[Nathan]

A solution is prepared by dissolving 15.0 g on NaOH in 150.0 mL of0.250 M nitric acid. Will the final solution be acidic, basic, or neutral? Calculate the concentrations of all the ions present in the solution after the reaction has occurred

[kingchemist]

No of moles = Volume in L x Molarity
So find the number of moles of HNO3 = w moles

Find the number of moles of NaOH using atomic masses. If I use Na=23, O=16 and H=1, then NaOH = 40g/mol. You use the correct values for your course.

No. of moles NaOH = mass of NaOH/Mass of 1 mole NaOH = 15.0/40 = x moles

Then write a balanced equation
HNO3     +       NaOH     ---->    NaNO3     +       H2O
1mol               1mol

So using w and x, you can decide which is the limiting reactant by referring to the 1:1 mole ratio, and then work out the number of moles of the excess reactant. If you take the solution to have a volume of 0.150 L and you know the number of moles of the excess reactant left, you can find the concentration
hey

[Nathan]

Can you explain more on the part of finding concentration since we know the 0.150L and excess reagent? Im not sure how to proceed from here
hey

[kingchemist]

Using simple atomic masses, 15.0g NaOH = 15/40 = 0.375 moles.

For acid, number of moles = 0.150 x 0.250 = 0.0375 moles of nitric acid.

Either
HNO3     +       NaOH     ---->    NaNO3     +       H2O
1mol               1mol
0.0375 mol     0.0375 mol
You actually have 10x this amount of NaOH so NaOH is massively in excess and HNO3 is limiting.

or
HNO3     +       NaOH     ---->    NaNO3     +       H2O
1mol               1mol
0.375 mol      0.375 mol
You would need 0.375 mol of HNO3 to react with all of the NaOH and you only have 0.0375 moles of acid.

Either way, the acid is limiting and controls the reaction.

The solution will clearly be alkaline at the end.

All acid will be used, 0.0375 moles of NaOH will be used up  and 0.375-0.0375 moles of NaOH will be left unused. You will also make 0.0375 moles of NaNO3 and make 0.0375 moles of H2O

hey

[Nathan]

But i thought we want to find out the concentration of the ion, not mole?
hey

[kingchemist]

All acid will be used, 0.0375 moles of NaOH will be used up  and 0.375-0.0375 moles of NaOH will be left unused. This is  0.3375 moles NaOH.
You will also make 0.0375 moles of NaNO3 and make 0.0375 moles of H2O

Using this, 1 mole of NaOH contain 1 mole of Na⁺ ions and 1 mole of OH^- ions
So 0.3375 moles of NaOH will contain ? mole of Na⁺ ions and ? mole of OH^- ions

Also 1 mole of NaNO₃ contain 1 mole of Na⁺ ions and 1 mole of NO₃^- ions
So 0.0375 mole of NaNO₃ contain ? mole of Na⁺ ions and ? mole of NO₃^- ions

I think you need to find the missing values yourself - it is easy, but I've done everything else for you. Next time I expect to see some working.
hey