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if 10.57 mL of an acetic Acid solution required 10.54 mL 0.9607 M NaOH to reach...

if 10.57 mL of an acetic Acid solution required 10.54 mL 0.9607 M NaOH to reach the endpoint, whats the concentration?

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Answer #1

Write the reaction between NaOH and acetic acid

CH3COOH + NaOH --- > CH3COONa + H2O

Mol ratio between NaOH and CH3COOH is 1 : 1

Step 1 calculate moles of NaOH

n NaOH = volume in L x molarity

0.01054*0.9607=0.01013 mol

Step 2 Calculate the moles of CH3COOH

n CH3COOH = moles of NaOH x 1 mol CH3COOH/ 1 mol NaOH

0.01013*1/1=0.01013 mol CH3COOH

Step 3 use volume of acid to calculate concentration

[CH3COOH]= 0.01013 mol / 0.01057 L

0.01013/0.01057=0.96 M

Concentration of acetic acid is 0.96 M

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