A solution of 0.200M NaOH (14.0 mL) is mixed with 42.4 mL of 0.170M HNO3. Assuming the final volume is the sum of the initial volumes, what is the molarity of Na+ cation and NO3- anion?

I do not know where to start to solved for the molarity of the ions. Please help !

I answered this under the science post.

To solve for the molarity of the Na+ cation and NO3- anion, we need to first determine the number of moles of each species present in the solution.

1. Start by calculating the number of moles of NaOH:

Moles of NaOH = Molarity of NaOH * Volume of NaOH solution
= 0.200 M * 0.014 L
= 0.0028 moles

2. Next, calculate the number of moles of HNO3:

Moles of HNO3 = Molarity of HNO3 * Volume of HNO3 solution
= 0.170 M * 0.0424 L
= 0.007208 moles

3. Now, we determine the limiting reagent by comparing the number of moles of NaOH and HNO3. The limiting reagent is the one that is completely consumed in the reaction, and the other reagent is in excess.

In this case, NaOH is the limiting reagent because it has fewer moles than HNO3.

4. Since NaOH is the limiting reagent, it reacts with an equal number of moles of HNO3 based on the balanced chemical equation:

NaOH + HNO3 → NaNO3 + H2O

The balanced equation tells us that 1 mole of NaOH reacts with 1 mole of HNO3 to form 1 mole of NaNO3.

5. Therefore, the number of moles of Na+ cation and NO3- anion in the final solution is equal to the number of moles of NaOH used in the reaction, which is 0.0028 moles.

To calculate the molarity of Na+ and NO3-, divide the number of moles by the total volume of the solution:

Molarity (Na+) = Moles of Na+ / Total Volume of the Solution
= 0.0028 moles / (0.014 L + 0.0424 L)
= 0.0525 M

Molarity (NO3-) = Moles of NO3- / Total Volume of the Solution
= 0.0028 moles / (0.014 L + 0.0424 L)
= 0.0525 M

Therefore, the molarity of Na+ cation and NO3- anion in the final solution is 0.0525 M.