A 10.0 mL sample of 3.00 M KOH(aq) is transferred to a 250.0 ml
volumetric flask and diluted to the mark. It was found that 38.5 ml of
this diluted solution was needed to react the stoichimetric point in a
titration of 10.0 mL of a phosphoric acid, H3PO4, solution. The reaction
is:
3 KOH(aq) + H3PO4(aq) ® K3PO4(aq) + 3 H2O(l)
a) Calculate the molarity of H3PO4 in the original solution.
The diluted solution is 3.00 x (10/250( = approx 0.12M.
mols KOH taken for titration will be 0.12M x 0.0385 L = o.00462 KOH
Convert mols KOH to mols H3PO4 using the coefficients in the balanced titration equation.
This problem is worded poorly. It is worded to make one think we want the original solution KOH BEFORE it is diluted
0.00462 x (1 mol H3PO4/3 mol KOH) = ?
Then M H3PO4 = mols H3PO4/L H3PO4 = ?
This problem is worded poorly. It is worded to make one think we want the concn of KOH BEFORE dilution BUT that isn't true since we alredy have that AND the problem specifically says H3PO4. But there is only one solution of H3PO4 and that IS the original.
To determine the molarity of H3PO4 in the original solution, we need to use the stoichiometry of the reaction and the volume information provided.
First, let's calculate the number of moles of KOH used in the titration:
Moles of KOH = Molarity of KOH x Volume of KOH
Moles of KOH = 3.00 M x 38.5 mL / 1000 mL/L
Moles of KOH = 0.1155 moles
Since the reaction is 3 KOH(aq) + H3PO4(aq) -> K3PO4(aq) + 3 H2O(l), we know that the moles of H3PO4 used in the titration are equal to the moles of KOH used.
Now we can calculate the molarity of H3PO4 in the original solution:
Molarity of H3PO4 = Moles of H3PO4 / Volume of H3PO4 (in liters)
Volume of H3PO4 = 10.0 mL / 1000 mL/L = 0.0100 L
Molarity of H3PO4 = 0.1155 moles / 0.0100 L
Molarity of H3PO4 = 11.55 M
Therefore, the molarity of H3PO4 in the original solution is 11.55 M.
To calculate the molarity of H3PO4 in the original solution, we can use the stoichiometry of the reaction and the volume of the solutions involved.
1. Start by converting the volume of the diluted KOH solution used in the titration to liters:
- 38.5 mL = 38.5/1000 = 0.0385 L
2. Use the stoichiometry of the reaction to determine the moles of H3PO4 reacted:
- From the balanced equation, the ratio of KOH to H3PO4 is 3:1.
- Since the volume of the H3PO4 solution used in the titration is the same as the volume of KOH used (10.0 mL), the moles of H3PO4 reacted are:
Moles of H3PO4 = 3/1 * Moles of KOH = 3/1 * Concentration of KOH * Volume of KOH used
Moles of H3PO4 = 3/1 * 3.00 M * 0.0385 L
3. Calculate the moles of H3PO4 in the original solution:
- The diluted solution in the volumetric flask was prepared by diluting a 10.0 mL sample, so the moles of H3PO4 in the diluted solution are equal to the moles in the original solution:
Moles of H3PO4 (original) = Moles of H3PO4 (diluted)
4. Determine the volume of the original H3PO4 solution:
- The volume of the original solution is the volume of KOH used in the titration, which is 10.0 mL.
5. Calculate the molarity of H3PO4 in the original solution:
Molarity of H3PO4 = Moles of H3PO4 (original) / Volume of H3PO4 (original)
Now we can calculate the molarity of H3PO4 in the original solution by substituting the values:
Molarity of H3PO4 = (3/1 * 3.00 M * 0.0385 L) / 10.0 mL
Note: It is important to ensure that all volumes are expressed in the same units (L or mL) for accurate calculations.