Question

petra titrates 50.0 ml of an orange juice sample with 65.0 ml of 0.20 m lioh to reach the final endpoint in the titration. what is the concentration of the citric acid in the orange juice sample?
h₃c₆h₅o₇ + 3lioh → li₃c₆h₅o₇ + 3h₂o
? m h₃c₆h₅o₇

Explanation:

Step1: Find moles of LiOH

Molarity formula: \( M = \frac{n}{V} \) (where \( M \) is molarity, \( n \) is moles, \( V \) is volume in liters).
Convert LiOH volume to liters: \( 65.0 \, \text{mL} = 0.0650 \, \text{L} \).
Moles of LiOH: \( n_{\text{LiOH}} = M \times V = 0.20 \, \text{M} \times 0.0650 \, \text{L} = 0.013 \, \text{mol} \).

Step2: Relate moles of LiOH to citric acid

From the reaction: \( \text{H}_3\text{C}_6\text{H}_5\text{O}_7 + 3\text{LiOH}
ightarrow \text{Li}_3\text{C}_6\text{H}_5\text{O}_7 + 3\text{H}_2\text{O} \), the mole ratio of \( \text{H}_3\text{C}_6\text{H}_5\text{O}_7 \) to \( \text{LiOH} \) is \( 1:3 \).
So moles of citric acid: \( n_{\text{citric acid}} = \frac{n_{\text{LiOH}}}{3} = \frac{0.013 \, \text{mol}}{3} \approx 0.004333 \, \text{mol} \).

Step3: Calculate concentration of citric acid

Volume of orange juice: \( 50.0 \, \text{mL} = 0.0500 \, \text{L} \).
Molarity of citric acid: \( M = \frac{n}{V} = \frac{0.004333 \, \text{mol}}{0.0500 \, \text{L}} \approx 0.0867 \, \text{M} \) (or more precisely, \( \frac{0.013/3}{0.0500} = \frac{0.20 \times 0.0650}{3 \times 0.0500} \approx 0.0867 \, \text{M} \)).

Answer:

\( \approx 0.087 \, \text{M} \) (or more accurately, \( \frac{0.20 \times 65.0}{3 \times 50.0} = \frac{13}{150} \approx 0.0867 \, \text{M} \), rounded to two decimal places or significant figures as needed)