Question

40.0 ml of an acetic acid of unknown concentration is titrated with 0.100 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 5.10. What was the concentration of the original acetic acid solution?(Ka(CH3COOH) = 1.8 × 10–5)

Answer 1

Answer:

0.5 M

Explanation:

We have to start with the reaction between NaOH and CH3COOH:

$CH_3COOH~+~NaOH->CH_3COONa~+~H_2O$

We will have a 1:1 ratio between the acid and the base. The next step then would be the calculation of the moles of NaOH and his convertion to moles of CH3COOH.

$20~mL~\frac{1~L}{1000~mL} ~=~0.02~L$

$mol~=~0.1~M*0.02~L=0.02~mol~NaOH$

$0.02~mol~NaOH~\frac{1~mol~CH_3COOH}{1~mol~NaOH}=~0.02~mol~CH_3COOH$

The final step is the calculation of the concentration of the acid.

$M=\frac{0.02~mol~CH_3COOH}{0.04~L}=~0.5~M$

Due to the Ka value we can use the acetic acid as a strong acid.