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)

The volume of NaOH needed to add is 12.35 mL

Explanation:

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

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:

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

where,

are the n-factor, molarity and volume of acid which is

are the n-factor, molarity and volume of base which is NaOH.

We are given:

Putting values in above equation, we get:

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

i believe it is c.) 1575 j.

2b2t

explanation:

2b2t