Question

The van't Hoff factor $({\text{i)}}$ for a dilute aqueous solution of the strong electrolyte barium hydroxide is:
A. $2$
B. $3$
C. $0$
D. $1$

Answer 1

An electrolyte dissolves into the solution and forms ions. The total number of ions represents the van't Hoff factor $({\text{i)}}$.
We can use the Formula:
$({\text{i)}}\,{\text{ = }}\dfrac{{{\text{observed value}}}}{{{\text{calculated}}\,\,{\text{value}}}}$

Step by step answer: The van't Hoff factor $({\text{i)}}$ express the degree of the dissociation of association of solute in the solution.
The formula of van't Hoff factor $({\text{i)}}$ is as follows-
$({\text{i)}}\,{\text{ = }}\dfrac{{{\text{observed value}}}}{{{\text{calculated}}\,\,{\text{value}}}}$
Observes value is the number of ions produced on dissolution and the calculated value is the number of ions before the dissolution.
In the case of an association, the value of van't Hoff factor $({\text{i)}}$ is less than one because the observed value will be less than the calculated value.
In the case of dissociation, the value of the van't Hoff factor $({\text{i)}}$ is greater than one because the observed value will be more than the calculated value.
The chemical formula of barium hydroxide is ${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}$. Barium hydroxide is a strong electrolyte that dissociates completely in water.
The dissolution of barium hydroxide is shown as follows:
${\text{Ba}}{\left( {{\text{OH}}} \right)_{\text{2}}}\mathop \to \limits^{{{\text{H}}_2}{\text{O}}} \,{\text{B}}{{\text{a}}^{2 + }}\left( {{\text{aq}}} \right)\,\, + \,{\text{2O}}{{\text{H}}^ - }\left( {{\text{aq}}} \right)$
The number of ions produced on dissolution is three, one is barium ion and two are hydroxide ions. The number of ions before dissolution is one.
So, the value of van't Hoff factor $({\text{i)}}$ is as follows:
$({\text{i)}}\,{\text{ = }}\dfrac{3}{{\text{1}}}$
$\Rightarrow ({\text{i)}}\,{\text{ = 3}}$

Therefore, option (B) ${\text{3}}$ is correct.

Note: A strong electrolyte dissociates into the solution the observed value of the number of ions will be greater from the calculated value of number of ions. The nonelectrolytes do not dissolve and dissociate in water, so the van't Hoff factor $({\text{i)}}$ for non-electrolytes is always one.