# How to find the added volume $V_e$

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:

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

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 $$pH_0+1.4=\mathrm{p}K_\mathrm{a}$$)

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) $$$$\alpha = \frac{10^{-2pH_1} + 10^{-(pKa + pH_1)}}{10^{-2pH_1} + 10^{-(pKa + pH_1)}} - 1 = 1 - 1 \rightarrow \alpha = 0$$$$ which translates into $$V_e = 0$$. This is fine, since if the $$pH$$ did not change, then we did not add any water after the first equilibrium.