Question

A basic solution of $\text{(OH) = 0}\text{.1M}$ is required from $\text{Ca(OH}{{\text{)}}_{2}}$ which is 40% ionized. Analytical molarity of $\text{Ca(OH}{{\text{)}}_{2}}$ is:
A. 0.04M
B. 0.4M
C. 0.25M
D. 0.125M
E. 1.25M

Answer 1

Hint: For this problem, as we know that one molecule of calcium hydroxide gives 2 molecules of basic ion i.e. hydroxyl ion when it will ionise completely. So, we will find the number of hydroxyl ions when it will ionise 40%.

Complete step by step answer:
- In the given question, we have to calculate the molarity or the concentration of calcium hydroxide by using the given data.
- As we know that ionisation is the process when the molecule dissociates to give its respective ions.
- Also, molarity is defined as the ratio of a number of moles of solute to the volume of the solution in litres and it has a unit of M. So, it can be calculated by using the formula:
$\text{Molarity = }\dfrac{\text{No}\text{. of moles of solute}}{\text{Volume (Litres)}}$
- Now, the concentration of the hydroxyl ion is given as 0.1M and also it is told that the calcium hydroxide ionises only 40%.
- So, firstly we know that the one molecules of calcium hydroxide will give 2 hydroxyl ions as shown below in the reaction:
$\text{Ca(OH}{{\text{)}}_{2}}\text{ }\to \text{ C}{{\text{a}}^{2+}}\ \text{+ 2O}{{\text{H}}^{-}}$
- So, if it will completely ionise to give o.1M hydroxyl ions, then the concentration of the calcium hydroxide required will be:
$\Rightarrow \dfrac{\text{0}\text{.1}}{2}\text{ = 0}\text{.05M}$
- But it is given that the calcium hydroxide will ionise only 40%, then the solution required will be:
$\Rightarrow \dfrac{\text{40}}{100}\text{ }\times \text{ 0}\text{.05 = 0}\text{.125M}$

So, the correct answer is “Option D”.

Note: Analytical molarity is different from that of the normal molarity because it helps in determining the preparation of the solution whose molarity is given and also its unit is M.