A solution made from pure Potassium hydroxide contained 1.85 g of Potassium
hydroxide in exactly 200 cm 3 of water. Using phenolphthalein indicator, titration of 20.0
cm 3 of this solution is carried out v/s sulphuric acid. 9.35 cm 3 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.
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
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