Question

A chemist weighs out 0.0865 g of sulfurous acid, H2SO3, which is a diprotic acid into a 250. mL of volumetric flask and dilutes to the mark with distilled water. She plans to titrate the acid with 0.1700 M NaOH solution. Calculate the volume of NaOH solution in milliliters the student will need to add to reach the final equivalence point. (Molar mass of sulfurous acid = 82.079 g/moL)

Answer 1

Answer: The volume of NaOH needed to add is 12.35 mL

Explanation:

To calculate the molarity of solution, we use the equation:

$\text{Molarity of the solution}=\frac{\text{Mass of solute}\times 1000}{\text{Molar mass of solute}\times \text{Volume of solution (in mL)}}$

Given mass of sulfurous acid = 0.0865 g

Molar mass of sulfurous acid = 82.079 g/mol

Volume of solution = 250 mL

Putting values in above equation, we get:

$\text{Molarity of sulfurous acid}=\frac{0.0865\times 1000}{82.079\times 250}\\\\\text{Molarity of sulfurous acid}=0.0042M$

To calculate the volume of base, we use the equation given by neutralization reaction:

$n_1M_1V_1=n_2M_2V_2$

where,

$n_1,M_1\text{ and }V_1$ are the n-factor, molarity and volume of acid which is $H_2SO_3$

$n_2,M_2\text{ and }V_2$ are the n-factor, molarity and volume of base which is NaOH.

We are given:

$n_1=2\\M_1=0.0042M\\V_1=250mL\\n_2=1\\M_2=0.1700M\\V_2=?mL$

Putting values in above equation, we get:

$2\times 0.0042\times 250=1\times 0.1700\times V_2\\\\V_2=\frac{2\times 0.0042\times 250}{1\times 0.1700}=12.35mL$

Hence, the volume of NaOH needed to add is 12.35 mL