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

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SUMMARY

The concentration of an unknown H3PO4 solution was determined to be 0.293 M after reacting with 110 mL of 0.120 M KOH. The initial approach incorrectly calculated the molarity by misapplying the mole-to-mole conversion and not properly accounting for the stoichiometry of the reaction H3PO4 + 3KOH → 3H2O + K3PO4. The correct method involves recognizing that 15 mL of H3PO4 corresponds to 0.015 L, and using the stoichiometric ratio of 1 mole of H3PO4 to 3 moles of KOH to find the accurate concentration.

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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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