A 100. mL of 0.250 M calcium nitrate solution is mixed with 400. mL of 0.100 M nitric acid solution. What is the final concentration of the nitrate ion?

Expert Answers
jeew-m eNotes educator| Certified Educator

`Ca(NO_3)_2 rarr Ca^(2+)+2NO_3^-`

Amount of `Ca(NO_3)_2` added  `= 0.25/1000xx100 `

Amount of `NO_3^-` produced `= 0.25/1000xx100xx2 = 0.05mol`

`HNO_3 rarr H^++NO_3^-`

Amount of `HNO_3` added  `= 0.1/1000xx400`

Amount of `NO_3^-` produced `= 0.1/1000xx400 = 0.04mol`

Total `NO_3^-` in final mix `= 0.05+0.04 = 0.09mol`

Volume in final mix `= 400+100 = 500ml`

`[NO_3^-] = 0.09/500xx1000 = 0.18M`

So the concentration of nitrate ions is 0.18M.