Find the required water volume to add so that the pH of two solutions will be the same

Question

Question.

We have two flasks: one contains 50 mL of 0.001 M HCl and the other 50 mL of 1M acetic acid. a) How much water should be added to the more acidic solution so that the pH of the two solutions be the same?

Attempt. So after finding both pH before adding the volume, we find that the first flask has a $\mathrm{p}H_1=-\log(10^{-3})=3$ and the second one can be found by considering $K_a(\text{acetic acid}) = \pu{1.8e-5}=\frac{x^2}{1-x}\implies x=\pu{4.25e-3 M}$. Hence $\mathrm{p}H_2=2.37$

Now here it comes the issues, following that both pH should be the same and we add a certain $V$ to the most acid one, which is the $2.37$ one: $$\mathrm{p}H_1=pH_2\implies 3=-\log(\frac{4.25\cdot 10^{-3}\cdot 0.05}{0.05+V})$$

Which if you solve this, you end up with $V\approx -0.0502L$ which doesn't make sense. And I'd rather follow a method similar to this where it starts from the equation $\mathrm{p}H_1=\mathrm{p}H_2$ since that would be how I would have reasoned through this in an exam

Answer 1

In the flask that requires diluting, we have:

$$\ce{AcOH <=> H+ + AcO^-}$$

Let:

$A$ represent $\ce{AcOH}$

$C$ represent $\ce{H+}$

$D$ represent $\ce{AcO^-}$

So that the reaction becomes:

$$\ce{A <=> C + D}$$

Let:

The subscript (1) represent "before diluting with water"

We have:

$$N_{A1}=N_{Ao}-x=0.05-x$$ $$N_{C1}=N_{Co}+x=x$$ $$N_{D1}=N_{Do}+x=x$$

The relationship between $K_a$ and $K_{n1}$ is:

$$K_a=K_{n1}\left(\frac{1}{V_1}\right)^{\Delta n}$$

Considering that for this reaction $\Delta n=1$, and solving for $K_{n1}$:

$$K_{n1}=K_a \times V_1=(\pu{1.8e-5})(0.05)=\pu{9e-7}$$

Using the expression for $K_{n1}$ and the equilibrium number of moles, we can solve for $x$:

$$K_{n1}=\frac{N_{C1}N_{D1}}{N_{A1}}=\frac{x^2}{0.05-x}=\pu{9e-7}$$

$$x=\pu{2.117e-4}$$

Now, we dilute the mixture by adding water, so the equilibrium will shift again.

Let:

The subscript (2) represent "after diluting with water"

We have:

$$N_{A2}=N_{A1}-y=0.05-x-y$$ $$N_{C2}=N_{C1}+y=x+y$$ $$N_{D2}=N_{D1}+y=x+y$$

Temperature is constant, so $K_a$ does not change, and we have:

$$K_a=K_{n2}\left(\frac{1}{V_2}\right)^{\Delta n}$$

Solving for $K_{n2}$:

$$K_{n2}=K_a \times V_2$$

Substituting in the expression for $K_{n2}$:

$$K_{n2}=K_a \times V_2=\frac{N_{C2}N_{D2}}{N_{A2}}=\frac{(x+y)^2}{0.05-x-y}$$

Since we're required to dilute the acetic acid until its pH reaches 3, that means that at equilibrium, the total moles of $H^+$ divided by the resulting total volume $V_2$ must be equal to $10^{-3}$:

$$\frac{x+y}{V_2}=10^{-3}$$

Solving for $V_2$:

$$V_2=\frac{x+y}{\pu{e-3}}$$

Substituting above:

$$K_a \times \frac{x+y}{10^{-3}}=\frac{(x+y)^2}{0.05-x-y}$$

We already know $x$, so we can solve this quadratic formula for $y$, and we keep the positive value:

$$y=\pu{6.724e-4}$$

We can then solve for $V_2$:

$$V_2=\frac{x+y}{10^{-3}}=\frac{\pu{8.841e-4}}{10^{-3}}=\pu{0.8841 L}=\pu{884.1 mL}$$

Finally, we calculate the volume of water we have to add:

$$\Delta V=V_2-V_1=884.1-50=\pu{834.1 mL}$$

In conclusion, diluting the acetic acid by adding approximately 834.1 mL of pure water will increase its pH to 3.