Answer:
14.3 mL
Explanation:
Assume the student used 0.113 g ascorbic acid and 0.0900 mol·L⁻¹ NaOH.
1. Balanced chemical equation.
The formula of ascorbic acid is H₂C₆H₆O₆ (MM = 176.12 g/mol).
However, for the balanced equation, let's write it as H₂A.
[tex]\rm H_{2}A + 2NaOH \longrightarrow Na_{2}A + 2H_{2}O[/tex]
2. Moles of ascorbic acid
[tex]\text{Moles of H$_{2}$A} =\text{0.113 g H$_{2}$A} \times \dfrac{\text{1 mmol H$_{2}$A}}{\text{0.176 12 mg H$_{2}$}A} = \text{0.6416 mmol H$_{2}$A}[/tex]
3. Moles of NaOH
The molar ratio is 2 mmol NaOH:1 mmol H₂A.
[tex]\text{Moles of NaOH}= \text{0.6416 mmol H$_{2}$A} \times \dfrac{\text{2 mmol NaOH}}{\text{1 mmol H$_{2}$A}} =\text{1.283 mmol NaOH}[/tex]
4. Volume of NaOH
[tex]V = \text{1.283 mmol NaOH}\times \dfrac{\text{1 mL NaOH}}{\text{0.0900 mmol NaOH}} = \textbf{14.3 mL NaOH}\\\\\text{The student will need $\large \boxed{\textbf{14.3 mL NaOH}}$}[/tex]
