A 0.981-gram sample of potassium hydrogen phthalate, KHC8H4O4, is dissolved in 100 mL of water. If 18.9 mL NaO

Question

A 0.981-gram sample of potassium hydrogen phthalate, KHC8H4O4, is dissolved in 100 mL of water. If 18.9 mL NaOH are required to reach the stoichiometric point, what is the molarity of the sodium hydroxide? the formula weight of KHC8H4O4 is 204.2 g/mol

Answer 1

n(KHC8H4O4) = 0.981 / 204.2 = 0.00480 mol.

The reaction is
NaOH(aq) + KHC8H4O4(aq) -> NaKC8H4O4(aq) + H2O(l)

Thus n(NaOH) = n(KHC8H4O4) = 0.00480 mol.
So [NaOH] = 0.00480 / 0.0189 = 0.254 mol / L.

Answer 2

Suggestions for you to try:

1. Convert grams of KHP to moles (you are given the MW)
2. Then relate moles KHP to moles NaOH (1:1 as I recall)
3. Figure out the moles of NaOH that would be neutralized by the amount of KHP added.
4. The problem tells you that 18.9 mL NaOH was used to reach
the stoic. end point. This is equal to the number of moles of NaOH you just calculated. Now take that info and get the NaOH
molarity.

Molarity = M = moles NaOH / Liters of NaOH solution
= (what you calc'd in 3) / 0.0189 L NaOH

Rock 'n roll. I mean, get to it.

Answer 3

500mL = 0.5L. M = moles per liter. So, x moles of KHP / 0.5L = 0.2000M solution Calculate x moles of KHP by multiplying 0.2 & 0.5. Then convert moles of KHP to grams (by multiplying by the molar mass of KHP.) The answer I got was 20.42g of KHP.