Question

Calculate total molarity and molarity of all ions present in $1litre$ solution containing $0.1M$ copper sulphate and $0.1M$ aluminium sulphate.

Answer 1

$0.1M$ $CuS{O}_4$ in $1L$ solution will give $0.1$ mol $C{u}^{2+}$,

$0.1$ mol $S{O}_4^{2-}$ in $0.1M$ $A{l}_2{\left(S{O}_3\right)}_3$ in $1L$ solution will give $0.2$ mol $A{l}^{3+}$ and $0.3$ mole $S{O}_4^{2-}$.

Total moles $=0.1+0.1+0.2+0.3=0.7$

$\therefore$ Total molarity $=\frac{0.7mol}{12}=0.7M$

$\left[C{u}^{2+}\right]=\frac{0.1molC{u}^{2+}}{1L}=0.1M$

$\left[A{l}^{3+}\right]=\frac{0.2molA{l}^{3+}}{1L}=0.2M$

$\left[S{O}_3^{2-}\right]=\frac{(0.1+0.3)molS{O}_4^{2-}}{1L}=0.4M$