Question

A solution containing 4.2 g of KOH and $Ca(OH{)}_2$ is neutralized by an acid. If it consumes 0.1 equivalent of acid, the % of KOH in the original sample is:

• A. 15%
• B. 25%
• C. 35%
• D. 45%

Answer 1

The correct option is C 35%

Let the amount of KOH in the solution be x, then amount of $CaO{H}_2$ is (4.2-x)

Equivalent mass of KOH = $\frac{56}{1}=56$

Equivalent mass of $Ca(OH{)}_2=\frac{74}{2}=37$

Gram equivalents of KOH + Gram equivalents of

$Ca(OH{)}_2$ = Gram equivalents of the acid

$\Rightarrow \frac{x}{56}+\frac{4.2-x}{37}=0.1$

$\Rightarrow 37x+235.2-56x=207.2$

$\Rightarrow 235.2-207.2=56x-37x$

$\Rightarrow 28=19x$

$\Rightarrow x=\frac{28}{19}=1.47$

Therefore, amount of KOH in the solution = 1.47

Percentage of KOH = $\frac{1.47}{42}\times 100=35$

Hence, Option C is correct.