## Chemistry (7th Edition)

1. Find the number of moles of $NaOH$: $\frac{0.200mol (NaOH)}{1L} \times 94.7mL\times \frac{1L}{1000mL} = 0.01894$ $mol (NaOH)$ 2. Use the stoichiometry of the reaction to find the number of moles of acetic acid: The coefficients indicate a ratio of 1 to 1. $0.01894mol(NaOH) \times \frac{1mol(CH_3CO_2H)}{1mol(NaOH)} = 0.01894mol(CH_3CO_2H)$ 3. Find the concentration, using the value for the volume of vinegar used. $\frac{0.01894mol(CH_3CO_2H)}{25mL} \times \frac{1000mL}{1L} = 0.758 mol/L (CH_3CO_2H)$