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Answer:
[tex]\large \boxed{\text{0.1767 mol/L}}$[/tex]
Explanation:
(a) Balanced equation
3NaOH + H₃PO₄ ⟶ Na₃PO₄ + 3H₂O
(b) Moles of NaOH
[tex]\text{Moles of NaOH} = \text{34.52 mL NaOH} \times \dfrac{\text{0.3840 mmol NaOH}}{\text{1 mL NaOH}} = \text{13.26 mmol NaOH}[/tex]
(c) Moles of H₃PO₄
The molar ratio is 1 mol H₃PO₄:3 mol NaOH.
[tex]\text{Moles of H$_{3}$PO}_{4} = \text{13.26 mmol NaOH} \times\dfrac{\text{ 1 mmol H$_{3}$PO}_{4}}{\text{3 mmol NaOH}}\\\\= \text{4.419 mmol H$_{3}$PO}_{4}[/tex]
(d) Molar concentration of H₃PO₄
[tex]c = \dfrac{\text{moles of solute}}{\text{litres of solution}}\\\\c = \dfrac{n}{V} = \dfrac{\text{4.419 mmol}}{\text{25.00 mL}} = \text{0.1767 mol $\cdot$ L$^{-1}$}\\\\\text{The molar concentration of the NaOH is $\large \boxed{\textbf{0.1767 mol/L}}$}[/tex]