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#?

1 Answer
Dec 4, 2015

Answer:

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?#