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Question: The normality of 0.025 M \[Ba{\left( {OH} \right)_2}\] is (A) 0.05 (B) 0.25 (C) 0.5 (D) 0.02...

The normality of 0.025 M Ba(OH)2Ba{\left( {OH} \right)_2} is
(A) 0.05
(B) 0.25
(C) 0.5
(D) 0.025

Explanation

Solution

You should know the definition of normality and equivalent weight. For that you should know the acidity or basicity of the compound. Lastly, you should substitute the values properly in the formula to calculate the right answer.

Formulae used:
1. Normality=mEq.wt.×VNormality = \dfrac{m}{{Eq.wt. \times V}} , where m is the weight of substance in g, and Eq.wt. is the equivalent weight of the substance, and V is volume in liters.
2. Eq.wt.=MAcidity/BasicityEq.wt. = \dfrac{M}{{Acidity/Basicity}} , where M is the molar mass (molecular weight) of the substance in g mol -1.
3. Molarity=mM×VMolarity = \dfrac{m}{{M \times V}}

Complete step-by-step solution: This question can be answered writing a simple formula that we will get at the end of the derivation, but I feel we should also know how to derive the specific formula we need to find out the answer. Let’s begin with the formula for normality.
Normality=mEq.wt.×VNormality = \dfrac{m}{{Eq.wt. \times V}} ………………...(1)
Substituting the expression of equivalent weight in above equation, we get
Normality=mMAcidity/Basicity×VNormality = \dfrac{m}{{\dfrac{M}{{Acidity/Basicity}} \times V}} ………………….(2)
Taking the acidity/basicity term in the numerator, we will get
Normality=(acidity/basicity)×mM×VNormality = (acidity/basicity) \times \dfrac{m}{{M \times V}} ……………...(3)
Comparing the right hand side of the equation above with the definition of molarity, we can easily replace the term mM×V\dfrac{m}{{M \times V}} by molarity.
Normality=(acidity/basicity)×MolarityNormality = (acidity/basicity) \times Molarity....(4)
The equation number four is the equation that you need to solve such equations. To use this equation, you need to understand the terms acidity and basicity first.
When we are talking about acids, we must use the term ‘basicity’ as we actually need to put ‘basicity of acids’ in the formula, which is simply the number of hydrogen ions (protons) in a molecule of acid. Similarly, ‘acidity of base’ is the number of hydroxyl ions in a molecule of a base. Since we have been asked to find the normality of a base Ba(OH)2Ba{\left( {OH} \right)_2} , we need to find its acidity first. The acidity of Ba(OH)2Ba{\left( {OH} \right)_2} is 2 since there are two hydroxide ions in one molecule of Ba(OH)2Ba{\left( {OH} \right)_2} . Let’s put the value of acidity as 2 and the value of molarity from the question as 0.025 M.
Normality=acidity×MolarityNormality = acidity \times Molarity
Normality=2×0.025Normality = 2 \times 0.025
Normality=0.05NNormality = 0.05N
And you have calculated your answer, and it is option (A).

Hence the correct option is (A).

Note: Always keep in mind that while using equation 4 for calculating the normality from molarity, you should correctly find out the value of acidity or basicity. You need to also consider that nowhere in the question it is mentioned that the value given 0.025 M is the molarity of the solution. It is up to you to understand that it is molarity by observing the unit ‘M’, which is used for molarity only. If you find ‘m’ somewhere as a unit, don’t take it as molarity, it is actually ‘molality’ which has a slightly different meaning than ‘molarity’.