# 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#?

##### 1 Answer

I got

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 **1** mol of **2** mols of

If you recall, **the** **does NOT depend on the volume of solution**. Therefore, simply calculate the total **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

#"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 *provided that the final volume is also assumed additive with respect to the initial volumes*.

So, you could just calculate the new molarities of

#("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")#