How do you calculate [`NO_3^(-)`] if 120 mL of 0.40 M `KNO_3` is mixed with 400 mL of 1.2 M `Pb(NO_3)_2`?

I got `"1.94 M NO"_3^(-)`.

First, it helps to write out each process:

`"KNO"_3(aq) -> "K"^(+)(aq) + "NO"_3^(-)(aq)`

`"Pb"("NO"_3)_2(aq) -> "Pb"^(2+)(aq) + 2"NO"_3^(-)(aq)`

Therefore, for each mol of `"KNO"_3`, we get 1 mol of `"NO"_3^(-)`. However, for each mol of `"Pb"("NO"_3)_2`, we get 2 mols of `"NO"_3^(-)`. Make sure you recognize that, because it seems to be a common oversight.

If you recall, the `bb"mol"` does NOT depend on the volume of solution. Therefore, simply calculate the total `"mol"`s of nitrate obtained from each source of nitrate, and then divide by the total volume.

`("0.40 M KNO"_3 xx "0.120 L")(("1 mols NO"_3^(-))/("1 mol KNO"_3)) = "0.048 mols NO"_3^(-)` from `"KNO"_3`

`("1.20 M Pb"("NO"_3)_2 xx "0.400 L")(("2 mols NO"_3^(-))/("1 mol Pb"("NO"_3)_2))`

`= "0.960 mols NO"_3^(-)` from `"Pb"("NO"_3)_2`

The total `"mol"`s is therefore:

`"0.048 mols" + "0.960 mols" = "1.008 mols NO"_3^(-)`

and the concentration is:

`color(blue)(["NO"_3^(-)]) = ("1.008 mols NO"_3^(-))/("0.120 + 0.400 L")`

`=` `color(blue)("1.94 M")`

Another way to do this is to recognize that since molarity is written as `"mols"/"L"`, these quantities have common denominators, `"mol"`s are additive, molarity is therefore also additive, provided that the final volume is also assumed additive with respect to the initial volumes.

So, you could just calculate the new molarities of `"NO"_3^(-)` and add them together.

`("0.120 L")("0.400 M") = (M_2)("0.120 + 0.400 L")`

`=> M_(2a) = "0.092 M NO"_3^(-)` from `"KNO"_3`

`(("2 mols NO"_3^(-))/("1 mol Pb"("NO"_3)_2))("0.400 L")("1.20 M") = (M_2)("0.120 + 0.400 L")`

`=> M_(2b) = "1.85 M NO"_3^(-)` from `"Pb"("NO"_3)_2`

Therefore:

`M_2 = color(blue)(["NO"_3^(-)])`

`= (n_("NO"_3^(-),a))/(V_"tot") + (n_("NO"_3^(-),b))/(V_"tot")`

`= M_(2a) + M_(2b)`

`= 0.092 + 1.85` `"M"`

`=` `color(blue)("1.94 M")`