4. Use the reaction: 3 NaOH + H3PO4  Na3PO4 + 3 H2O a. How many mL of 0.225 M NaOH will react with 4.568 gram H3PO4? 54 b. How many mL of 0.385 M H3PO4 react with 50.0 mL of 0.404 M NaOH? c. What is the maximum amount of Na3PO4 that is formed from the reaction of 25.00 mL of 0.1050 M NaOH and 15.00 mL of 0.08650 M H3PO4? d. What is the molarity of Na3PO4 in the solution prepared in c?

a. 621.5 mL

b. 17.5 mL

c. 0.875 moles of salt

d. 0.022 M

Explanation:

3 moles of sodium hydroxide react to 1 mol of phosphoric acid in order to produce 1 mol of sodium phosphate and 3 moles of water. The reaction is:

3 NaOH + H₃PO₄  →  Na₃PO₄  +  3H₂O

a. Ratio is 1:3. We convert the mass of the acid, to moles:

4.568 g / 97.994 g/mol = 0.0466 moles

1 mol of acid react to 3 moles of hydroxide, therefore

0.0466 moles will react to (0.0466 . 3) /1 = 0.140 moles

Molarity (M) → mmol / mL

We convert mol to mmol → 0.140 mol . 1000 mmol / 1mol = 140 mmol

mL = mmol / M → 140 mmol / 0.225M = 621.5 mL

b. Ratio is 3:1. We need to determine, the moles of base.

50 mL . 0.404 M = 20.2 mmol → 0.0202 mol of NaOH

3 moles of hydroxide react to 1 mol of acid.

20.2 mmol of hydroxide would react to (20.2 . 1) / 3 = 6.73 mmol

mL = mmol / M → 6.73 mmol / 0.385M = 17.5 mL

c. First of all, we need to find out the limiting reactant:

25 mL . 0.1050M = 2.625 mmoles of base

15mL . 0.08650M = 1.2975 mmoles of acid.

1 mol of acid need 3 moles of base, for reaction

1.2975 mmol of acid would need (1.2975 . 3) /1 = 3.8925 mmol of base

We do not have enough base, so this is the limiting. (we need 3.8925 mmoles and we only have 2.625 mmoles)

So now, we propose:

3 moles of base can produce 1 mol of salt

2.625 mmoles of base would produce (2.625 . 1) /3 = 0.875 mmoles

d. To find out molarity, we need the total volume of solution:

25 mL + 15 mL = 40 mL

0.875 mmoles of salt / 40mL of solution = 0.022 M