# How to find the pH of acetic acid after adding sodium hydroxide?

$$3\cdot 10^{-2}\ \mathrm{mol}$$ of $$\ce{NaOH}$$ are added to a $$1\,\mathrm{l}$$ solution of $$2\cdot10^{-2}~\mathrm{M}\ \ce{CH3COOH}$$. Find the $$\mathrm{pH}$$ of the solution.

I don't understand why my solution is wrong:

I said this, $$\ce{NaOH + CH3COOH -> H2O + CH3COONa}$$, so I'm left with $$2\cdot10^{-2}~\mathrm{mol}$$ of $$\ce{CH3COONa}$$ and $$1\cdot10^{-2}\ \mathrm{mol}$$ of $$\ce{NaOH}$$.

Now, from those $$1\cdot10^{-2}\ \mathrm{mol}$$ of $$\ce{NaOH}$$ I get the same moles of $$\ce{OH-}$$ (from dissociation).

Then, $$\ce{CH3COONa -> Na+ + CH3COO-}$$, and $$\ce{CH3COO- + H2O <=> CH3COOH + OH-}$$

Setting up the equilibrium equation: $$K_h = \frac {\text{Products}}{\text{Reactants}}$$, I get (where $$K_h= K_\mathrm{w}/K_\mathrm{a}, K_\mathrm{a} = 1.8\cdot 10^{-5}$$):

$$K_h=\frac {(x)(1\cdot 10^{-2}+x)}{x-1\cdot10^{-2} }=5.55\cdot 10^{-10}$$

Which yields $$[\ce{OH-}]= x= 1.11\cdot 10^{-9}$$ and so $$\mathrm{pH}=5.04$$ which is wrong. I know the answer should be around 12 from the answer sheet.

What am I doing wrong?

• Note that $\ce{Ac}$ is $\ce{CH3-C=O}$ and not $\ce{CH3COO}$. – Jan Jun 30 '16 at 17:51
• @Jan I don't understand why my question's been change, the question is with HAc, not HOAc... – YoTengoUnLCD Jun 30 '16 at 17:52
• I seriously doubt that you attempted to titrate acetaldehyde with sodium hydroxide, especially since the carbonyl proton of acetaldehyde is not acidic. Acetic acid is $\ce{HOAc}$, by definition. – Jan Jun 30 '16 at 18:07
• @Jan I suppose in my country it's commont to use HAc for acetic acid (yes, this was supposed to be acetid acid with sodium hydroxide)... – YoTengoUnLCD Jun 30 '16 at 18:11
• @YoTengoUnLCD. In the USA, it's also common to abbreviate acetic acid as HAc. – Dr. J. Jul 22 '18 at 12:12

If you add $3 \cdot 10^{-2}~\mathrm{mol}$ $\ce{NaOH}$ to $1~\mathrm{L}$ of $2 \cdot 10^{-2}~\mathrm{M}$ $\ce{AcOH}$ you end up with a solution that contains $2 \cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{AcO-}$ and $1\cdot 10^{-2}~\mathrm{mol~L^{-1}}$ $\ce{OH-}$. So the pH is 12.
The pH of the solution is imposed by the the concentration of $\ce{OH-}$ from $\ce{NaOH}$: $\ce{[OH^{-}]}=10^{-2} \mathrm{M}$, since the quantity of $\ce{OH-}$ produced from the reaction of $\ce{OAc-}$ with water is negligible in front of $10^{-2} \mathrm{M}$: $$\ce{OAc- + H2O <=>AcOH + OH-}$$ The equilibrium constant is:$$K=\frac{K_w}{K_a}=10^{-9.3}$$
So, $\mathrm{pOH}=2$, i.e $\mathrm{pH}=12$