Answer to Question #136836 in General Chemistry for taylor cline

Question

Acetic acid (HC2H3O2) is an important ingredient of vinegar. A sample of 50.0 mL of a commercial vinegar is titrated against a 1.00 M NaOH solution. What is the concentration (in M) of acetic acid present in the vinegar if 5.75 mL of the base is needed for the titration?

Answer 1

/CH3COOH + NaOH = CH3COONa + H2O

N_CH3COOH * V_{solution CH3COOH} = N_NaOH * V_{solution NaOH}

N = C/f_eq, f_eq = 1, N=C.

C_CH3COOH * V_{solution CH3COOH} = C_NaOH * V_{solution NaOH}

C_CH3COOH = (C_NaOH * V_{solution NaOH})/V_{solution CH3COOH}

V_{solution NaOH} = 5,75 mL = 5,75 * 10^-3 L

V_{solution CH3COOH} = 50,0 ml = 50,0 * 10^-3 L

C_CH3COOH = (1,00 * 5,75 *10^-3)/(50,0 * 10^-3) = 0,115.