Consider the titration of 100.0 ml of 0.200 M acetic acid(Ka=1.8×10−5)with 0.100 M KOH. After...

osula9a

osula9a

Answered

2022-01-09

Consider the titration of 100.0 ml of 0.200 M acetic acid (Ka=1.8×105) with 0.100 M KOH. After adding the subsequent volumes of KOH, determine the pH of the resulting solution. 50.0 mL

Answer & Explanation

Fasaniu

Fasaniu

Expert

2022-01-10Added 46 answers

We have a titration of 100.0 mL (0.100 L) of 0.200 M acetic acid (Ka=1.8105) 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(0.150L)
First, let us calculate the number of moles of acetic acid
nCH3COOH=0.200M0.100L=0.02mol
And the number of moles of KOH
nKOH=0.100M0.050L=0.005mol
2) KOH is strong base, and it will dissociate completely into K+ and OH. Therefore, we are adding 0.005 mol of OH ions. When we add strong base into weak acid solution, the reaction that will occur is
CH3COOH+OHCH3COO+H20
Therefore, when 0.005 mol of OH is added, 0.005 mol of acetic acid is consumed (CH3COOH), and 0.005 mol of it's conjugate base is produced (CH3COO). So, after KOH has beed added, we end up with a buffer solution that contains (0.02mol0.005mol) 0.015 mol of acetic acid, and 0.005 mol of it's conjugate base.
OHCH3COOHCH3COO+H2OInitial(mol)0.0050.0200/Change(mol)0.0050.005+0.005/Equilibrium(mol)00.0150.005/
Step 2
3) Now, let us calculate the concentration of CH3COOH and CH3COO

[CH3COOH]=Number of moles of acetic acidTotal volume of a solution=0.015mol0.150L=0.100M
[CH3COO]=Number of moles of conjugate baseTotal volume of a solution=0.005mol0.150L=0.033M
4) Let us calculate the pH using Henderson-Hasselbalch equation
pH=pKa+log([co

Jonathan Burroughs

Jonathan Burroughs

Expert

2022-01-11Added 37 answers

Step 1
Given:
100.0 mL of 0.200 M acetic acid (Ka=1.8×105) titrated with 0.100 M KOH.
The reaction is
HC2H3O2H++C2H3O2
[H+]=[C2H3O2]
The expression for Ka is
Ka=[H+][C2H3O2][HC2H3O2]
Based on the balanced chemical equation
[H+]=[C2H3O2]
However, since 50 mL of KOH is added, the initial concentration of acetic acid and acetate will be reduced
[HC2H3O2]=100mL(0.200M)50.0mL(0.100M)100+50=0.100
[C2H3O2]=50mL(0.100M)100mL+50mL=0.0333
Let x=be the amount of H+ produced at equilibrium
At equilibrium, the concentrations are:
[H+]=x
[C2H3O2]=0.0333+x
[HC2H3O2]=0.100x
Equating it with the equilibrium constant
1.8×105=[x][0.0333+x]0.100x
[H+]=5.39×103
pH=log[H+]=log(5.39×103)=4.26

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