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How would you find the pH of an acid dissolved in water? Would you need to take the fact that the pH of water is already 7 into account and go on from there?

Say, for example, that you add 0.0500 mL of 0.00100 M HCl into 50 L of pure water, what will be the pH of the new solution?

I made the mistake of just finding the moles of the HCl placed into the pure water, and finding the new concentration by dividing by the litres of water present. This gave me a pH of 9, which does not seem to make any sense as acids are meant to lower pH!

Some help would be wonderful!

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closed as off-topic by Mithoron, Mathew Mahindaratne, Buck Thorn, Todd Minehardt, airhuff Oct 14 at 19:03

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In this solution, there are two sorts of $\ce{H+}$ ions, those coming from added $\ce{HCl}$, whose concentration is $10^{-9}\ \mathrm M$, and those coming from the water which are unknown, and so described by $x$. But the concentration of $\ce{OH-}$ is also $x$. So $K_\mathrm w$ may be written: $$K_\mathrm w = (x + 10^{-9})x = 10^{-14}$$ If you solve this equation, you find: $$x = 0.999\times10^{-7}$$ The total concentration of $\ce{H+}$ ion is $1.0001\times10^{-7}$; $\mathrm{pH} = 6.999$.

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Most of $\mathrm{pH}$ calculations make simplifying assumptions and one must determine if they are justified for the particular case.

Simplification for strong acids $$\mathrm{pH}=-\log{[\mathrm{acid}]}$$

implies $\pu{[\mathrm{acid}] >> 10^{-7} mol/L}$.

Because water dissociation constant $K_\mathrm{w}=\ce{[H+][OH-]}=\pu{10^{-14} mol^2/L^2}$,

neutral water contains $\pu{10^{-7} mol/L}$ of $\ce{H+}$.

That means the very most $\ce{H+}$ ion must come from acid dissociation and not from water dissociation.


If OTOH very most $\ce{H+}$ are from water dissociation as in your case, another simplification says $\mathrm{pH}=7$, no matter how much of acid you added.


For general case: $$c_{\mathrm{acid}}=\ce{[A-]}=\ce{[H+]} - \ce{[OH-]} = \ce{[H+]} - \frac{K_\mathrm{w}}{\ce{[H+]}}$$

$${\ce{[H+]}}^2 - \ce{[H+]}\cdot c_{\mathrm{acid}} - K_\mathrm{w}=0$$

$$\ce{[H+]} = \frac{c_\mathrm{acid} + \sqrt{{ c_\mathrm{acid}}^2 + 4\cdot K_\mathrm{w}}}{2}$$

The limit case for $c_\mathrm{acid} << \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = \sqrt{K_\mathrm{w}}$$ $$\mathrm{pH}=\frac 12 \mathrm{p}K_\mathrm{w}$$

The limit case for $c_\mathrm{acid} >> \sqrt{K_\mathrm{w}}$ is $$\ce{[H+]} = c_\mathrm{acid}$$

$$\mathrm{pH} = -\log{(c_\mathrm{acid})}$$


Simplification for weak acids $$\mathrm{pH}=0.5 \left(\mathrm{p}K_\mathrm{a}-\log {\ce{[acid]}}\right)$$

implies negligible water dissociation and also negligible relative acid dissociation $\ce{[HA]} >> \ce{[A-]}$.

So one can assume:

  • The acid concentration $\ce{[HA]}$ is not affected by acid dissociation.
  • $\ce{[A-]} = \ce{[H+]}$
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