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.
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.
1 answer
1 answer