
\[10\] millimoles of a di-acidic base exactly neutralize \[100\] mL of acid. Then the normality of that acid is:
A.\[0.2\] N
B.\[0.1\] N
C.\[0.4\] N
D.\[0.5\] N
Answer
585.9k+ views
Hint: The number of ionisable hydroxide ions which are present in one molecule of base is referred to as the acidity of the base. A di-acidic base on ionization will give two hydroxide ions. Normality is equal to the gram equivalents of a solute present per liter of solution.
Complete step by step answer:
In order to determine the normality of acid we have to determine the gram equivalent of the base. The given base is diacidic but the acid can be both monobasic and dibasic in nature. We have to separately calculate for monobasic and dibasic acid.
Given,
Number of moles of base is equal to \[10\] millimoles = \[10 \times {10^{ - 3}}\] mol
The acidity of the base is \[2\].
The number of equivalence of base will be = number of moles of base X acidity of base
= \[10 \times {10^{ - 3}} \times {\text{ }}2\]
= \[2 \times {10^{ - 2}}\]
Given the diacidic base has completely neutralized the base. Therefore, the number of equivalents of base is equal to the number of equivalences of acid.
The volume of acid used for neutralization is \[100\]mL. The volume of the acid in liters will be
\[V = \dfrac{{100}}{{1000}} = 0.1L\]
Hence the normality of the acid will be as follows:
\[N = \dfrac{{equivalence{\text{ }}of{\text{ }}acid}}{{volume{\text{ }}of{\text{ }}acid}}\]
\[N = \dfrac{{2 \times {{10}^{ - 2}}mol}}{{0.1L}}\]
\[N = 0.2\].
\[10\] Millimoles of a diacidic base will neutralize \[10\] millimoles of dibasic acid or \[20\] millimoles of a monobasic acid. Hence for a dibasic acid, \[10\] millimoles are present in \[100\] mL solution and for a monobasic acid, \[20\] millimoles are present in \[100\] mL solution. So the normality for dibasic or monobasic acid will be \[0.2\]N.
Note:
Normality should not be confused with morality. Molarity is also a measurement of the concentration of a solution. It is calculated as the moles of solute and not gram equivalents of solute present per liter of solution.
Complete step by step answer:
In order to determine the normality of acid we have to determine the gram equivalent of the base. The given base is diacidic but the acid can be both monobasic and dibasic in nature. We have to separately calculate for monobasic and dibasic acid.
Given,
Number of moles of base is equal to \[10\] millimoles = \[10 \times {10^{ - 3}}\] mol
The acidity of the base is \[2\].
The number of equivalence of base will be = number of moles of base X acidity of base
= \[10 \times {10^{ - 3}} \times {\text{ }}2\]
= \[2 \times {10^{ - 2}}\]
Given the diacidic base has completely neutralized the base. Therefore, the number of equivalents of base is equal to the number of equivalences of acid.
The volume of acid used for neutralization is \[100\]mL. The volume of the acid in liters will be
\[V = \dfrac{{100}}{{1000}} = 0.1L\]
Hence the normality of the acid will be as follows:
\[N = \dfrac{{equivalence{\text{ }}of{\text{ }}acid}}{{volume{\text{ }}of{\text{ }}acid}}\]
\[N = \dfrac{{2 \times {{10}^{ - 2}}mol}}{{0.1L}}\]
\[N = 0.2\].
\[10\] Millimoles of a diacidic base will neutralize \[10\] millimoles of dibasic acid or \[20\] millimoles of a monobasic acid. Hence for a dibasic acid, \[10\] millimoles are present in \[100\] mL solution and for a monobasic acid, \[20\] millimoles are present in \[100\] mL solution. So the normality for dibasic or monobasic acid will be \[0.2\]N.
Note:
Normality should not be confused with morality. Molarity is also a measurement of the concentration of a solution. It is calculated as the moles of solute and not gram equivalents of solute present per liter of solution.
Recently Updated Pages
Master Class 11 Computer Science: Engaging Questions & Answers for Success

Master Class 11 Business Studies: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 English: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Biology: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

Discuss the various forms of bacteria class 11 biology CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

State the laws of reflection of light

Explain zero factorial class 11 maths CBSE

