Question

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?

Answer 1

Answer: The molarity of KOH solution is 0.066 M.

Step-by-step 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}=(0.4877g)/(204.22g/mol)=0.0024mol`

The chemical reaction for the formation of chromium oxide follows the equation:

`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 =
`(1)/(1)* 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 = 36.21 mL = 0.03621L (Conversion factor: 1L = 1000 mL)

Putting values in above equation, we get:

`\text{Molarity of KOH }=(0.0024mol)/(0.03621L)=0.066M`

Hence, the molarity of KOH solution is 0.066 M.