Question

The ionic strength of solution obtained by mixing 100 mL of 0.1 M ${Al}_2{\left(S{O}_4\right)}_3$ and 100 mL of 0.2 M ${Na}_{2}S{O}_4$ is :

• A. 2.1
• B. 1.05
• C. 0.21
• D. 0.105

Answer 1

Answer: (B)

1.05

As we know, ionic strength is $\frac{1}{2}\Sigma c{Z}^2$.
$\left[{Al}_2{\left(S{O}_4\right)}_3\right]$ conc. after mixing $=\frac{0.1\times 100}{200}=0.05M$
$\left[{Al}_2{\left(S{O}_4\right)}_3\right]\to 2A{l}^{+3}+3S{O}_4^{-2}$
$\left[{Na}_{2}S{O}_4\right]$ conc. after mixing $=\frac{0.2\times 100}{200}=0.1M$
$\left[{Na}_{2}S{O}_4\right]\to 2N{a}^{+}+S{O}_4^{-2}$
$I.S.=\frac{1}{2}\Sigma c{Z}^2$
$=0.05\times 2\times 3^2+0.05\times 3\times 2^2+0.1\times 2\times 1^2+0.1\times 1\times 2^2$
$=\frac{2.1}{2}=1.05$