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: Recall the titration formula

In a titration, the moles of acid and base are related by the stoichiometry of the reaction. The formula is \(n_1M_1V_1 = n_2M_2V_2\), where \(n\) is the stoichiometric coefficient, \(M\) is the molarity, and \(V\) is the volume. For the reaction \(\ce{H3C6H5O7 + 3LiOH -> Li3C6H5O7 + 3H2O}\), the stoichiometric coefficient of \(\ce{H3C6H5O7}\) (\(n_1\)) is 1 and for \(\ce{LiOH}\) (\(n_2\)) is 3. Let \(M_1\) be the molarity of \(\ce{H3C6H5O7}\), \(V_1 = 50.0\space mL\), \(M_2 = 0.20\space M\), and \(V_2 = 65.0\space mL\).

Step2: Rearrange the formula to solve for \(M_1\)

From \(n_1M_1V_1 = n_2M_2V_2\), we can solve for \(M_1\) as \(M_1=\frac{n_2M_2V_2}{n_1V_1}\).

Step3: Substitute the values

Substitute \(n_1 = 1\), \(n_2 = 3\), \(M_2 = 0.20\space M\), \(V_2 = 65.0\space mL\), and \(V_1 = 50.0\space mL\) into the formula:
\(M_1=\frac{3\times0.20\space M\times65.0\space mL}{1\times50.0\space mL}\)
First, calculate the numerator: \(3\times0.20\times65.0 = 39\)
Then, divide by the denominator: \(\frac{39}{50.0}= 0.078\space M\)

Answer:

\(0.078\)