightarrow Na_2SO_4 + 2H_2O$.
Number of millimoles of $NaOH = 10 \, mL \times 0.1 \, M = 1 \, mmol$.
Number of millimoles of $H_2SO_4 = 10 \, mL \times 0.05 \, M = 0.5 \, mmol$.
Since $2 \, mmol$ of $NaOH$ reacts with $1 \, mmol$ of $H_2SO_4$,$1 \, mmol$ of $NaOH$ will react with $0.5 \, mmol$ of $H_2SO_4$.
Both reactants are completely consumed,resulting in a neutral solution.
Therefore,the $pH$ of the solution is $7$.