# Homework Help: Stoichiometry, concentration of solutions: Ba(OH)2 and H3PO4

1. Mar 25, 2013

### 9781133886563

1. The problem statement, all variables and given/known data

What volume of 0.0521 M Ba(OH)2 is required to react completely with 24.50 mL of 0.141 M H3PO4? Phosphoric acid is a triprotic acid.

Answer: volume of 0.0521 M Ba(OH)2 required = 99.5 mL

2. The attempt at a solution
1. I begin by balancing the equation: 3Ba(OH)2 + 2H3PO4 → Ba3(PO4)2 + 6H2O

2. I use M = Molarity = mol solute / L solution to find the mols of H3PO4... mol = 0.141 × 24.50 mL = 3.45 mol

3. According to the mole ratio... 3 × 3.45 mol of Ba(OH)2 = 10.36 mol of Ba(OH)2 needed.

4. M = mol/L ... mL = 10.36 mol / .0521 M = 199.0 mL of Ba(OH)2 needed, which is a wrong answer.

These are the steps I am taking. I must be making a mistake since I know that the answer is 99.5 mL.

2. Mar 25, 2013

### Saitama

Check this again.

3. Mar 26, 2013

### Staff: Mentor

Note that your answer is exactly twice too large (which points in exactly the same direction Pranav-Arora alredy signaled).

4. Mar 26, 2013

### 9781133886563

I see it now. The mole ratio is 3/2.

Thank you both.