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What is the pH of the solution if $\pu{30.58 cm3}$ of hydrochloric acid ($\ce{HCl}$) with a concentration of $\pu{0.2855 mol dm-3}$ and $\pu{29.41 cm3}$ of potassium hydroxide with a concentration of $\pu{0.2971 mol dm-3}$ are mixed into a $\pu{250 cm3}$ flask ($\ce{KOH}$) solution, and then fill the flask after the temperature has returned? $K_\mathrm{w}=1.00 \times 10^{-14}$

The acceptable pH given is 8.85 but I got 9.45. Could someone share the right way to do this?

My methodology:
n(HCl) = 0,00873059 mol  < n(KOH) = 0,000873711
remaining n(KOH) = 0, 000 006 821 mol
c(KOH) = n/ V  = 0,000027284 mol/dm3
pOH = -log(0,...27284)
pH = 14 - 4,45 = 9,45
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    $\begingroup$ So how do you get 9.45? Did you note that the flask is filled so you dilute the mixture after combining the stocks? $\endgroup$ Oct 24, 2022 at 14:02
  • $\begingroup$ I calculated the remaining amount of substance (KOH) and then used 250cm3 as the volume of the substance to get the concentration ( c = n/V ). $\endgroup$
    – LagSurfer
    Oct 24, 2022 at 14:12

1 Answer 1

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Several issues are there, but the big one is the book is wrong. They apparently forgot to divide by the $250$ mL flask volume.

First the potassium hydroxide addition was not properly transcribed, it should read $0.008737711$ mol with two consecutive 7's. This gives a net potassium hydroxide addition after tge acid of $0.000007121$ mol. Dividing this by the filled flask volume of 250 mL then gives $0.000028484$ mol/L.

When I was in graduate school I remember having to take an entrance exam for the Ph.D. program. The exam was of course directed more to critical thinking than computing power, and one of its questions asked for the pH of various hydrochloric acid solutions at ambient temperature -- including $10^{-8}$ molar, where if you weren't careful you found a basic solution with pH 8! I did catch that; ever since, I learned to include water autoionization, which gives a correct answer. So we render the net potassium hydroxide concentration as actually the difference between hydroxide and hydrogen ion concentrations:

$\ce{[OH^-]}-\ce{[H^+]}=0.000028484$

$\ce{[OH^-]}×\ce{[H^+]}=1.00×10^{-14}$

Solving the difference equation for the hydroxide ion concentration and plugging that into the product equation leads to the quadratic equation

$\ce{[H^+]}^2+0.000028484\ce{[H^+]}-1.00×10^{-14}=0$

with the positive root

$\ce{[H^+]}=3.51×10^{-10}\text{ mol/L}$

and with the logarithm, $pH=9.45$.

So apart from a minor transcribing error, you were correct. Clearly it is the book that was wrong.

The difference in pH between the answers is $0.60$ unit, whose powers of $10$ is close to $4$. If you (or in this case, the book) forgot to divide by the volume of $250$ mL above, or equivalently assumed a one-liter flask, you would have gotten the equations

$\ce{[OH^-]}-\ce{[H^+]}=0.000007121$

$\ce{[OH^-]}×\ce{[H^+]}=1.00×10^{-14}$

from which the reader can then calculate the book's incorrect $pH=8.85$. It looks like this question (unwittingly) tested critical thinking, too!

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