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
anor277
Dec 4, 2015

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?

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