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

Step-by-step explanation:

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

`\text{Molarity of the solution}=\frac{\text{Mass of solute}* 1000}{\text{Molar mass of solute}* \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}=(0.0865* 1000)/(82.079* 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* 0.0042* 250=1* 0.1700* V_2\\\\V_2=(2* 0.0042* 250)/(1* 0.1700)=12.35mL`

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