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

[9781133886563]

Homework Statement

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.

 

[Saitama]

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

Check this again.

 

[Borek]

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

 

[9781133886563]

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

Thank you both.

 

[DeeNos]

It seems that you may have made a mistake in step 3 of your solution. The mole ratio for Ba(OH)2 and H3PO4 is 3:2, meaning that for every 3 moles of Ba(OH)2, 2 moles of H3PO4 are needed. Therefore, the correct calculation would be 2 x 3.45 mol = 6.9 mol of Ba(OH)2 needed. Plugging this into the M = mol/L equation, we get mL = 6.9 mol / 0.0521 M = 132.4 mL. However, since the question asks for the volume of 0.0521 M Ba(OH)2, we need to divide this answer by 2 to get the final answer of 99.5 mL. It is always important to double check your mole ratios when using stoichiometry to avoid mistakes.