Question

A 100 ml, 0.1 M ${{K}_{2}}S{{O}_{4}}$ is mixed with 100 ml, 0.1 M $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$ solution, the resultant solution is _____________
(a)- 0.1 N
(b)- 0.4 N
(c)- 0.2 M $SO_{4}^{2-}$ ions
(d)- 0.1 M $A{{l}^{3+}}$ ions

Answer 1

Hint: First, you have to find the number of moles in each of the following solutions that must be added. Then add the number of moles, to find the total number of moles. Then add the volume to find the total volume. Now, we have the number of moles and volume, so we can easily find the molarity and then the normality.

Complete step-by-step answer:We are give two solution, i.e., 100 ml, 0.1 M ${{K}_{2}}S{{O}_{4}}$ and 100 ml, 0.1 M $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$. We can find the number of moles by multiplying the given molarity with the given volume and then divide it by 1000.
The moles in 100 ml, 0.1 M ${{K}_{2}}S{{O}_{4}}$, will be:
$\dfrac{0.1\text{ x 100}}{1000}=0.01$ moles.
Since in ${{K}_{2}}S{{O}_{4}}$, two electrons are involved, so the gram equivalent will be 0.02.
The moles in 100 ml, 0.1 M $A{{l}_{2}}{{(S{{O}_{4}})}_{3}}$, will be:
$\dfrac{0.1\text{ x 100}}{1000}=0.01$ moles.
Since the aluminium has a +3 oxidation state, so the gram equivalent will be 0.03.
The total number of moles will be = 0.01 +0.01 = 0.02 moles.
Total volume of the solution will be: 100 + 100 = 200 ml = 0.2 L
Now, we have the total number of moles and volume of the solution in liters. We can calculate the molarity as:
$Molarity=\dfrac{0.02}{0.2}=0.1M$
Total number of gram equivalence will be = 0.02 + 0.03 = 0.05.
The normality will be:
$Normality=\dfrac{0.05}{0.2}=0.25N$
So, the molarity of the solution is 0.1 M and the normality will be 0.25 N.

Therefore, the correct answer is an option (d).

Note: You can also solve this question, by using the formula:
${{M}_{1}}{{V}_{1}}+{{M}_{2}}{{V}_{2}}={{M}_{3}}({{V}_{1}}+{{V}_{2}})$
According to the question, the values will be:
$0.1\text{ x 100 }+\text{ 0}\text{.1 x 100}={{M}_{3}}(100+100)$
${{M}_{3}}=0.1M$