Question

A solution of the primary standard potassium hydrogen phthalate (KHP), KHC8H4O4 , was prepared by dissolving 0.4877 g of KHP in about 50 mL of water. Titration of the KHP solution with a KOH solution of unknown concentration required 23.22 mL to reach a phenolphthalein end point. What is the concentration of the KOH solution?

Answer 1

Answer: The concentration of KOH solution is 0.103 M.

Explanation:

To calculate the number of moles, we use the equation:

$\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}$

Given mass of KHP = 0.4877 g

Molar mass of KHP = 204.22 g/mol

Putting values in equation 1, we get:

$\text{Moles of KHP}=\frac{0.4877g}{204.22g/mol}=0.0024mol$

The chemical reaction for the reaction of KHP with KOH follows:

$KHC_8H_4O_4(aq.)+KOH\rightarrow K_2C_8H_4O_4(aq.)+H_2O(l)$

By Stoichiometry of the reaction:

1 mole of KHP reacts with 1 mole of KOH.

So, 0.0024 moles of KHP will react with = $\frac{1}{1}\times 0.0024=0.0024mol$ of KOH.

• To calculate the molarity of KOH, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Moles of solute}}{\text{Volume of solution (in L)}}$

We are given:

Moles of KOH = 0.0024 moles

Volume of solution = 23.22 mL  = 0.02322L      (Conversion factor:  1L = 1000 mL)

Putting values in above equation, we get:

$\text{Molarity of KOH }=\frac{0.0024mol}{0.02322L}=0.103M$

Hence, the molarity of KOH solution is 0.103 M.