Answer to Question #253930 in Physical Chemistry for Pratiksha

Question

Q1) A solution made from pure Potassium hydroxide contained 1.85 g of Potassium 

hydroxide in exactly 200 cm3

 of water. Using phenolphthalein indicator, titration of 20.0 

cm3

 of this solution is carried out v/s sulphuric acid. 9.35 cm3

 of sulphuric acid solution is 

required for complete neutralisation. [atomic masses: K= 39, O = 16, H = 1]

(a) write the equation for the titration reaction.

(b) calculate the molarity of the Potassium hydroxide solution.

(c) calculate the moles of Potassium hydroxide neutralised.

(d) calculate the moles of sulphuric acid neutralised.

(e) calculate the molarity of the sulphuric acid.

Answer 1

2 KOH + H₂SO₄ → K₂SO₄ + 2 H₂O

Molar mass of KOH= 56.1056

1.85/56.1056= 0.033mol

0.033/0.2= 0.1649mol/L