# Question #9c009

Well, $\left[N {a}^{+}\right]$ has a $0.400 \cdot m o l \cdot {L}^{-} 1$ concentration..........
The concentration of $N {a}_{2} S {O}_{4}$ was specified to be $0.200 \cdot m o l \cdot {L}^{-} 1$. Because the formula contains two equiv of sodium ion, $\left[N {a}^{+}\right] = 2 \times 0.200 \cdot m o l \cdot {L}^{-} 1 = 0.400 \cdot m o l \cdot {L}^{-} 1$ WITH RESPECT TO SODIUM IONS. Do you see what's going on?
What is the concentration of nitrate ions in $0.100 \cdot m o l \cdot {L}^{-} 1$ $C a {\left(N {O}_{3}\right)}_{2} \left(a q\right)$? Is it different from $\left[C {a}^{2 +}\right]$??