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

0.5 M

Step-by-step 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~(1~L)/(1000~mL) ~=~0.02~L`

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

`0.02~mol~NaOH~(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=(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.