1)A solution prepared by adding 0.400 mol of acetic acid (pKa = 4.74) and 0.400 mol of sodium acetate to 100.0 mL of water. The pH of this buffer solution is initially 4.74.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution
2) A solution prepared by adding 0.400 mol of acetic acid (pKa = 4.74) to 100.0 mL of water and then titrating with strong base until a buffer with a 1:1 buffer ratio is produced. The pH of this buffer solution is initially 4.74.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution
3) 100.0 mL of pure water.
Predict the final pH when 55.0 mL of 1.10 M NaOH is added to solution
For 1) I got 4.87 but I am having difficulties with the other two.
1. Working in millimols. mmols NaoH added = 55 x 1.1 = 60.5
......HAc + OH^- ==> Ac^- + H2O
I.....400...0.........400
add........60.5...............
C..-60.5..-60.5.......+60.5
E...339.5...0........460.5
Convert the E line (in mmols) to M with M = mmols/total mL. Total mL = 100 + 55 = 155. Plug that into the Henderson-Hasselbalch equation and solve for pH. You are correct at 4.87
2.
Done the same way but you start with 400 mmols HAc and titrate exactly half of it. So you will form 200 mmols Acetate and you will have 200 mmols HAc remaining. Starting with those numbers you want to add 60.5 mmols NaOH and go from there. I get 5.01 for pH.
3. This one is a lot simpler because there is no buffer.
1.1 x 55 mL = 60.5 mmols NaOH added. (OH^-) = 60.5 mmols/155 mL = approx 0.4 but that's a close estimate. Convert that to pOH and then to pH. I get about 13.6
Two things here. You should see that with a buffered solution (both 1 and 2) there is little change in adding relatively high amounts of NaOH BUT with pure water with a pH = 7, the pH changes drastically (from 7 to 13.6) when the same amount of NaOH is added.
The second thing is, how did I know that the 400 mmols HAc would be end up at 200. First you know that it must be half and half between HAc and Ac^- but if you don't reason that out, you can do it this way. This is what you have in the titration.
.......HAc + OH^- ==> Ac^- + H2O
I......400...0.........0......0
add..........x.................
C......-x...-x.........+x
E.....400-x..0..........x
Now the problem tells you that you want the ratio to be 1:1 (which means you want 400-x = x)
Solve for x = 200 mmols NaOH must be added. That will be 200/1.1 = 181.82 mL of that 1.1 M NaOH solution so the final volume of THAT solution will be 100 mL + 181.81 mL + 55 mL = ?. I love these problems.
To predict the final pH in each of these scenarios, we can use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the pKa and the concentration of the acid and its conjugate base:
pH = pKa + log ([A-] / [HA])
where [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the acid.
1) In this case, we initially have a buffer solution with a 1:1 ratio of acid (acetic acid) and its conjugate base (sodium acetate). The pKa of acetic acid is 4.74. We are asked to predict the final pH when 55.0 mL of 1.10 M NaOH is added.
First, let's calculate the concentration of acetic acid and sodium acetate after adding the 55.0 mL of 1.10 M NaOH. Since the volume of the solution changes due to the addition of NaOH, we need to consider the new total volume.
Initial volume = 100.0 mL
Volume of NaOH added = 55.0 mL
Final volume = 100.0 mL + 55.0 mL = 155.0 mL = 0.155 L
To calculate the new concentration of acetic acid:
Initial moles of acetic acid = 0.400 mol
Moles of acetic acid neutralized by NaOH = (55.0 mL) x (1.10 mol/L) = 60.5 mmol = 0.0605 mol
Final moles of acetic acid = 0.400 mol - 0.0605 mol = 0.3395 mol
Concentration of acetic acid = (final moles of acetic acid) / (final volume) = 0.3395 mol / 0.155 L
To calculate the new concentration of sodium acetate:
Initial moles of sodium acetate = 0.400 mol
Moles of sodium acetate formed by the reaction of NaOH with acetic acid = 0.0605 mol
Final moles of sodium acetate = 0.400 mol + 0.0605 mol = 0.4605 mol
Concentration of sodium acetate = (final moles of sodium acetate) / (final volume) = 0.4605 mol / 0.155 L
Now we can substitute these values into the Henderson-Hasselbalch equation:
pH = 4.74 + log ([0.4605] / [0.3395])
Calculating this expression gives a pH of approximately 4.87, which matches your result.
2) In this scenario, we are preparing a buffer solution by adding acetic acid to water and then titrating with strong base until a 1:1 buffer ratio is achieved. The initial pH is given as 4.74.
Since there is no conjugate base initially, we consider the moles of acetic acid that react with NaOH to form sodium acetate. Following the same steps as in the first scenario, we find that the final moles of acetic acid and sodium acetate will be the same: 0.0605 mol.
Now we can calculate the concentration of acetic acid and sodium acetate:
Concentration of acetic acid = (0.3395 mol) / (0.155 L)
Concentration of sodium acetate = (0.4605 mol) / (0.155 L)
Substituting these values into the Henderson-Hasselbalch equation gives:
pH = 4.74 + log ([0.4605] / [0.3395])
Evaluating this expression, we find a final pH of approximately 4.90.
3) In this case, we have a solution of pure water. Pure water is neutral and has a pH of 7.0. When 55.0 mL of 1.10 M NaOH is added, a strong base, it will react with the water as follows:
NaOH + H2O -> Na+ + OH- + H2O
The addition of the strong base will result in an increase in hydroxide ion concentration ([OH-]), causing the solution to become basic (pH > 7). However, to calculate the final pH, we need to consider the initial concentration of hydroxide ions.
Initial concentration of hydroxide ions (before adding NaOH) = 10^-(pH) = 10^-(7) = 1 x 10^(-7) M
Moles of hydroxide ions added = (55.0 mL) x (1.10 mol/L) = 60.5 mmol = 0.0605 mol
Final concentration of hydroxide ions = (initial concentration) + (moles of OH- added) / (final volume) = (1 x 10^(-7) M) + (0.0605 mol) / (0.155 L)
Finally, we can calculate the final pH using the equation:
pOH = -log ([OH-])
pH = 14 - pOH
Substituting the value of [OH-] into the equation, we can determine the final pH.