How to find the added volume $V_e$

Question

We have a volume $V_A$ of a solution of $\ce{CH3COOH}$ acid, its molar concentration is $C_A$ and it has $pH_0=3,4$, we add a volume of $V_e$ from distilled water to it, in this case it has $pH_0'=3,7$, then we titrate it with an aqueous solution of $\ce{(Na^+, OH^-)}$, its molar concentration is $C_B$, and its volume is $V_B$

let's consider that $\ce{CH3COOH}$ acid whose molar concentration $C_A'$ is weak and has a $pH$ that achieves: $pH=\frac{1}{2}(\mathrm{p}K_\mathrm{a}-logC_A')$ and we have $\mathrm{p}K_\mathrm{a}=4,8$

How can I prove that: $V_e=α.V_A$, where $α$ is a constant whose expression is required to be specified in terms of $pH_0$ and $pH_0'$

here is what i have tried:

\begin{equation} \text{starting with these relations: $$}\begin{cases} *\frac{C_A}{C_A'}=F \\ *\frac{V_A+V_e}{V_A}=F ⇒ V_e=V_A(F-1) \\ *C_A'=10^{\mathrm{p}K_\mathrm{a}-2pH_0'} \end{cases} \end{equation}

I came to a conclusion That: $V_e=V_A(C_A.10^{2pH_0'-(pH_0+1,4)}-1)$ (from what's given we find that $pH_0+1.4=\mathrm{p}K_\mathrm{a}$)

Answer 1

The reaction of interest is: $$\ce{HA (aq)} \rightarrow \ce{H+ (aq)} + \ce{A- (aq)}$$

First Equilibrium Let $X_1$ be the moles that have reacted of $\ce{HA}$ and let me name your $C_A$ by simply $C_0$. By stoichiometry, in equilibrium we will have \begin{align} n_{HA} = C_0V_A - X_1 &\rightarrow C_{HA} = \frac{n_{HA}}{V_A} = \frac{C_0V_A - X_1}{V_A} = C_0 - Y_1 \\ n_{H^+} = X_1 &\rightarrow C_{H^+} = \frac{X_1}{V_A} = Y_1 \\ n_{A^-} = X_1 &\rightarrow C_{A^-} = \frac{X_1}{V_A}= Y_1 \end{align} where we defined $ Y_1 := X_1/V_A $. However, this is data, because $Y_1 = 10^{-pH_1}$. The law of mass action states that $$ K_a = \frac{Y_1^2}{C_0 - Y_1} \rightarrow C_0 = \frac{Y_1^2}{K_a} + Y_1 = \frac{Y_1^2 + K_aY_1}{K_a} \tag{1}$$

Second Equilibrium Upon adding water a new equilibrium will be reached. Since $pH_2 > pH_1$, the reaction has proceeded in reverse, which makes sense. Let $X_2$ be the moles that have been formed of $\ce{HA}$. By stoichiometry, in equilibrium we will have \begin{align} n_{HA} = (C_0V_A - X_1) + X_2 &\rightarrow C_{HA} = \frac{(C_0V_A - X_1) + X_2}{V_A + V_e} = \frac{C_0V_A}{V_A + V_e} - Y_2 \\ n_{H^+} = X_1 - X_2 &\rightarrow C_{H^+} = \frac{n_{H^+}}{V_A + V_e} = \frac{X_1 - X_2}{V_A + V_e} = Y_2 \\ n_{A^-} = X_1 - X_2 &\rightarrow C_{A^-} = \frac{n_{A^-}}{V_A + V_e} = \frac{X_1 - X_2}{V_A + V_e} = Y_2 \end{align} where we defined $ Y_2 := (X_1 - X_2)/(V_A + V_e) $. However, this is data, because $Y_2 = 10^{-pH_2}$. The law of mass action states that $$ K_a = \frac{Y_2^2}{\dfrac{C_0V_A}{V_A + V_e} - Y_2} \rightarrow \frac{C_0V_A}{V_A + V_e} = \dfrac{Y_2^2}{K_a} + Y_2 = \dfrac{Y_2^2 + K_aY_2}{K_a} \tag{2} $$

Solving for the asked ratio Now we divide, member by member, Eq. (1) and Eq. (2) \begin{align} \dfrac{C_0}{\dfrac{C_0V_A}{V_A + V_e}} &= \frac{Y_1^2 + K_aY_1}{Y_2^2 + K_aY_2} \\ \dfrac{V_A + V_e}{V_A} &= \frac{Y_1^2 + K_aY_1}{Y_2^2 + K_aY_2} \\ 1 + \dfrac{V_e}{V_A} &= \frac{Y_1^2 + K_aY_1}{Y_2^2 + K_aY_2} \\ V_e &= \bigg( \frac{Y_1^2 + K_aY_1}{Y_2^2 + K_aY_2} - 1 \bigg) V_A \end{align} and going back to our definitions of $Y_1$ and $Y_2$ \begin{align} V_e &= \bigg( \frac{10^{-2pH_1} + 10^{-pKa} 10^{-pH_1}}{10^{-2pH_2} + 10^{-pKa}10^{-pH_2}} - 1 \bigg) V_A \rightarrow \boxed{\alpha = \frac{10^{-2pH_1} + 10^{-(pKa + pH_1)}}{10^{-2pH_2} + 10^{-(pKa + pH_2)}} - 1} \tag{3} \\ \end{align}

Notes:

1. I could get a result, but is still function of the acid equilibrium constant.
2. Not sure about the information given of the titration.
3. If $pH_2 = pH_1$, by Eq. (3) \begin{equation} \alpha = \frac{10^{-2pH_1} + 10^{-(pKa + pH_1)}}{10^{-2pH_1} + 10^{-(pKa + pH_1)}} - 1 = 1 - 1 \rightarrow \alpha = 0 \end{equation} which translates into $V_e = 0 $. This is fine, since if the $pH$ did not change, then we did not add any water after the first equilibrium.