Answer to Question #254580 in Physical Chemistry for saloni

Question

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

hydroxide in exactly 200 cm* of water. Using phenolphthalein indicator, titration of 20.0

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

required for complete neutralisation. (atomic masses: K= 39, 0 = 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 g/mol

1.85/56.1056 = 0.033 mol

0.033/0.2= 0.165mol/L