Question

The ionic strength of a solution containing $0.008MAlC{l}_3$ and $0.005MKCl$ is

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

Answer 1

Answer: (B)

Ionic strength is given as,
$I=\frac{1}{2}\left[C_1{Z}_1^2+C_2{Z}_2^2+C_3{Z}_3^2+....\right]$
$\begin{array}{ccccc}& AlC{l}_3\to & A{l}^{3+}+& 3C{l}^{-}& \\ Initially& 0.008& 0& 0& \\ Finally& 0& 0.008& 0.024& \left(C\right)\\ & & \left(3\right)& \left(1\right)& \left(Z\right)\end{array}$
$\begin{array}{ccccc}& KCl\to & K^{+}+& C{l}^{-}& \\ Initially& 0.005& 0& 0& \\ Finally& 0& 0.005& 0.005& \left(C\right)\\ & & \left(1\right)& \left(1\right)& \left(Z\right)\end{array}$
$\therefore I=\frac{1}{2}[0.008(3{)}^2+0.024(1{)}^2+0.005(1{)}^2+0.005(1{)}^2$
$=\frac{1}{2}[0.072+0.024+0.005+0.005]=0.053$