Question

1. 50.00 mL of 0.1000 M propanoic acid (CH3CH2COOH – Ka = 1.34 X 10-5) is titrated with 0.2000 M KOH. Calculate the pH at the following points in the titration: 1) Initial pH – no KOH has been added. 2) 5.00 mL of KOH has been added. 3) 12.50 mL of KOH has been added. 4) At the equivalence point. (Calculate the volume of KOH to reach the equivalence point & identify a good indicator.) 5) Provide a sketch of the titration showing the volume & pH at significant points.

Answer 1

1.) [CH₃CH₂COOH]=0.1000 M and K_a = 1.34 X 10^-5.

k_a=[CH₃CH₂COO^-][H⁺]/[CH₃CH₂COOH]=[H⁺]²/0.10=1.34 X 10^-5.

So, [H⁺]=(1.34 X 10^-6)^1/2=1.158*10^-3. p^H=-log₁₀[H⁺]=2.936.

2.) 50.00 mL of 0.1000 M propanoic acid=5/1000 mole=5 mmole.

5.00 mL of 0.2000 M NaOH=1/1000 mole=1 mmole.

1 mmole NaOH reacts with 1 mmole propanoic acid to form 1 mmole salt (CH₃CH₂COONa) of the acid. Total volume after addition of 5.00 mL KOH is 50+5=55 ml. Moles of acid remains unreacted is (5-1) mmole=4 mmole.

[CH₃CH₂COOH]=4 mmole*1000 ml/55 ml=0.0727 M.

[CH₃CH₂COONa]=1 mmole*1000 ml/55 ml=0.0182 M.

Let's take, [H⁺]=x M.

         CH₃CH₂COOH <-> H⁺ +CH₃CH₂COO^-

I            0.0727                        0.0182

C             -x                     x          +x

E          0.0727-x             x        0.0182+x

k_a=[CH₃CH₂COO^-] [H⁺]/[CH₃CH₂COOH]=(0.0182+x)x/( 0.0727-x)=1.34*10^-5. So, x=0.5353*10^-4.

p^H=-log₁₀[H⁺]=4.27

3.) Similar to above calculation, one can have

[CH₃CH₂COOH]=2.5 mmole*1000 ml/62.5 ml=0.04 M.

[CH₃CH₂COONa]=2.5 mmole*1000 ml/62.5 ml=0.04 M.

x~1.34*10^-5. p^H=-log₁₀[x]=4.87

4.) At equivalance point, 5 mmole acid reacts with 5mmole base. So the required volume of NaOH is 25 ml.

The total volume of the reaction mixture is 50+25=75 ml.

So, the concentration of salt is 5 mmole/75 ml=0.0667 M.

CH₃CH₂COO^-+H₂O <-> CH₃CH₂COOH+OH^-

k_b=k_w/k_a=10^-14/1.34*10^-5=[OH^-][CH₃CH₂COOH]/[CH₃CH₂COO^-]=[OH^-]²/0.0667 M.

So, [OH^-]=0.223*10^-4. p^H=14-p^OH=14+log₁₀[OH^-]=9.35