A 25.00-mL sample of an unknown H₃PO₄ solution is titrated with a 0.150 M NaOH solution. The equivalence point is reached when 22.97 mL of NaOH solution is added. What is the concentration of the unknown H₃PO₄ solution? The neutralization reaction is H₃PO₄(aq) + 3 NaOH(aq) → 3 H₂O(l) + Na₃PO₄(aq)

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Hey everyone. So today we're being asked to find the concentration of H three P. 04 or phosphoric acid in the solution. Now this is a tight tray shin example But it's not your typical titrate example because there's one stark difference than a usual one and that is phosphoric acid. Now phosphoric acid is a tri protic acid, which means when it reacts with the base It will provide three acidic hydrogen. And this is important because even though there is a 123 moller ratio between acid and base between phosphoric acid and sodium hydroxide, There are actually three hydrogen is that can participate in a reaction at once. So let's write that down H three p. 04 and we'll write that blue H three p- is try pro tech. Let me clear that up a bit. It is try product and we're trying to figure out what is the concentration of H three P 04 phosphoric acid. Used to reach the equivalence point where the molds of acid more. 8 3 pm four is equal to the moles of the base malls of N. A. O. H. So with that in mind let's go ahead and write down our variables, We know that we have a volume for the acid of milliliters. 20 points or Middle l a volume or a polarity for the base of zero, What is it? 0.15 Molar and the volume of the base of .54 ml. So let's write out of the equation. We have M. A. V. A. For more clarity and volume of the acid would be equal to the polarity. The base that is the volume of the base. Let me write that a little clear. Huh? That is volume. And keep in mind because it's try product Every reaction for the acid will happen three times theoretically. It's a little weird to explain. But now rearranging we get that M. A. Is equal to majority of the base times the volume of the base divided by three times the volume of the acid. And we already have these values. So let's go ahead and plug them in. This will be 0.150 moller times 21. ml divided by three because it's try product Times 20 ml. And this will give us a final value Of 0.04385 Molar. Therefore the unknown concentration of phosphoric acid in our solution that was used to reach the equivalent point equivalence point sorry, 0.4385 molar. I hope this helps. And I look forward to seeing you in the next one

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Key Concepts

Titration

Stoichiometry

Molarity

A 25.00-mL sample of an unknown H3PO4 solution is titrated with a 0.150 M NaOH solution. The equivalence point is reached when 22.97 mL of NaOH solution is added. What is the concentration of the unknown H3PO4 solution? The neutralization reaction is H3PO4(aq) + 3 NaOH(aq) → 3 H2O(l) + Na3PO4(aq)

Verified Solution

Key Concepts

Titration

Stoichiometry

Molarity