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What is the pH of a solution of $3.4 \times 10^{-10}$ moles of $\ce{HCl}$ in a volume of $\pu{150.7L}$?

This is dilute so I expect pH to be high.

$$\frac{3.4 \times 10^{-10}\ \mathrm{mol}}{150.7\ \mathrm L} = 2.26 \times 10^{-12} \frac{\mathrm{mol}}{\mathrm L}$$

Then pH is 11.65. Is this correct?

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    $\begingroup$ What's the pH of pure water? If you add an amount (however small) of strong acid to pure water, what do you expect the new pH will be? $\endgroup$ – Nicolau Saker Neto Apr 27 '15 at 1:04
  • $\begingroup$ It should be ~7 $\endgroup$ – user15831 Apr 27 '15 at 1:07
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Like you guessed, the pH should be around 7, since the concentration of HCl is very low. Obviously, it can't be 11.65, since it's still an acid. The reason for this discrepancy is that, for low concentrations, you must consider the self-ionization of water in order to get the correct concentration of $\ce{H3O+}$ ions.

Consider the equilibrium:

$$\ce{2H2O <=> H3O+ + OH-}$$ $$K_w = \ce{[H3O+][OH^{-}]} = 1 \times 10^{-14} \tag 1$$

and also the dissociation of the acid:

$$\ce{H2O + HCl -> H3O+ + Cl-}$$

Then $$c_0 = \ce{[Cl^{-}]} = 2.26 \times 10^{-12} \tag 2 $$

and

$$\ce{[H3O+] = [Cl^{-}] + [OH^{-}]} \Rightarrow \ce{[OH^{-}] = [H3O+] - [Cl^{-}]} \tag 3$$

If we replace $(2)$ in $(3)$ and that in $(1)$, we end up with

$$ K_w = \ce{[H3O+]} (\ce{[H3O+]} -c_0) $$

$$ \ce{[H3O+]} (\ce{[H3O+]} -c_0) - K_w = 0$$

using $ x = \ce{[H3O+]}$

$$x^2 - c_0x - Kw = 0$$

This can be easily solved. Taking only the positive value of $x$:

$$ x = \frac{c_0 + \sqrt{c_o^2 + 4K_w}}{2}$$

Inserting the values of $x$ and $K_w$, the result is:

$$ x = 1.000023 \times 10^{-07} = \ce{[H3O+]}$$

So

$$pH = -\log\ce{[H3O+]} = -\log{1.000011 \times 10^{-07}} = 6.999995 $$

Which is not surprising at all, but is the calculated answer.

For a less diluted solution of $\ce{HCl}$ with concentration of $1 \times 10^{-8} \frac{mol}{L}$, which still can't be calulated directly, the result is $pH = 6.98$.

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  • $\begingroup$ Thanks but I don't get how you came up with Equation 2 relating [H3O+] to [OH-] and [Cl-]. $\endgroup$ – user15831 Apr 27 '15 at 10:49
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    $\begingroup$ I think you meant equation 3. That's what we call charge balance. The only species in our system are $\ce{H2O}$, which we may consider with constant concentration because of how high it is, and the mentioned ions. The total system charge must be zero, so the concentration of cations has to equal the concentration of anions. Another way to look at it is noting that you have two sources of $\ce{H3O+}$, the water self-ionization and the acid's dissociation, and each one of these processes will lead to corresponding concentrations of their counter ions. $\endgroup$ – Molx Apr 27 '15 at 12:39
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As you may have noticed, your answer of 11.65 is counter-intuitive as it is higher than 7, meaning that dilute hydrochloric acid is somehow more BASIC than water.

The key here is accounting for the $1.0 \times 10^{-7} M$ H+ ions, or $1.507 \times 10^{-5}$ moles of H+ ions for 150.7L, already present in the water due to self-ionization. An addition of $3.4 \times 10^{-10}$ more moles of H+ ions would be practically insignificant (5 orders of magnitude difference). Hence, the pH would be near 7.

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    $\begingroup$ Just for curiosity, an analytical solution can be found for such a dilute acid, but the hydrogen ion concentration is the same as in pure water to at least six significant digits (one part in a million), so for all practical purposes the resulting pH is identical to pure water. $\endgroup$ – Nicolau Saker Neto Apr 27 '15 at 2:09

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