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

  • $\begingroup$ Adrian: The ionization of a weak acid changes with concentration so you cannot simply dilute and change the pH in a linear manner. You know the total amount of acid in the 50mL; calculate the concentration needed to give a pH of 3 using Ka. then figure the amount of water for the amount of acid. $\endgroup$
    – jimchmst
    Commented Oct 12, 2022 at 21:50
  • $\begingroup$ @jimchmst okay thanks for telling me why it is wrong. Now for the new method $3=pH_2$, you said we know the total amount of acid on 50 mL which I'm guessing you are talking about 1M? And how do I find the concentration needed using Ka? You mean to solve for y in the equation $Ka=\frac{\frac{y^2}{(0.05+V)^2}}{\frac{1-y}{0.05+V}}$? but we still have 2 variables $\endgroup$
    – Adrian
    Commented Oct 12, 2022 at 22:00
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    $\begingroup$ $$K_a=\frac{\pu{E-6}}{c - \pu{e-3}}$$ and the volume to add: $$V=50(\frac{1}{c}-1)\pu{mL}$$ $\endgroup$
    – Poutnik
    Commented Oct 13, 2022 at 4:31

1 Answer 1


In the flask that requires diluting, we have:

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


$A$ represent $\ce{AcOH}$

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

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

So that the reaction becomes:

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


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 $x$:



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


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 $10^{-3}$:


Solving for $V_2$:


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:


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.

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    $\begingroup$ Is not it rather complicated? $\endgroup$
    – Poutnik
    Commented Oct 13, 2022 at 4:46
  • $\begingroup$ Perhaps long, but the hardest part is solving the two quadratic equations, which I wouldn't say is complicated at the level this material is usually taught. $\endgroup$
    – Sam202
    Commented Oct 13, 2022 at 4:59
  • $\begingroup$ I mean, compared to 2 simple equations in the other comment. $\endgroup$
    – Poutnik
    Commented Oct 13, 2022 at 5:03
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    $\begingroup$ To a surprise to all of you, this expanded post does help more. Thank you Sam, it is now crystal clear $\endgroup$
    – Adrian
    Commented Oct 13, 2022 at 6:37
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    $\begingroup$ @Poutnik Your solution is definitely more practical, but the average high-school or undergraduate student (considering this question was asked in relation to an upcoming exam) won't see it as "trivially" as you do. I personally think members should not be discouraged from posting correct solutions out of fear of being considered "too complicated" by others, especially since that can be very subjective. $\endgroup$
    – Sam202
    Commented Oct 13, 2022 at 7:40

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