Question

use this equation for the next two problems. h₃po₄(aq) + 3naoh(aq) → na₃po₄(aq) + 3h₂o(l)

1. how much (in ml) 0.20 m h₃po₄ is needed to react with 100 ml of 0.10 m naoh?
2. if 36.0 ml of h₃po₄ reacts exactly with 80.0 ml of 0.500 m naoh, what is the concentration of the phosphoric acid?

Explanation:

for Problem 9:

Step1: Calculate moles of NaOH

Moles of NaOH = \( M_{NaOH} \times V_{NaOH} = 0.10\ \text{M} \times 0.100\ \text{L} = 0.010\ \text{mol} \)

Step2: Relate moles of NaOH to H₃PO₄

From reaction: 1 mol H₃PO₄ reacts with 3 mol NaOH.
Moles of H₃PO₄ = \( \frac{0.010\ \text{mol NaOH}}{3} = 0.00333\ \text{mol} \)

Step3: Calculate volume of H₃PO₄

\( V_{H_3PO_4} = \frac{\text{Moles of H}_3\text{PO}_4}{M_{H_3PO_4}} = \frac{0.00333\ \text{mol}}{0.20\ \text{M}} = 0.01665\ \text{L} = 17\ \text{mL} \)

for Problem 10:

Step1: Calculate moles of NaOH

Moles of NaOH = \( M_{NaOH} \times V_{NaOH} = 0.500\ \text{M} \times 0.0800\ \text{L} = 0.0400\ \text{mol} \)

Step2: Relate moles of NaOH to H₃PO₄

Moles of H₃PO₄ = \( \frac{0.0400\ \text{mol NaOH}}{3} = 0.01333\ \text{mol} \)

Step3: Calculate concentration of H₃PO₄

\( M_{H_3PO_4} = \frac{\text{Moles of H}_3\text{PO}_4}{V_{H_3PO_4}} = \frac{0.01333\ \text{mol}}{0.0360\ \text{L}} = 0.370\ \text{M} \)

Answer:

Problem 9: 17 mL
Problem 10: 0.370 M