Potassium hydrogen phthalate (molar mass = 204.2g/mol) is used to standardize sodium hydroxide. If 26.37 mL of NaOH(aq) is required to titrate 0.7719g KHP to the equivalence point, what is the concentration of the NaOH(aq)

Question

(26.37 mL NaOH)(x) = (1 mol KHP/.7719g KHP)(204.2g KHP/mol KHP)

10.03M NaOH

Answer 1

I think it would be easier to do this in steps. I think you are confusing yourself in trying to do it all in one step.
mols KHP = 0.7719 g x (1 mol/204.2 g) = ?? (which is the inverse of what you have written. Look at the units for the right side of what you wrote. mols cancels with mols and g cancels with g so the right hand side of your equation, which should come out in moles, is unitless).
mols NaOH = same thing (that's when the indicator tells you that mols of one = moles of the other---to be more exact when the equivalents of one = equivalents of the other).
mols NaOH = M x L.
You know mols NaOH and L, calculate M.
I don't get anything close to 10 M.

Answer 2

.0143 mol