Question

A solution of sodium hydroxide is standardized against potassium hydrogen phthalate (KHP, formula weight 204.2 g/mol). From the following data, calculate the molarity of the NaOH solution: mass of KHP 1.794 g; buret reading before titration 0.23 mL; buret reading after titration 32.47 mL. Answer in units of M.

Answer 1

Answer: The molarity of NaOH solution is 0.273 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 = 1.794 g

Molar mass of KHP = 204.2 g/mol

Putting values in above equation, we get:

$\text{Moles of KHP}=\frac{1.794g}{204.2g/mol}=0.0088mol$

The chemical reaction for the reaction of KHP and NaOH follows

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

By Stoichiometry of the reaction:  

1 mole of KHP reacts with 1 mole of NaOH.  

So, 0.0088 moles of KHP will react with = $\frac{1}{1}\times 0.0088=0.0088mol$ of NaOH

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 NaOH = 0.0088 moles

Volume of solution = [32.47 - 0.23] mL = 32.24 mL = 0.03224 L      (Conversion factor:  1 L = 1000 mL)

Putting values in above equation, we get:

$\text{Molarity of NaOH }=\frac{0.0088mol}{0.03224L}=0.273M$

Hence, the molarity of NaOH solution is 0.273 M.