How much sodium hydroxide, at what concentration, do I need to add to 500 mL of 0.1 N HCL to obtain pH 6.8. How do you calculate this?

pH = -log(H^+)
M HCl = N HCl
mols = M x L.
After you know mols HCl you have, then make a solution of NaOH and calculate how much of that must be added to equal the HCl. Post your work if you need further asssistance.

mols of HCl= 0.1X.5L=0.05 mols of HCl

Therefore, number of mols of 2N NaOH needed to obtain pH 6.8:

0.05 mols HCL/2 = 25 mL

Is this right?

No. I may have contributed to your answer. I may have misled you. I should have said earlier to calculate the amount of NaOH needed to make the HCl solution pH = 6.8. Where is the calculation for (H^+)if the pH is 6.8. Please note while you do this that pH of 6.8 is VERY VERY close to a neutral solution of pH = 7 and some precautions may be necessary. I haven't worked the problem and they may not be necessary but you need to keep that in mind.

If I were shooting for pH 7, would my previous answer be accurate?

yes. Iat would be exactly 7.0.
0.1 M x 0.5L = 0.05 mols HCl.
Adding 25 mL 2M NaOH is
2M x 0.025 L = 0.05 mols NaOH.
What you want to do is to back off very slightly with the NaOH (you might try something like 24.95 mL which I calculate to give 0.00019 M or a pH of 3.7. Inching closer, 24.99 mL would leave H^+ of 4E-5 for pH 4.4. The problem here is two fold.
1. Reading more accurately than to the nearest 0.01 mL is impossible AND
2. a pH of 6.8 is 1.58E-7 in H^+ and that is very close to the H^+ in pure water (1E-7). That means that you need to take the ionization of water into account and that is much more difficult to do.

If you want to try getting closer, I suggest you use a weaker solution of NaOH, say 0.1 M.

To calculate the amount of sodium hydroxide (NaOH) needed to obtain pH 6.8 in a solution of 500 mL 0.1 N HCl, you would follow these steps:

Step 1: Calculate the mols of HCl in the solution.
- The concentration of HCl is given as 0.1 N, which means 0.1 moles of HCl are present in 1 liter (1000 mL) of the solution.
- In 500 mL of the solution, the mols of HCl can be calculated as follows:
Mols HCl = N HCl × Volume of solution (in liters)
= 0.1 N × 0.5 L
= 0.05 moles of HCl

Step 2: Determine the amount of NaOH needed.
- Since NaOH is a strong base, it reacts with HCl in a 1:1 ratio, meaning 1 mol of NaOH reacts with 1 mol of HCl.
- To neutralize the 0.05 moles of HCl, you would need 0.05 moles of NaOH.
- If you have a 2 M NaOH solution, you can calculate the volume of NaOH needed as follows:
Volume NaOH = M NaOH × Mols NaOH
= 2 M × 0.05 mol
= 0.1 L (or 100 mL)

Therefore, you would need to add 100 mL of 2 M NaOH to the 500 mL of 0.1 N HCl solution to obtain pH 6.8.

However, please note that targeting a pH of 6.8 is very close to neutral (pH 7), and some precautions may be necessary. Additionally, achieving a pH of exactly 6.8 may be challenging due to the limitations of equipment precision and the complexity of accounting for the ionization of water. If you are aiming for a more precise pH value, you may need to adjust the volumes of NaOH and HCl more subtly or consider using a weaker solution of NaOH, such as 0.1 M.