Question

A solution containing 4.2g of KOH and Ca(OH)2 is neutralised by an acid. It consumes 0.1 equivalent of acid, calculate the percentage composition of the sample

Answer 1

Answer:

A solution has 4.2 g of KOH and Ca(OH)₂.

Let the amount KOH in the solution be “x” gm then the amount of Ca(OH)₂ be “(4.2 - x)” gm.

Molar mass of KOH = 56 g/mol

Molar mass of Ca(OH)₂ = 74 g/mol

According to the question, since neutralisation is taking place, therefore, we can write for the reaction i.e.,

Equivalents of KOH + Equivalents of Ca(OH)₂ = Equivalents of H₂SO₄

∴ [1 * moles of KOH] + [2 * moles of Ca(OH)₂] = [2 * moles of H₂SO₄]

⇒ x/56 + 2(4.2 - x)/74 = 0.1 ……[since 0.1 eqivalent of H₂SO₄ is consumed ]  

⇒ x/56 + (4.2 - x)/37 = 0.1

Or, 37x + 235.2 – 56x = 207.2

Or, 19 x = 28

Or, x = 1.47 gm

And, (4.2-x) = 2.73 g

∴ Composition of KOH is 1.47 gm and Ca(OH)₂ is 2.73 gm in the solution.

Finally,  

% composition of KOH = [1.47/4.2] * 100 = 35%

And,

% composition of Ca(OH)₂ = [2.73/4.2] * 100 = 65%

Thus, the % composition of KOH and Ca(OH)₂ in the solution is 35% & 65% respectively.