Solution Stoichiometry?

Question

regarding:
0.269g of Nitric Acid (HNO3) are added to 36.3mL of 1.18M Nitric Acid solution. Assuming there is no change in volume, calcuate the final concentrations of ions.
-Would the [H+]=1.18M ; [NO3-]=1.18M?
-I think the molarity of .269g HNO3 is equal to .149M, but the problem says that Nitric Acid has Molarity of 1.18M, which is where the confusion kicks in. Am I supposed to find some strange correlation between these numbers, or just disregard the difference? Anyway, is my answer at all on target?

ALSO-

What are the concentrations of ions in 0.015M Potassium Sulfate (K2SO4)?
It couldn't be as simple as... [K+]=0.03M ; and [SO4 2-]=0.015M...could it?

please and thank you!

Answer 1

Initially you have 36.3 mL of Nitric acid at a concentration of 1.18 mole/Liter.

The molar mass of HNO3 is 63.01 g/mol.

In the inital solution you have

63.01 g/mol * 1.18 mol/L * 36.3 mL * 0.001 L/mL

=2.699 grams of HNO3

If you add 0.269 grams, you get

2.699 g + 0.269 g = 2.968 grams of HNO3 in the same 36.3 mL of solution. The number of moles is

2.968 g / 63.01 g/mol

and the number of moles per liter (molarity) is

(2.968 g / 63.01 g/mol) / (36.3 mL * 0.001 L/mL)

= 0.0471 mol / 0.0363 L

1.298 M or 1.30 M with rounding. (Notice that this is 1.18 + 0.118)

As for the K2SO4 solution: Yes. You got it.