What are the final concentrations of all the ions when following are mixed?
`50 mL of 0.12M Fe(NO_(3))_(3), 100 mL of 0.10M FeCl_(3)` and `100mL of 0.26M Mg(NO_(3))_(2)`

Explanation

Milli-mole of `Fe(NO_(3))_(3)= 50xx0.12=6`
`("milli-mole"=MxxV_(mL))`
`:.` Milli-mole of `Fe^(3+)=6`
Milli-mole of `NO_(3)^(-)=3xx6=18`
`:.`Milli-mole of `FeCl_(3)= 100xx0.1=10`
Milli-mole of `Fe^(3+)=10`
`:.` Milli-mole of `Mg(NO_(3))_(2)= 100xx0.26=26`
Mill-mole of `Mg^(2+)= 26`
Milli-mole of `NO_(3)^(-)= 2xx26= 52`
On mixing these three, total volume becomes `250mL`
Thus `[Fe^(3+)]=(6+10)/(250)= 0.064M`
`[Ion=("milli-mole")/(V_("in"mL))]`
`[NO_(3)^(-)]=(18+52)/(250)= 0.28M`
`[Cl^(-1)]=(30)/(250)=0.12M`
`[Mg^(2+)]=(26)/(250)= 0.104M`