Question

uppose you are standardizing a sodium hydroxide solution with K H P (molar mass=204.2 g/mol) according to the equation K H P + N a O H ⟶ H 2 O + N a K P You prepare the standard solution from 0.294 g of K H P in 250.0 mL of water. You then require 7.42 mL of N a O H solution to complete the titration. What is the concentration of the N a O H solution?

Answer 1

Answer: The concentration of NaOH solution is 0.194 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 = 0.294 g

Molar mass of KHP = 204.22 g/mol

Putting values in above equation, we get:

$\text{Moles of KHP}=\frac{0.294g}{204.22g/mol}=0.00144mol$

The chemical reaction follows the equation:

$KHC_8H_4O_4(aq.)+NaOH\rightarrow KNaC_8H_4O_4(aq.)+H_2O(l)$

By Stoichiometry of the reaction:

1 mole of KHP reacts with 1 mole of NaOH.

So, 0.00144 moles of KHP will react with = 0.0144 of KOH.

To calculate the molarity of NaOH, 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 = 0.00144 moles

Volume of solution = 7.42ml  = 0.00742L      (Conversion factor:  1L = 1000 mL)

Putting values in above equation, we get:

$Molarity=\frac{0.00144mol}{0.00742L}=0.194M$

Hence, the molarity of NaOH solution is 0.194 M