I have estimated all of these calculation so it will be necessary for you recalculate all of them.
I assume you mean to calculate the molarity of the H3PO4 solution but that isn't clear in the problem. mols KOH initially = M x L = about 0.02
You diluted this to 250 mL and drew a 32.8 mL aliquot so mols in the 32.8 mL is
about 0.02 x 32.8/250 = about 0.002,
Convert mols KOH to mols H3PO4 using the coefficients in the blanced equation you gave. That's about 0.002 mols KOH x (1 mols H3PO4/3 mol KOH) = about 0.002/3 - approx 0.0007 mols H3PO4.
So concn H3PO4 = mols H3PO4/L H3PO4 = about 0.0007/0.006.9 = ?
If instead of M H3PO4 you wanted M KOH in the diluted solution it is M = mols KOH/LKOH = abouat 0.02/0.250 = ?
Part B. You have M H3PO4. You want to know grams H3PO4 in the 6.9 mL
mols H3PO4 = M x L = ?
mols H3PO4 = grams/molar mass. You know mol and molar mass, substitute and solve for grams.
Post your work if you get stuck.
PART A)
Suppose that 6.9 mL of 2.5 M KOH(aq)
is transferred to a 250 mL volumetric flask
and diluted to the mark. It was found that
32.8 mL of this diluted solution was
needed to reach the stoichiometric point in a
titration of 6.9 mL of a phosphoric acid solution
according to the reaction
3 KOH(aq) + H3PO4(aq) →
K3PO4(aq) + 3 H2O(ℓ)
Calculate the molarity of the solution.
Answer in units of M.
PART B)
What mass of H3PO4 is in the sample of the
acid solution?
Answer in units of g.
1 answer
1 answer