Question

Consider a weak electrolyte, $A{B}_3$ having a concentration of 0.1 M in solution. If the degree of dissociation is 0.2, Calculate the ionisation constant $K_i$ of $A{B}_3$.

• A. $2.4\times {10}^{-9}$
• B. $1.4\times {10}^{-2}$
• C. $3.4\times {10}^{-7}$
• D. $5.4\times {10}^{-5}$

Answer 1

The correct option is D $5.4\times {10}^{-5}$

$A{B}_3$ dissociates as:

$A{B}_3(aq.)\rightleftharpoons A^{3+}(aq.)+3B^{-}(aq.)Initial:C00Equilibrium:C(1-\alpha )C\alpha 3C\alpha$

$K_i=\frac{\left[A^{3+}\right][B^{-}{]}^3}{\left[A{B}_3\right]}{K}_i=\frac{C\alpha \times (3C\alpha {)}^3}{C(1-\alpha )}=\frac{\left(27\alpha \right)(C\alpha {)}^3}{(1-\alpha )}{K}_i=\frac{27\times 0.2\times (0.1\times 0.2{)}^3}{(1-0.2)}{K}_i=5.4\times {10}^{-5}$