Question

What concentration of `NO_3`- results when 697 mL of .515 M `NaNO_3` is mixed with 635 mL of .785 M `Ca(NO_3)_2`?

Answer 1

There is a volume of `697` `+` `635` `mL`. There are `0.857* mol` of nitrate ion.

Explanation:

Moles of `NaNO_3` `=` `0.697*Lxx0.515*mol*L^-1` `=` `0.359*mol.`

Moles of `Ca(NO_3)_2` `=` `0.635*Lxx0.785*mol*L^-1` `=` `0.498*mol.` Therefore there are `0.996` `mol` nitrate from the calcium salt. (Why? Because each mole of calcium nitrate contributes 2 mol nitrate anion.)

In total there are thus `0.857` `mol` nitrate ion dissolved in `1.332*L` of solution. I have assumed that the volumes of solution are additive, which is entirely reasonable.

`[NO_3^-]` `=` `(0.857*mol)/(1.332*L)` `=` `?? mol*L^-1?`