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

  • Thread starter land_of_ice
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In summary, the concentration of the unknown H3PO4 solution can be calculated by using the mole to mole conversion of 3 moles of KOH to every 1 mole of H3PO4 and the volume of the acid solution. The correct concentration is .293 Molarity, not the previously calculated .12 Molarity.
  • #1
land_of_ice
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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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  • #2
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.
 
  • #3
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.

--
 

1. What is solution stoichiometry?

Solution stoichiometry is the branch of chemistry that deals with calculations involving the amount of substances in a solution and their relationships with one another.

2. How is solution stoichiometry different from regular stoichiometry?

Solution stoichiometry involves calculations in a solution, while regular stoichiometry deals with the amounts of substances in a chemical reaction. Solution stoichiometry also takes into account the properties of solvents and solutions, while regular stoichiometry focuses on the amounts of reactants and products.

3. What is the difference between molarity and molality?

Molarity is the concentration of a solution measured in moles of solute per liter of solution, while molality is the concentration of a solution measured in moles of solute per kilogram of solvent. Molarity takes into account the volume of the solution, while molality takes into account the mass of the solvent.

4. How do you calculate the concentration of a solution?

The concentration of a solution can be calculated using the formula: concentration (in mol/L) = moles of solute / liters of solution. This can also be written as M = n/V, where M is the molarity, n is the number of moles of solute, and V is the volume of the solution in liters.

5. What are some common units used in solution stoichiometry?

Some common units used in solution stoichiometry include moles (mol), liters (L), grams (g), and milliliters (mL). Other units that may be used include molarity (M) and molality (m).

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