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Oduvanchick [21]
3 years ago
5

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 ad

ded, 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)
Chemistry
1 answer:
kirill115 [55]3 years ago
6 0

Answer:

0.5 M

Explanation:

We have to start with the <u>reaction</u> between NaOH and CH3COOH:

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

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

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 <u>concentration of the acid</u>.

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 <u>strong acid</u>.

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