# Calculating the pH of a dilluted aqueous KOH solution

Calculate $\mathrm{p}\ce{H}$ of an aqueous solution of $0.01\:\mathrm{nM}~\ce{KOH}$ in water.

I think that water also has $\ce{OH-}$ ions in concentration of $10^{-7}\:\mathrm{M}$ (since $\mathrm{p}\ce{H}=7$)

So the total concentration of $\ce{OH-}$ after mixing the solution with water would be:

$0.01\:\mathrm{nM} + 10^{-7}\:\mathrm{M} = 10^{-11}\:\mathrm{M} + 10^{-7}\:\mathrm{M} \implies 1.0001 \times 10^{-7}\:\mathrm{M}$

and given that $\mathrm{pH} = 14 - \mathrm{p}\ce{OH}$

$\implies 14 - \log(1.0001\times10^{-7})$

$\implies \mathrm{pH} \approx 7$

Does this make sense?

• Thanks, I am glad to hear! Has been some time since I had chemistry in high school and couldn't find any example on the internet. Glad that I didn't provide any false information :) – user4603 Jun 5 '14 at 1:36
• Since this is my first post here on stack chemistry, I am wondering how you guys handle questions like this. Shall I delete it, or can someone just post a "yes" answer that I accept - so that it isn't kept marked as "unanswered". (maybe someone might find this question useful as future reference, maybe it is so trivial that it isn't necessary?) – user4603 Jun 5 '14 at 1:43
• ok, I added actual answer – DavePhD Jun 5 '14 at 1:54

Yes, that's right, the concentration of $\ce{{}^{-}OH}$ from dissociation of water will be much greater than the amount added from $0.01\:\mathrm{nM}~\ce{KOH}$.

$0.01\:\mathrm{nM}~\ce{KOH}$ will not significantly the affect $\mathrm{p}\ce{H}$ of water.

Your result is correct but I don't agree with your method. Your method would fail under certain circumstances. Which circumstances you might ask? When the pH of the solution is actually between 6.8 and 7.2. In these cases, your assumption that water auto-ionizes to create exactly $\ce{[HO^-]} = 1.0\times10^{-7} M$ will have significant figure implications).

$\ce{[HO^-]} = 1.0\times10^{-7}$

This is not true when you add any amount of acid or base to water because the addition of either (or both) will disrupt the equilibrium that exists in water (referred to as $\ce{K_{w}}$, the auto-protolysis of water, or the auto-ionization of water):

$\ce{2H_2O <=> HO^- + H_3^+O}$

$\ce{K_{w}=[HO^-][H_3^+O]}=1.0\times10^{-14}$

All you know about your system is that $\ce{[HO^-]_{water} = [H_3^+O]_{water}}$. This is in fact true for any water solution; as we can see from the auto-ionization reaction above, there exists a 1:1 ratio between $\ce{[HO^-]}$ and $\ce{[H_3^+O]}$.

You also know that because $\ce{KOH}$ dissociates completely in water solution, $\ce{[HO^-]_{KOH}} = 1.0 \times 10^{-11} M$

Therefore, the correct way to solve this system would be using $\ce{K_{w}}$. Work below:

$\ce{K_{w}=\sum{[HO^-]\sum[H_3^+O]}}=1.0\times10^{-14}$

$\ce{=([HO^-]_{water}}+\ce{[HO^-]_{KOH})([H_3^+O]_{water})}$

Because $\ce{[HO^-]_{water} = [H_3^+O]_{water}}$, we can make this substitution:

Let $x=\ce{[HO^-]_{water} = [H_3^+O]_{water}}$

Therefore,

$\ce{=(x}+\ce{[HO^-]_{KOH})(x)} = 1.0\times10^{-14}$

And you can solve for x from here. Keep in mind what x equals!

• Sorry, I am unfamiliar with the Latex markup here and had to keep refreshing to see the effects of my syntax. – Dissenter Jun 5 '14 at 2:53
• You have a preview under your edit window, there you can see how your post will look like. – Martin - マーチン Jun 5 '14 at 2:54
• You should also use \times ($\times$) when you want to multiply terms. – LDC3 Jun 5 '14 at 2:55
• I never noticed that it changed! You're right; I'll just scroll down next time to see my changes. – Dissenter Jun 5 '14 at 2:55
• @LDC3 or even better \cdot($\cdot$) as $\times$ refers to a vector product – Martin - マーチン Jun 5 '14 at 2:57