Question

Find the equivalent weight of potassium hydroxide.
(K=39, O=16, H=1)
A. 66$\mu$
B. 78$\mu$
C. 56$\mu$
D. 43$\mu$

Answer 1

Equivalent weight is molecular mass of substance divided by its Acidity/Basicity/Valency/charge/ number of electrons changed.

Complete step by step answer: First of all we have to understand about equivalent weight. The equivalent weight of an element is the mass which displaces or reacts with 1 gram of hydrogen, 8 grams of oxygen and 35.5 grams of chlorine. Equivalent weight and molecular weight both are different things. Molecular mass of a substance is the average relative mass of its molecule as compared with an Atom of carbon – 12 isotopes taken as 12. Molecular mass can be calculated by adding atomic masses of all the atoms present in one molecule of the substance.
For example: Molecular mass of ${H_2}S{O_4}$= 2$\times$atomic mass of H + atomic mass of sulphur + 4$\times$atomic mass of S
or =$2 \times 1 + 32 + 4 \times 16 = 98.0\mu$
Whereas equivalent weight or calculated in various methods. Equal weight of an acid can be calculated as molecular mass of acid divided by basicity. Here basicity is the number of replaceable ${H^ + }$ ions present in acid.
Mathematically we can write equivalent weight of acid as
$\dfrac{{Molecular{\text{ Weight}}}}{{Basicity}}$
Equal and weight of base can be calculated as molecular mass of base divided by acidity of base. Here acidity is the number of replaceable $O{H^ - }$ ions present in a base.
Mathematically we can write it as
$\dfrac{{{\text{Molecular Weight}}}}{{Acidity}}$
Equivalent weight of salt can be calculated as molecular mass of salt divided by valency or charge on the salt
Or mathematically it is $\dfrac{{Molecular{\text{ Weight}}}}{{Valency}}$
Now we have to calculate the equivalent weight of K0H. KOH is a base. Since it has only one replaceable $O{H^ - }$ hence its acidity is 1. Now we have to calculate the molecular weight of KOH by putting Values of atomic masses.
This implies that molecular mass of K0H=
$1 \times Atomic{\text{ Mass of K + 1}} \times Atomic{\text{ Mass of O + 1}} \times {\text{Atomic Mass of H}}$
Or= $1 \times 39 + 1 \times 16 + 1$
= $56\mu$
So we got molecular mass of K0H= $56\mu$
Now find equivalent weight of base by using formula= $\dfrac{{Molecular{\text{ Mass}}}}{{Acidity}}$
Putting values in above equation, we get equivalent weight= $\dfrac{{56}}{1} = 56\mu$
Hence equal and weight of K0H is $56\mu$
So, the correct answer is “Option C”.

Note: For compounds like$N{a_2}C{O_3}$ whose we don’t know whether it is acidic or basic, their equivalent weight can be calculated as molecular mass divided by charge on Anion or Cation. Since cation here$N{a^{2 + }}$ and 2 atoms of Na are present there charge 2 divided becomes 2.