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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.

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  • $\begingroup$ Is there something wrong about my question? I'm ready to change things. $\endgroup$ – Anonymous1967 Nov 17 '18 at 22:48
  • $\begingroup$ You need to consider pKa of acid in calculation. $\endgroup$ – Mithoron Nov 17 '18 at 23:34
  • $\begingroup$ Oh, do I need to use the Henderson-Hasselbach equation? $\endgroup$ – Anonymous1967 Nov 18 '18 at 0:10
  • $\begingroup$ 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. $\endgroup$ – Mithoron Nov 18 '18 at 0:52
  • $\begingroup$ Why $\ce{NaOH}$? I don't see any $\ce{NaOH-molecule}$. Do you mean $\ce{KOH}$? $\endgroup$ – Anonymous1967 Nov 18 '18 at 11:21
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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 $[\ce{OH-}]$.

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

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