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…

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?\nh₃c₆h₅o₇ + 3lioh → li₃c₆h₅o₇ + 3h₂o\n? m h₃c₆h₅o₇

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?\nh₃c₆h₅o₇ + 3lioh → li₃c₆h₅o₇ + 3h₂o\n? m h₃c₆h₅o₇

Answer

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} \rightarrow \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)