Question

Reshma and Shubha were working together in the chemistry laboratory. They wanted to find out the exact molarity of a solution of sodium hydroxide (NaOH), prepared by dissolving exactly one gram of sodium hydroxide pellets in water and diluting the resultant solution to 250 mL in a volumetric flask. Reshma weighed exactly one gram of solid sodium hydroxide using a chemical balance ; but while transferring it to a beaker, she spilled some amount of the weighed solid on the floor but she still went ahead with the preparation. Shubha was unaware of this spillage. She titrated the sodium hydroxide solution prepared by Reshma with 10 mL solution of 0.05 M dibasic acid having formula (C₂H₃O₂)₂. The constant titre reading for the titration obtained by Shubha-was 11.3 mL. 

(i) Calculate the amount of sodium hydroxide spilled on the floor. 

(ii) How many molecules of each of the reactants were present in the titration performed by Shubha ?

Answer 1

Eq. of NaOH = Eq. of dibasic acid.

N₁V₁ = N₂V₂

N₁ × 11.3 = 10 ml ×0.05 × 2

Normality of NaOH = 0.088 N

amount of NaOH present in 250 ml NaOH solution.

(I) Amount of NaOH spilled on floor is = 1 gm – 0.88 gm = 0.12 gm NaOH spilled on floor.

(II) Molecules of NaOH present during titration

Molecules of Dibasic acid present during titration = 3.0115 x 10²⁰ molecules