1. Suppose that a 3.00 mL aliquot of the product mixture was titrated with 0.25

Question

1.       Suppose that a 3.00 mL aliquot of the product mixture was titrated with 0.25 M NaOH and required 35.07 mL to reach the endpoint.

a)      What is the concentration of acetic acid [CH3COOH] in the product mixture?

b)      What is the concentration of the ester [CH3COOC3H7] in the product mixture?

Suppose that a 3.00 mL aliquot of the product mixture was titrated with 0.25 M NaOH and required 35.07 mL to reach the endpoint. What is the concentration of acetic acid [CH3COOH] in the product mixture? What is the concentration of the ester [CH3COOC3H7] in the product mixture?

Explanation / Answer

CH3COOH+ NaOH ---> CH3COONa + H2O

so balancing the number of moles,

3*x=0.25*35.07

or x=2.9225 M

b)concnetration of the ester formed=(3*2.9225)/(3+35.07)

=0.23 M

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