Question

1. if 36.0 ml. of $h_{3}po_{4}$ reacts exactly with 80.0 ml. of 0.500 m naoh, what is the concentration of the phosphoric acid?

answer
(with correct significant figures and units)
image of hand done
calculations
$2hcl(aq)+zn(s)\to zncl_{2}(aq)+h_{2}(g)$

Explanation:

Step1: Write balanced reaction

$$\text{H}_3\text{PO}_4 + 3\text{NaOH} ightarrow \text{Na}_3\text{PO}_4 + 3\text{H}_2\text{O}$$

Step2: Calculate moles of NaOH

$$n_{\text{NaOH}} = M \times V = 0.500\ \text{mol/L} \times 0.0800\ \text{L} = 0.0400\ \text{mol}$$

Step3: Find moles of H₃PO₄

$$n_{\text{H}_3\text{PO}_4} = \frac{n_{\text{NaOH}}}{3} = \frac{0.0400\ \text{mol}}{3} \approx 0.01333\ \text{mol}$$

Step4: Compute concentration of H₃PO₄

$$M_{\text{H}_3\text{PO}_4} = \frac{n}{V} = \frac{0.01333\ \text{mol}}{0.0360\ \text{L}} \approx 0.370\ \text{M}$$

Answer:

0.370 M