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

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

hope this : )