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.

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pstnonsonjoku pstnonsonjoku · Answer 1 · 2021-10-24T09:44:46+02:00

The concentration of H2SO4 is 0.176 M. The number of moles of H2SO4 is 0.0165 moles.

The equation of the reaction is;

2KOH(aq) + H2SO4(aq) ------> K2SO4(aq) + H2O(l)

The number of moles of KOH = 1.85 g/56 g/mol = 0.033 moles

concentration = number of moles /volume

volume = 200 cm3 or 0.2 L

concentration of KOH = 0.033 moles/0.2 L = 0.165 M

From the titration formula;

CAVA/CBVB =na/nb

CA = concentration of acid

VA = volume of acid

CB = concentration of base

VB = volume of base

na = number of moles of acid

nb = number of moles of base

Making CA the subject of formula;

CA = CBVBna/VAnb

CA = ?

CB = 0.165 M

VA = 9.35 cm3

VB = 20.0 cm3

na = 1

nb =2

CA = CBVBna/VAnb

CA = 0.165 M × 20.0 cm3 × 1/9.35 cm3 × 2

CA = 0.176 M

The concentration of sulphuric acid is 0.176 M

From the reaction equation;

2 moles of KOH reacts with 1 mole of H2SO4

0.033 moles of KOH reacts with 0.033 moles × 1 mole/2 moles

= 0.0165 moles of H2SO4

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