# 0.040mol of (NH4)2Ni(SO4)2•6H2O is dissolved in water to give 200cm3 of aqueous solution. What is the concentration, in mol dm–3, of ammonium ions?

May 11, 2017

$\text{Concentration"="Moles of solute"/"Volume of solution} \ldots \ldots = 0.40 \cdot m o l \cdot {\mathrm{dm}}^{-} 3.$
$\text{Concentration with respect to the nickel salt}$
$= \frac{0.040 \cdot m o l}{0.200 \cdot {\mathrm{dm}}^{3}} = 0.20 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$
But given the formulation of the salt, (NH_4)_2[Ni (SO_4)_2]*6H_2O, there are 2 equiv of ammonium per equiv of salt. And thus $\left[N {H}_{4}^{+}\right] = 0.40 \cdot m o l \cdot {\mathrm{dm}}^{-} 3$