Posted by amanda1012 on .
A solution of the primary standard potassium hydrogen phthalate, KHC8H4O4, was prepared by dissolving 0.4877 g of potassium hydrogen phthalate in about 50 mL of water. Titration with a KOH solution required 36.21 mL to reach a phenolphthalein end point. What is the molarity of the KOH solution?
0.0024 M KHC8H4O4 was calculated.
I keep getting the wrong answer
KHC8H4O4 + KOH ==>K2C8H4O4 + H2O
mols phthalate - grams/molar mass = ?
mols KOH = same (look at the coefficients in the balanced equation.)
M KOH = mols KOH/L KOH
I get about 0.066 M.
Post your work and I'll find the error.
Oh I was doing 1:2 ratio instead of 1:1... That is where I went wrong! When I recalculated it with 1:1 I got 0.066628 M
My answer is 0.06595. I used 204.22 for the molar mass of the phthalate. To get that value you had to use something like 198 for the molar mass of the phthalate. There is either a typo or you used the wrong molar mass.