Courses
Courses for Kids
Free study material
Offline Centres
More
Store

# Find the equivalent weight of calcium hydroxide? $\left( {Ca = 40,O = 16,H = 1} \right)$(A) $23u$ (B) $37u$ (C) $32u$ (D) $56u$

Last updated date: 10th Sep 2024
Total views: 428.4k
Views today: 13.28k
Verified
428.4k+ views
Hint: Equivalent weight is the mass of one equivalent, that is the mass of a given substance which will combine with or displace a fixed quantity of another substance. The equivalent weight of an element is the mass which combines with or displaces $1.008grams$ of hydrogen, or $35.5grams$ of chlorine.

The molecular formula of calcium hydroxide is $Ca{\left( {OH} \right)_2}$
Molecular mass of $Ca{\left( {OH} \right)_2} = 40 + \left( {16 + 1} \right) \times 2$
$= 40 + \left( {17 \times 2} \right)$
$= 74u$
The formula/expression for the equivalent weight of $Ca{\left( {OH} \right)_2}$ is given as follows:
$Equivalent{\text{ }}weight = \dfrac{{molecular{\text{ }}mass}}{{no.{\text{ }}of{\text{ }}replaceable{\text{ }}O{H^ - }ions}}$
$Ca{\left( {OH} \right)_2} \rightleftharpoons C{a^{2 + }} + 2O{H^ - }$
From this reaction, we can say that there are $2$ replaceable $O{H^ - }$ groups.
Substitute the values in the above expression:
Equivalent weight $= \dfrac{{74u}}{2} = 37u$

So, the correct answer is Option B.

Note:
Equivalent weight unlike molecular weight is proportional mass of chemical entities which combine or displace other chemical entities.
The idea of equivalent mass to compare chemically different elements. Atoms combine with each other to form chemical compounds, such that the elements are always present in definite proportions by mass and this property can be used to make chemically different molecules with different mass, equal.
Equivalent mass is determined experimentally and has dimensions and units of mass. Equivalent mass concept is of greater importance in analytical chemistry because in order to follow primary standards the compounds with greater equivalent mass are preferred in order to reduce errors in calculation.
It is also widely used in polymer chemistry during ion-exchange reaction where one equivalent of ion exchange polymer will exchange with one mole of singly charged ions.
Equivalent weight =$\dfrac{{molecular{\text{ }}weight}}{{N{\text{ }}factor}}$
Normality can be find out form the equivalent weight form the following formula:
$Normality = \dfrac{{no.{\text{ }}of{\text{ }}equivalent{\text{ }}weight}}{{litres{\text{ }}of{\text{ }}solution}}$