# What is the pH of a solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH?

What is the pH of a solution made by mixing 10.00 mL of 0.10 M acetic acid with 10.00 mL of 0.10 M KOH? The $$K_a =1.8 × 10^{-5}$$ for $$\ce{CH_3CO_2H}$$. Assume that the volumes of the solutions are additive.

I've done the following: Write the chemical reaction.

$$\ce{CH_3COOH + KOH <=> CH_3COOK + H_2O}$$

Then I calculated the number of moles of the reactants.

$$\ce n({CH_3COOH})= c \cdot V = 0.001 \text{mol}$$

$$\ce n({KOH})= c \cdot V = 0.001\text{mol}$$

Then I assume the mistake lies here:

I thought because 1 mmol of $$\ce{KOH}$$ reacts with 1mmol of $$\ce{CH_3COOH}$$ this will form 1 mmol of $$\ce{CH_3COOK}$$ which leads to no formation of $$\ce{OH^-ions}$$. Because there is no excess. Another thing I thought is that $$\ce{OH^-}= 1 \text{mmol}$$. So I could calculate $$\ce{pOH= 3}$$ but to no avail because of the answer I later saw that is shown below.

The final outcome is $$\ce{pH = 8,72}$$.

A hint of where my thought process doesn't make sense would be appreciated? Thanks in advance.

• Is there something wrong about my question? I'm ready to change things. – Anonymous196 Nov 17 '18 at 22:48
• You need to consider pKa of acid in calculation. – Mithoron Nov 17 '18 at 23:34
• Oh, do I need to use the Henderson-Hasselbach equation? – Anonymous196 Nov 18 '18 at 0:10
• No, for a moment I thought there was more NaOH. Find concentration of resulting salt and plug it into equilibrium - yes you ignored it, acetate is weak base - and calculate pH. – Mithoron Nov 18 '18 at 0:52
• Why $\ce{NaOH}$? I don't see any $\ce{NaOH-molecule}$. Do you mean $\ce{KOH}$? – Anonymous196 Nov 18 '18 at 11:21

Part A: The conceptually part: Scientific concepts and conception

1-"I thought because 1 mmol of KOH reacts with 1mmol of CH3COOH this will form 1 mmol of CH3COOK"

yes ,the reaction reached at the equivalent point, the weak acetic acid in mixture has been fully neutralized by the stong($$\ce{KOH}$$) base (no unreacted acid or base present anymore), the solution will only contain $$\ce{H2O}$$ and the basic salt ($$\ce{ CH3COOK}$$),so the concentration of the salt : $$[\ce{CH3COOK}] =\frac{\text{moles of salt}}{\text{Total volume}} =\frac{1\text{m}mol}{(10+10)\text{ml}}=0.05\text{M}$$ 2-"which leads to no formation of OH−ions. Because there is no excess."

No, the result, at the equivalence point, will be the same as dissolving $$(\ce{CH3COOK})$$ in water at the same concentration(0.05$${M}$$) as you have at the equivalence point. the salt ($$\ce{ CH3COOK}$$) is electrically neutral substance formed by cation $$\ce{K+}$$ (an acid)and anion$$\ce{ CH3COO−}$$ (a base), completely dissociated or ionized in an aqueous solution as : $$\ce{CH3COOK -> K^+_{(aq)} + CH3COO^−{(aq)}}$$ a) Since $$\ce{K+}$$ is the conjugate acid of a strong base, it won't be strong enough to react with water; $$\ce{K+}$$ actually spectator ion.

b) Meanwhile, since $$\ce{ CH3COO−}$$ is the conjugate base of a weak acid, and therefore strong enough to be able to hydrolyze and accept ions$$\ce{ H+}$$ from water, so water act as an acid leaving a hydroxide ion $$\ce{OH−}$$ as :

$$\ce{CH3COO^− +H2O <=>CH3COOH + OH^−}$$

Part B:: Calculating the approximate pH of $$\pu{0.05 mol L-1} \ce{CH3COOK}$$ with neglecting water autoionization:

The equilibrium equation of the hydrolysis of the conjugate base $$\ce{CH3COO-}$$:

$$K_\mathrm{b}=\frac{K_\mathrm{w}}{K_\mathrm{a}} =\frac{[\ce{OH-}][\ce{CH3COOH}]}{[\ce{CH3COO-}]}$$

Assuming :$${[\ce{OH-}]=[\ce{CH3COOH}]}$$, $${[\ce{CH3COO-}]=[\ce{CH3COOK}]_0}$$

Substiute $$K_\mathrm{a}\ and\ [\ce{CH3COO^-}]$$ in the equilibrium equation and solve for $$[\ce{OH-}]$$.

$$[\ce{OH-}] = 5.27\times{10^{-6}} , \mathrm{pOH}=5.278 , \mathrm{pH}=8.72$$