The distinctive odor of vinegar is due to acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) which reacts with sodium hydroxide in the following fashion: $$\mathrm{CH}_{3} \mathrm{COOH}(a q)+\mathrm{NaOH}(a q) \longrightarrow$$ If \(3.45 \mathrm{~mL}\) of vinegar needs \(42.5 \mathrm{~mL}\) of \(0.115 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point in a titration, how many grams of acetic acid are in a \(1.00-\) qt sample of this vinegar?

We are given the balanced chemical reaction equation: $$CH_3COOH_{(aq)} + NaOH_{(aq)} \longrightarrow CH_3COONa_{(aq)} + H_2O_{(l)}$$

We are given that 42.5 mL of 0.115 M NaOH is needed to reach the equivalence point in a titration. To determine the number of moles of NaOH used, we can multiply the volume of the NaOH solution by its molarity (concentration). Make sure to convert the volume from mL to L (divide by 1000): $$n_{NaOH} = 42.5 \times 10^{-3}\,L \times 0.115\,M = 0.0048875\,mol$$

Since the stoichiometry of the balanced chemical equation tells us that one mole of acetic acid reacts with one mole of NaOH, the number of moles of acetic acid present in the vinegar is equal to the number of moles of NaOH used in the titration. Thus, we have: $$n_{CH_3COOH} = n_{NaOH} = 0.0048875\,mol$$

The volume of the vinegar titrated is given as 3.45 mL. To find the concentration (molarity) of acetic acid in the vinegar, we can use the equation: $$M = \frac{n}{V}$$ where \(M\) is the molarity, \(n\) is the amount of moles, and \(V\) is the volume. Make sure to convert the volume from mL to L (divide by 1000): $$M_{CH_3COOH} = \frac{0.0048875\,mol}{3.45 \times 10^{-3}\,L} = 1.4174\,M$$

We have found the molarity of acetic acid in the vinegar sample to be 1.4174 M. Now, we need to determine the grams of acetic acid in a 1.00-quart sample. We need to convert the volume from quarts to liters (1 qt = 0.946353 L): $$V_{sample} = 1.00\,qt \times 0.946353\,\frac{L}{qt} = 0.946353\,L$$ Next, we can convert the molarity of acetic acid into the moles of acetic acid present in a 1.00-quart sample: $$n_{CH_3COOH_{sample}} = 1.4174\,M \times 0.946353\,L = 1.3416\,mol$$ Lastly, we can convert the moles of acetic acid to grams by multiplying the moles by the molar mass of acetic acid (\(MM_{CH_3COOH} = 60.05\,\frac{g}{mol}\)): $$m_{CH_3COOH_{sample}} = 1.3416\,mol \times 60.05\,\frac{g}{mol} = 80.56\,g$$ Therefore, there are 80.56 grams of acetic acid in a 1.00-quart sample of this vinegar.