Question

9. A solution of NaOH is standardized with potassium acid phthalate (KHP), KHC,H,O, molar mass = 204 g/mol. If 2.550 g of KHP is titrated with 54.50 mL of the NaOH solution, what is the molarity of NaOH?

please show me how to answer this

Answer 1

NaOH(aq) + KHC8H4O4(aq) = KNaC8H4O4(aq) + H2O(l)

Given that mass of KHP = 2.550 g

Number of moles = amount in g / molar mass

= 2.550 g / 204 g/ mole

= 0.0125 moles KHP

Now calculate the moles of NaOH which recats with KHP as follows:

0.0125 moles KHP * 1 mole NaOH /1 Mole KHP

= 0.0125 moles NaOH

Molarity = number of moles / volume in L

= 0.0125 moles NaOH/54.50 ml *1 L/1000 ml

= 0.2294 M