Mix Examples-Ionic Equilibrium Questions in English

(D) The concentration of $H^{+}$ ions is calculated using the formula: $[H^{+}] = \frac{\text{Total moles of } H^{+}}{\text{Total volume of solution}}$.
Let the volume of each solution be $V \ L$.
Moles of $H^{+}$ from $HCl = M \times V \times n = 0.1 \times V \times 1 = 0.1V \ mol$.
Moles of $H^{+}$ from $H_2SO_4 = M \times V \times n = 0.2 \times V \times 2 = 0.4V \ mol$.
Total moles of $H^{+} = 0.1V + 0.4V = 0.5V \ mol$.
Total volume of the mixture $= V + V = 2V \ L$.
Concentration of $H^{+} = \frac{0.5V}{2V} = 0.25 \ M$.

(B) The dissociation constant of water at $STP$ $(298 \ K)$ is $K_w = [H^{+}][OH^{-}] = 10^{-7} \times 10^{-7} = 10^{-14}$.
The relation between standard free energy change and equilibrium constant is $\Delta G^{\circ} = -RT \ln K_w = -2.303 RT \log K_w$.
Given $R = 1.987 \ cal \ K^{-1} mol^{-1}$,$T = 298 \ K$,and $K_w = 10^{-14}$.
$\Delta G^{\circ} = -2.303 \times 1.987 \times 298 \times \log(10^{-14})$.
$\Delta G^{\circ} = -2.303 \times 1.987 \times 298 \times (-14)$.
$\Delta G^{\circ} = 2.303 \times 1.987 \times 298 \times 14 \approx 19091 \ cal / mol$.

(B) At the start,the solution contains only the weak acid $HX$. The concentration $C = 0.2 \ M$. The $pH$ of a weak acid is given by $pH = 0.5(pK_a - \log C)$. Substituting the values: $pH = 0.5(3.3 - \log 0.2) = 0.5(3.3 - (-0.699)) \approx 0.5(4.0) = 2.0$.
$(b)$ When $10 \ mL$ of $0.2 \ M$ $NaOH$ is added to $20 \ mL$ of $0.2 \ M$ $HX$,we reach the half-equivalence point because the moles of $NaOH$ added $(10 \ mL \times 0.2 \ M = 2 \ mmol)$ are exactly half the initial moles of $HX$ $(20 \ mL \times 0.2 \ M = 4 \ mmol)$. At the half-equivalence point,the concentration of the acid $[HX]$ equals the concentration of its conjugate base $[X^-]$. According to the Henderson-Hasselbalch equation,$pH = pK_a + \log([X^-]/[HX])$. Since $[X^-] = [HX]$,$pH = pK_a = 3.3$.

(C) To determine the pH order,we calculate the millimoles (mmol) of $H^+$ and $OH^-$ ions in each mixture:
$A$. $Ca(OH)_2$ is a strong base. $OH^-$ mmol $= 10 \times 0.2 \times 2 = 4$. $HCl$ is a strong acid. $H^+$ mmol $= 25 \times 0.1 = 2.5$. Excess $OH^- = 4 - 2.5 = 1.5 \text{ mmol}$. The solution is basic,so pH $> 7$.
$B$. $H_2SO_4$ is a strong acid. $H^+$ mmol $= 10 \times 0.01 \times 2 = 0.2$. $Ca(OH)_2$ is a strong base. $OH^-$ mmol $= 10 \times 0.01 \times 2 = 0.2$. Since $H^+$ mmol $= OH^-$ mmol,the solution is neutral,so pH $= 7$.
$C$. $H_2SO_4$ is a strong acid. $H^+$ mmol $= 10 \times 0.1 \times 2 = 2$. $KOH$ is a strong base. $OH^-$ mmol $= 10 \times 0.1 = 1$. Excess $H^+ = 2 - 1 = 1 \text{ mmol}$. The solution is acidic,so pH $< 7$.
Comparing the pH values: $C$ (acidic) $< B$ (neutral) $< A$ (basic). Therefore,the increasing order is $C < B < A$.

(C) For the first ionization step of the weak dibasic acid: $H_{2}A \rightleftharpoons H^{+} + HA^{-}$.
The equilibrium constant expression is $K_{a1} = \frac{[H^{+}][HA^{-}]}{[H_{2}A]}$.
Given that $0.1 \text{ mol}$ of $H_{2}A$ is dissolved in $1 \text{ L}$ of solution,the initial concentration $[H_{2}A] = 0.1 \text{ M}$.
The solution also contains $0.1 \text{ M}$ $HCl$,which is a strong acid and dissociates completely,providing $[H^{+}] = 0.1 \text{ M}$.
Since $K_{a1} = 8.1 \times 10^{-8}$ is very small,the dissociation of $H_{2}A$ is negligible. Therefore,we can assume $[H_{2}A] \approx 0.1 \text{ M}$ and $[H^{+}] \approx 0.1 \text{ M}$ at equilibrium.
Substituting these values into the expression: $8.1 \times 10^{-8} = \frac{(0.1)[HA^{-}]}{0.1}$.
Solving for $[HA^{-}]$,we get $[HA^{-}] = 8.1 \times 10^{-8} \text{ M}$.