Question

The ionic strength of a solution containing $0.008\, M\, AlCl_{3}$ and $0.005\, M\, KCl$ is

• A. $0.134\, M$
• B. $0.053\, M$
• C. $0.106\, M$
• D. $0.086\, M$

Answer 1

Answer: (B)

Ionic strength is given as,
$I = \frac{1}{2} \left[C_{1} Z^{2}_{1} + C_{2}Z^{2}_{2} + C_{3}Z^{2}_{3} +....\right]$
$\begin{matrix}&AlCl_{3} \to& Al^{3+} +&3Cl^{-}&\\ Initially&0.008&0&0&\\ Finally&0&0.008&0.024&\left(C\right)\\ &&\left(3\right)&\left(1\right)&\left(Z\right)\end{matrix}$
$\begin{matrix}&KCl \to& K^{+} +&Cl^{-}&\\ Initially&0.005&0&0&\\ Finally&0&0.005&0.005&\left(C\right)\\ &&\left(1\right)&\left(1\right)&\left(Z\right)\end{matrix}$
$\therefore I = \frac{1}{2}[0.008(3)^{2} +0.024(1)^{2} +0.005(1)^{2} + 0.005(1)^{2}$
$= \frac{1}{2} [0.072 + 0.024 + 0.005 +0.005] = 0. 053$