Answer:
0.5 M
Explanation:
We have to start with the reaction between NaOH and CH3COOH:
[tex]CH_3COOH~+~NaOH->CH_3COONa~+~H_2O[/tex]
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.
[tex]20~mL~\frac{1~L}{1000~mL} ~=~0.02~L[/tex]
[tex]mol~=~0.1~M*0.02~L=0.02~mol~NaOH[/tex]
[tex]0.02~mol~NaOH~\frac{1~mol~CH_3COOH}{1~mol~NaOH}=~0.02~mol~CH_3COOH[/tex]
The final step is the calculation of the concentration of the acid.
[tex]M=\frac{0.02~mol~CH_3COOH}{0.04~L}=~0.5~M[/tex]
Due to the Ka value we can use the acetic acid as a strong acid.
