Question

37. A solution containing 6.5 g of KOH and Ca(OH)2 is neutralized by an acid. If it consumes 0.15 equivalents of acid, calculate composition of sample in solution.

Answer 1

Answer:

......................

Answer 2

Answer:

Let mass of KOH be present in mixture =a g

and Mass of Ca(OH)

2

=(4.2−a)g

Eq. mass of KOH=56; Eq. mass of Ca(OH)

2

=

2

74

=37

g equivalent of KOH+g equivalent of Ca(OH)

2

=g equivalent of the acid

56

a

+

37

(4.2−a)

=0.1

or 37a−56a=0.1×56×37−4.2×56

or 19a=28

a=

19

28

=1.47

Mass of KOH in the sample =1.47 g

Percentage of KOH=35

and Percentage of Ca(OH)

2

=100−35=65.

Explanation:

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