 osula9a

2022-01-09

Consider the titration of 100.0 ml of 0.200 M acetic acid $\left({K}_{a}=1.8×{10}^{-5}\right)$ with 0.100 M KOH. After adding the subsequent volumes of KOH, determine the pH of the resulting solution. 50.0 mL Fasaniu

Expert

We have a titration of 100.0 mL (0.100 L) of 0.200 M acetic acid $\left({K}_{a}=1.8\cdot {10}^{-5}\right)$ with 0.100 M KOH.
(b)
Let us calculate the pH of a solution, after 50.0 mL (0.050 L) of KOH has been added.
1) The total volume of a solution is now $100.0mL50.0mL=150.0mL\left(0.150L\right)$
First, let us calculate the number of moles of acetic acid
${n}_{C{H}_{3}COOH}=0.200M\cdot 0.100L=0.02mol$
And the number of moles of KOH
${n}_{KOH}=0.100M\cdot 0.050L=0.005mol$
2) KOH is strong base, and it will dissociate completely into ${K}^{+}$ and $O{H}^{-}$. Therefore, we are adding 0.005 mol of $O{H}_{-}$ ions. When we add strong base into weak acid solution, the reaction that will occur is
$C{H}_{3}COOH+O{H}^{-}⇒C{H}_{3}CO{O}^{-}+{H}_{2}0$
Therefore, when 0.005 mol of $O{H}^{-}$ is added, 0.005 mol of acetic acid is consumed $\left(C{H}_{3}COOH\right)$, and 0.005 mol of it's conjugate base is produced $\left(C{H}_{3}CO{O}^{-}\right)$. So, after KOH has beed added, we end up with a buffer solution that contains $\left(0.02mol-0.005mol\right)$ 0.015 mol of acetic acid, and 0.005 mol of it's conjugate base.
$\overline{)\begin{array}{ccccc}& O{H}^{-}& C{H}_{3}COO{H}^{-}& ⇒C{H}_{3}CO{O}^{-}+& {H}_{2}O\\ Initial\left(mol\right)& 0.005& 0.020& 0& /\\ Change\left(mol\right)& -0.005& -0.005& +0.005& /\\ Equilibrium\left(mol\right)& 0& 0.015& 0.005& /\end{array}}$
Step 2
3) Now, let us calculate the concentration of $C{H}_{3}COOH$ and $C{H}_{3}CO{O}^{-}$


4) Let us calculate the pH using Henderson-Hasselbalch equation Jonathan Burroughs

Expert

Step 1
Given:
100.0 mL of 0.200 M acetic acid $\left({K}_{a}=1.8×{10}^{-5}\right)$ titrated with 0.100 M KOH.
The reaction is
$H{C}_{2}{H}_{3}{O}_{2}⇌{H}^{+}+{C}_{2}{H}_{3}{O}_{2}^{-}$
$\left[{H}^{+}\right]=\left[{C}_{2}{H}_{3}{O}_{2}^{-}\right]$
The expression for ${K}_{a}$ is
${K}_{a}=\frac{\left[{H}^{+}\right]\left[{C}_{2}{H}_{3}{O}_{2}^{-}\right]}{\left[H{C}_{2}{H}_{3}{O}_{2}\right]}$
Based on the balanced chemical equation
$\left[{H}^{+}\right]=\left[{C}_{2}{H}_{3}{O}_{2}^{-}\right]$
However, since 50 mL of KOH is added, the initial concentration of acetic acid and acetate will be reduced
$\left[H{C}_{2}{H}_{3}{O}_{2}\right]=\frac{100mL\left(0.200M\right)-50.0mL\left(0.100M\right)}{100+50}=0.100$
$\left[{C}_{2}{H}_{3}{O}_{2}^{-}\right]=\frac{50mL\left(0.100M\right)}{100mL+50mL}=0.0333$
Let $x=be$ the amount of ${H}^{+}$ produced at equilibrium
At equilibrium, the concentrations are:
$\left[{H}^{+}\right]=x$
$\left[{C}_{2}{H}_{3}{O}_{2}^{-}\right]=0.0333+x$
$\left[H{C}_{2}{H}_{3}{O}_{2}\right]=0.100-x$
Equating it with the equilibrium constant
$1.8×{10}^{-5}=\frac{\left[x\right]\left[0.0333+x\right]}{0.100-x}$
$\left[{H}^{+}\right]=5.39×{10}^{-3}$
$pH=-\mathrm{log}\left[{H}^{+}\right]=-\mathrm{log}\left(5.39×{10}^{-3}\right)=4.26$

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