Matt P asked in Science & MathematicsChemistry · 5 years ago

# How many mLs of 0.0005 M H3PO4 are needed to neutralize 75 mL of 0.006 M Ba(OH)2?

The answer key I have here says 600 mL, but I think the answer is 1.35 L. Is this right?

Relevance
• 5 years ago

Balanced equation :

3Ba(OH)2 + 2H3PO4 → Ba3(PO4)2 + 6H2O

3mol Ba(OH)2 reacts with 2 mol H3PO4

Mol Ba(OH)2 in 75mL of 0.006M solution = 75/1000*0.006 = 4.5*10^-4mol

This will react with (4.5*10^-4) *2/3 = 3*10^-4 mol H3PO4

1000mL of H3PO4 solution contains 0.0005 mol H3PO4

Volume that contains 3*10^-4 mol H3PO4 = (3*10^-4) / 0.0005 *1000 = 600mL

• Matt P5 years agoReport

Doh, I forgot to write a balanced eq.... I was trying to do it by the number of H+/OH- ions in each. Thanks!

• BB
Lv 7
5 years ago

3Ba(OH)2 (aq) + 2H3PO4 (aq) ---> Ba3(PO4)2 (aq) + 6H2O (l)

0.075 L * 0.006 mol/L Ba(OH)2/L = 0.00045 moles Ba(OH)2.

0.00045 moles Ba(OH)2 is neutralised by 0.0003 moles H3PO4.

1000 mL 0.0005 M H3PO4 contains 0.0005 moles H3PO4.

0.0003 / 0.0005 x 1000 mL = 600 mL.