Question

A $0.1M$ basic solution is required from $Ca{\left( {OH} \right)_2}$ which is $40\%$ ionized. Analytical molarity of $Ca{\left( {OH} \right)_2}$ should be
A) $0.125M$
B) $0.25M$
C) $0.4M$
D) $0.5M$

Answer 1

In chemistry, the concentration of the solution is very important. The strength of the solution is very important for inhale solution and medicinal chemistry. There are different ways of representing the concentration of solutions. There are molality, molarity, normality formality, mole fraction, mass percentage, volume percentage, mass by volume and parts per million.
Formula used:
The concentration of [$O{H^ - }$] is dependent on the ionized percentage and molarity of the solution.
$[O{H^ - }] = C\alpha$
Here, the concentration of the solution is C.
Ionized percentage is represented by $\alpha$.
The analytical molarity of the solution depends on the concentration of the solution and basicity of the solution.
Analytical molarity $= \dfrac{{\text{the concentration of the solution}}}{{basicity{\text{ }}of{\text{ }}the{\text{ }}solution}}$

Complete step by step answer:
The given data is, $0.1M$ basic solution is required from $Ca{\left( {OH} \right)_2}$ which is $40\%$ ionized.
Concentration is calculated as,
The concentration of [$O{H^ - }$] is dependent on the ionized percentage and molarity of the solution.
$[O{H^ - }] = C\alpha$
Here, the concentration of the solution is C.
Ionized percentage is represented by ${{\alpha }}$ is {\text{40% }}.
Molarity of the solution [$O{H^ - }$] is $0.1M$.
$[O{H^ - }] = C\alpha$
$0.1M = C\dfrac{{40}}{{100}}$
$C = \dfrac{{0.1M \times 100}}{{40}}$
$C = \dfrac{{10}}{{40}} = \dfrac{1}{4} = 0.25M$
Analytical molarity of $Ca{\left( {OH} \right)_2}$ is calculated as,
The analytical molarity of the solution depends on the concentration of the solution and basicity of the solution.
The concentration of the solution is $C = 0.25M$.
The basicity of the solution is $2$
Analytical molarity $= \dfrac{{\text{the concentration of the solution}}}{{basicity{\text{ }}of{\text{ }}the{\text{ }}solution}}$
Analytical molarity$= \dfrac{{0.25M}}{2} = 0.125M$
According to the above calculation, we conclude $0.1M$ basic solution is required from $Ca{\left( {OH} \right)_2}$which is $40\%$ ionized. Analytical molarity of $Ca{\left( {OH} \right)_2}$ should be $0.125M$.

Hence, option (A) is the correct answer.

Note:
The molarity of the solution depends on the number moles of the solute and the volume of the solution in litres. The molarity of the solution is equal to the ratio of the number of moles of the solute to the volume of the solution in litres. The symbol of molarity is M.
$Molarity = \dfrac{{\text{number of moles of the solute}}}{{\text{volume of the solution litre}}}$
Moles are defined as the given mass of the molecule is divided by the molecular mass of the molecule.
$moles\, = \dfrac{{\text{mass of molecule}}}{{\text{molecular weight of the molecule}}}$
The molecular weight of the molecule is dependent on the atomic weight of the atom present in the molecule. The molecular weight of the molecule is equal to the sum of the molecular weight of the atom and the number of the respective atom in the molecule.
$\text{Molecular Weight} = \text{number of atoms} \times \text{atomic weight the atom}$