15.0 mL of a 1:5 diluted wine sample plus 20 mL deionized water and 3 drops of phenolphthalein were mixed and titrated to completion using 23.86 mL of 0.0784 M NaOH solution. The density of the original wine sample was 1.095 g/mL. A) What is the molarity of the acid in the original wine sample? B) What is the percent by mass of acid in the original wine sample?With complete solutions.
Answer (1)
Answer:Here's how to solve this problem step-by-step. We'll assume the wine's acidity is due to tartaric acid (C₄H₆O₆), a common acid in wine, which is diprotic (meaning it donates two protons per molecule). If a different acid is assumed, the final answer will change. A) Molarity of the acid in the original wine sample: Step 1: Moles of NaOH used: - Moles of NaOH = (Volume of NaOH in Liters) × (Molarity of NaOH) - Moles of NaOH = (23.86 mL × (1 L / 1000 mL)) × 0.0784 mol/L - Moles of NaOH = 0.001868 mol Step 2: Moles of tartaric acid in the diluted sample: Since tartaric acid is diprotic, it reacts with NaOH in a 1:2 mole ratio. - Moles of tartaric acid (diluted) = 0.001868 mol NaOH × (1 mol tartaric acid / 2 mol NaOH) - Moles of tartaric acid (diluted) = 0.000934 mol Step 3: Molarity of tartaric acid in the diluted sample: - Total volume of diluted sample = 15.0 mL + 20.0 mL = 35.0 mL = 0.035 L - Molarity of tartaric acid (diluted) = Moles of tartaric acid (diluted) / Total volume of diluted sample (L) - Molarity of tartaric acid (diluted) = 0.000934 mol / 0.035 L - Molarity of tartaric acid (diluted) = 0.0267 M Step 4: Molarity of tartaric acid in the original wine sample: The wine sample was diluted 1:5. This means the original wine was 5 times more concentrated. - Molarity of tartaric acid (original) = Molarity of tartaric acid (diluted) × 5 - Molarity of tartaric acid (original) = 0.0267 M × 5 - Molarity of tartaric acid (original) = 0.1335 M Answer A: The molarity of the acid in the original wine sample is approximately 0.1335 M. B) Percent by mass of acid in the original wine sample: Step 1: Mass of tartaric acid in 1 mL of original wine: - Mass of tartaric acid = (Molarity of tartaric acid (original) × Molar mass of tartaric acid) - Mass of tartaric acid = (0.1335 mol/L × 150.09 g/mol) (Molar mass of C₄H₆O₆) - Mass of tartaric acid = 20.04 g/L = 0.02004 g/mL Step 2: Mass of 1 mL of original wine: Given density of original wine = 1.095 g/mL Step 3: Percent by mass of tartaric acid: - % by mass = (Mass of tartaric acid in 1 mL / Mass of 1 mL of wine) × 100% - % by mass = (0.02004 g / 1.095 g) × 100% - % by mass = 1.83% Answer B: The percent by mass of acid in the original wine sample is approximately 1.83%. Important Note: This calculation assumes that the only acid present in the wine is tartaric acid and that it is the only substance reacting with the NaOH. Real wine is a complex mixture, and other acids may be present. This calculation provides a reasonable approximation, but may not be perfectly accurate.
