posted by Ruby .
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.