How much 3.78 M H₃PO₄ is required to neutralize 39 mL of 0.71 M Ba(OH)₂?
have to do this for chemistry and I don't understand it, please help!
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have to do this for chemistry and I don't understand it, please help!
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Verified answer
2 H₃PO₄ + 3 Ba(OH)₂ → Ba₃(PO4)₂ + 6 H2O
(39 mL) x (0.71 M Ba(OH)₂) x (2 mol H₃PO₄ / 3 mol Ba(OH)₂) / (3.78 M H₃PO₄) =
4.9 mL H₃PO₄
[Sir Mordred has failed to account for the fact that H₃PO₄ and Ba(OH)₂ don't react in equimolar amounts.]