A .682 g sample of an unknown weak monoprotic acid, HA, was dissolved in sufficient water to make 50mL of solution and was titrated with a .135 M NaOH solution. After the addition of 10.6mL of base, a pH of 5.65 was recorded. The equivalence point was reached after the addition of 27.4mL of the .135M NaOH.
3. Calc. molarity of the unreacted HA remaining in solution at pH of 5.65.
4. Calc. the [H3O+] at pH of 5.65.
5. Calc the value of the dissociation constant Ka of the acid.
6. Find the pK range of an indicator that would be appropriate for this titration.
I know that molarity is mol/L but how do I find mols of the unreacted HA?
Can you please guide me through the rest of this problem because I am so confused??
I am a little lost on this problem, too, as you didn't post the first two questions and I can't find them. I think the second was to determine the molar mass of the acid, HA. I think the first was to determine mols of something. Anyway, could you post those two questions and your answers to them. Let's make sure those numbers are correct before proceeding to the remaining part of the problem.
Okay.
1. Calc. the # of moles of acid in the original solution.
I got .003699 mol acid
2. Calc. the molar mass of the acid.
I got 184.374 g/mol
I have to go to a funeral so, I may not respond back for awhile, but if you could help me anyway I would greatly appreciate
ok. The equation is
HA + NaOH ==> NaA + HOH
mols HA initially = 0.003699
At pH 5.65, 10.6 mL of the 0.135 M NaOH have been added. How many mols is that?
0.0106 x 0.135 = 0.001431. So how many mols HA are left? That must be what you had at the beginning minus the mols at pH 5.65. I find 0.002268 mol HA remaining. What molarity is that? The volume was 50 mL at the beginning of the titration and we added 10.6 mL so the final volume at pH 5.65 is 60.6 mL. Therefore, (HA) at that point is mols/L.
4). Calculate (H^+) at pH = 5.65. I assume you can do this. The answer I get is 2.238E-6.
5). Calculate Ka. You can do this one of two ways. Use the Henderson-Hasselbalch equation.
pH = pKa + log [(base)/(acid)]
5.65 = pKa + log [(A^-)/(HA)]
You have (A^-) = 0.001431 mols/0.0606 L
You have (HA) = 0.002268 mols/0.0606 L
Solve for pKa, then Ka. OR just use the Ka expression.
Ka = (H^+)(A^-)/(HA).
You know H^+ from question 4.
(A^-) and (HA) are as above in the HH equation. Solve for Ka. I found 1.41E-6 if I didn't make an error.
Check my work carefully. I have been in a hurry this afternoon and I've worked on scratch paper in bits and pieces. It's easy to make a typo or an error under those conditions. I assume you can look in a chart to find the appropriate indicator for question 6. To do that you will need to calculate the pH at the equivalence point and that is due to the hydrolysis of the salt at the equivalence point.
A^- + HOH ==> HA + OH^-
Kw/Ka = (HA)(OH^-)/(A^-)
You have Kw, Ka, and (A^-). (HA)=(OH^-) so that becomes (OH^-)^2. Solve for (OH^-)
0 answers