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$3\cdot 10^{-2}\ \mathrm{mol}$ 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?

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  • $\begingroup$ Note that $\ce{Ac}$ is $\ce{CH3-C=O}$ and not $\ce{CH3COO}$. $\endgroup$
    – Jan
    Jun 30, 2016 at 17:51
  • $\begingroup$ @Jan I don't understand why my question's been change, the question is with HAc, not HOAc... $\endgroup$
    – YoTengoUnLCD
    Jun 30, 2016 at 17:52
  • $\begingroup$ 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. $\endgroup$
    – Jan
    Jun 30, 2016 at 18:07
  • $\begingroup$ @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)... $\endgroup$
    – YoTengoUnLCD
    Jun 30, 2016 at 18:11
  • $\begingroup$ @YoTengoUnLCD. In the USA, it's also common to abbreviate acetic acid as HAc. $\endgroup$
    – Dr. J.
    Jul 22, 2018 at 12:12

2 Answers 2

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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.

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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 $\mathrm{pH}=12$

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