Question

A solution containing $\text{4}\text{.2 g}$ of $\text{KOH}$ and $\text{Ca}{{\left( \text{OH} \right)}_{2}}$ is neutralized by an acid. If it consumes $\text{0}\text{.1}$ gram- equivalents of the acid, calculate the composition of the sample.

Answer 1

This problem requires knowledge on the normality of the solution. The normality can be defined as the number of gram-equivalents of that substance present in one litre of the solution. We shall calculate gram equivalents of the bases required and thus their percentage composition.

Complete step by step answer:
As given in the problem, the total mass of the bases, $\text{KOH}$and$\text{Ca}{{\left( \text{OH} \right)}_{2}}$is = $\text{4}\text{.2 g}$
Let the mass of$\text{KOH}$= x g
Therefore the mass of $\text{Ca}{{\left( \text{OH} \right)}_{2}}$= $\left( \text{4}\text{.2 - x} \right)$grams.
The equivalent weight of a base is defined as the molecular weight of the base divided by the acidity of the base. Here $\text{KOH}$ is a monoacidic base and $\text{Ca}{{\left( \text{OH} \right)}_{2}}$ is a diacidic base. Therefore the equivalent weight of $\text{KOH}$ is the same as the molecular weight of it = $\left[ 39+16+1 \right]=56$ grams, whereas the equivalent weight of $\text{Ca}{{\left( \text{OH} \right)}_{2}}$= \dfrac{\left[ 40+\left\\{ \left( 16+1 \right)\times 2 \right\\} \right]}{2}=\dfrac{74}{2}=37grams
As per the problem, for complete neutralization,
Gram equivalents of$\text{KOH}$ + Gram equivalents of$\text{Ca}{{\left( \text{OH} \right)}_{2}}$= Gram equivalents of the acid
$\dfrac{\text{x}}{\text{56}}\text{-}\dfrac{\left( \text{4}\text{.2 - x} \right)}{\text{37}}\text{=0}\text{.1}$,
$\text{or, }\left( \text{37a - 56a} \right)\text{ = }\left( \text{0}\text{.1 }\\!\\!\times\\!\\!\text{ 56 }\\!\\!\times\\!\\!\text{ 37} \right)\text{ - }\left( \text{4}\text{.2 }\\!\\!\times\\!\\!\text{ 56} \right)$
$\text{or, 19 a = 28}\Rightarrow \text{a = }\dfrac{28}{19}=1.47$
Hence the mass of the potassium hydroxide in the sample= $1.47$grams.
The amount of calcium hydroxide present in the sample = $\left( \text{4}\text{.2 - 1}\text{.47} \right)=2.73$ grams.
Therefore, the percentage of potassium hydroxide present in the sample is:
$\dfrac{1.47}{4.2}\times 100=35$
So the percentage of calcium hydroxide present on the sample = $\dfrac{2.73}{4.2}\times 100=65$
The composition of the sample is 35% $\text{KOH}$and 65%$\text{Ca}{{\left( \text{OH} \right)}_{2}}$.

Note:
The normality of the solution is used to express the concentration of the solution. There are other units to express the concentration of the solution as well. The molarity of the solution being equal to the number of moles of the solute present per litre of the solution, and the molality of the solution being equal to the number of moles of the solute present per kg of the solvent.