Question

A chemistry student weighs out 0.306 g of citric acid (H3C6H5O7), a triprotic acid, into a 250 ml volumetric flask and dilutes to the mark with distilled water. He plans to titrate the acid with 0.1000 M NaOH solution. Calculate the volume of NaOH solution the student will need to add to reach the final equivalence point. Be sure your answer has the correct number of significant digits.

Answer 1

Answer:

$\boxed{\text{47.8 mL of NaOH}}$

Explanation:

For simplicity, let's write the formula of citric acid as H₃A.

1. Balanced chemical equation.

$\rm H _{3} A + 3NaOH \longrightarrow Na _{3} A + 3H _{2} O$

2. Moles of H₃A

$\text{Moles of H _{3} A} =\text{ 0.306 g H _{3} A} \times \dfrac{\text{1 mol H _{3} A}}{\text{192.12 g H _{3} A }} = 1.593 \times 10^{-3} \text{ mol H _{3} A}$

3. Moles of NaOH.

$\text{Moles of NaOH} = 1.593 \times 10^{-3} \text{ mol H _{3} A} \times \dfrac{\text{3 mol NaOH} }{\text{1 mol H _{3} A}}\\= 4.778 \times 10^{-3}\text{ mol NaOH}$

4. Volume of NaOH

$V = 4.778 \times 10^{-3}\text{ mol NaOH}\times \dfrac{\text{1 L NaOH}}{\text{0.1000 mol NaOH}} = \text{0.047 78 L NaOH} =\textbf{47.8 mL NaOH}\\\\\text{The student will have to use }\boxed{\textbf{47.8 mL of NaOH}}$