Solution stoichometry, attempted ANSWER IS WRONG, can anyone fix answer

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The discussion centers on calculating the concentration of an unknown H3PO4 solution that reacts with KOH. Initially, the approach taken involved converting 15 mL of H3PO4 to liters and using the concentration of KOH, but this led to an incorrect answer. The correct method requires recognizing that the 15 mL is the volume of the acid, not the base, and applying the stoichiometric relationship from the balanced equation. The correct concentration of the H3PO4 solution is determined to be 0.293 M. Accurate stoichiometric calculations are crucial for determining molarity in acid-base reactions.
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A 15.0 mL sample of an unknown H3PO4 solution requires 110 mL of 0.120 M KOH to completely react with the H3PO4 according the following reaction.

H3PO4 + 3KOH --------> 3H20 + K3PO4

What was the concentration of the unknown H3PO4 solution?

______________________________________…

This SEEMED correct way to start problem, as it turns out this is not how to set up the problem, (attempt at doing the problem) : 15mL converted that to L , got 0.015 L, then multiplied that by .12 moles of the solute KOH divided by 1 Liter of the solution H3PO4, used the mole to mole conversion 3 moles of KOH to every 1 mole of H3PO4 and got the wrong answer. It's supposed to be
.293 Molarity.

?
 
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land_of_ice said:
A 15.0 mL sample of an unknown H3PO4 solution requires 110 mL of 0.120 M KOH to completely react with the H3PO4 according the following reaction.

H3PO4 + 3KOH --------> 3H20 + K3PO4

What was the concentration of the unknown H3PO4 solution?

______________________________________…

This SEEMED correct way to start problem, as it turns out this is not how to set up the problem, (attempt at doing the problem) : 15mL converted that to L , got 0.015 L, then multiplied that by .12 moles of the solute KOH divided by 1 Liter of the solution H3PO4, used the mole to mole conversion 3 moles of KOH to every 1 mole of H3PO4 and got the wrong answer. It's supposed to be
.293 Molarity.

?

You have to do the three dissociation reactions separately.
 
land_of_ice said:
15mL converted that to L , got 0.015 L, then multiplied that by .12 moles of the solute KOH

15 mL is the volume of acid, not base.

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