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if 9.76 mL of an acetic Acid solution required 9.80 mL of 1.0962 M NaOH to...

if 9.76 mL of an acetic Acid solution required 9.80 mL of 1.0962 M NaOH to reach the endpoint, whats the concentration of the Acetic Avid Solution?

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Answer #1

We know the relationship , N1V1 = N2V2 and Normality, N = n*M Where n is replacable hydrogen/ OH and M is molarity

In the case of Acetic acid ( CH3COOH), one replacable hydrogen is present so N = 1* M =M

In the case of NaOH, one replacable OH is present so N = 1*M = M

From question, Volume of acetic acid, V1 (suppose) = 9.76 ml = 0.00976 L

Concentration of acetic acid in terms of Normality , N1(suppose) = ?

Volume of NaOH , V2 (Suppose) = 9.80 ml = 0.00980 L

Concentration of NaoH in terms of normaliy, N2 = 1.0962

Put all the know values in relationship , N1V1 = N2V2

N1 * 0.00976 L = 1.0962 * 0.00980 L

N1 = (1.0962 * 0.00980 L) / 0.00976 L

N1 = 1.1 N = 1.1 M of acetic acid

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