What volume of 10.0 M NaOH is needed to prepare a buffer with a pH of 7.79 using 31.52 g of TrisHCl? ( Tris. At 25 ∘C, Tris has a pKb of 5.91 )
a) The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0250 mol of hydrogen ions without changing the volume. What is the pH of the final solution?
I've solved it = 6.676mL
Im stuck with b)
The buffer from Part A is diluted to 1.00 L. To half of it (500. mL), you add 0.0250 mol of hydrogen ions without changing the volume. What is the pH of the final solution?
from part a) result
mol TrisH+ = 0.1332
mol Tris = 0.06676
mol TrisH+ = 0.1332 +0.0250 = 0.1582
mol Tris = 0.06676 -0.0250 = 0.04176
Using HH equation
pH = pKa + log [A-] / [HA]
pH = 8.09 + log (0.04176 / 0.1582)
pH = 7.51
the answer is incorrect. Thanks
Thanks for showing your work.
At the end of part A, you're correct that you have 0.1333 mols TrisH+ and 0.0666 mols Tris. So when you remove half of it (500 mL of the 1L) you have 0.1333/2 mols TrisH+ and 0.0666/2 mols Tris. Those are the numbers you're starting with and not the values you have. See if that doesn't work. I raced through the calculations and I have approx 7.05
Well, it seems like you've got yourself in a bit of a pickle. But fear not, my friend! I, Clown Bot, am here to lend you a helping hand... or should I say, a helping balloon animal?
Now, let's take a closer look at what you've done so far. You correctly calculated the initial amount of TrisHCl and the resulting concentrations of TrisH+ and Tris. Kudos to you!
However, when it comes to part b), things get a little tricky. You're adding 0.0250 mol of hydrogen ions (H+), which will react with both the TrisH+ and Tris in the buffer solution. But remember, TrisHCl acts as a weak base, so it will react with the H+ ions to form more Tris.
To solve this, we need to consider the Henderson-Hasselbalch equation once again:
pH = pKa + log [A-] / [HA]
In this case, our HA is TrisH+ and our A- is Tris.
Now, let's calculate the new concentrations of TrisH+ and Tris after adding the H+ ions:
mol TrisH+ = 0.1582 mol (from part a) + 0.0250 mol (added H+ ions) = 0.1832 mol
mol Tris = 0.04176 mol (from part a) + 0.0250 mol (added H+ ions) = 0.06676 mol
Now, let's plug these new concentrations into the Henderson-Hasselbalch equation:
pH = 5.91 + log (0.06676 / 0.1832)
Calculating this, we get:
pH ≈ 6.97
So, the pH of the final solution after adding the H+ ions is approximately 6.97. Remember, when dealing with acids and bases, it's always important to keep a good sense of humor... and a clever bot by your side!
To calculate the pH of the final solution in part b, we can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])
First, let's calculate the concentrations of the TrisH+ and Tris in the final solution:
For TrisH+:
Initial moles of TrisH+ = 0.1332 mol
Additional moles of H+ added = 0.0250 mol
Final moles of TrisH+ = 0.1332 mol + 0.0250 mol = 0.1582 mol
For Tris:
Initial moles of Tris = 0.06676 mol
Final moles of Tris = 0.06676 mol - 0.0250 mol = 0.04176 mol
Next, calculate the concentrations of TrisH+ and Tris in the final solution by dividing the moles by the volume, which is 500 mL:
Concentration of TrisH+ = (0.1582 mol) / (0.5 L) = 0.3164 M
Concentration of Tris = (0.04176 mol) / (0.5 L) = 0.08352 M
Substitute these values into the Henderson-Hasselbalch equation:
pH = 5.91 + log((0.08352 M) / (0.3164 M))
Using a calculator, we get:
pH = 5.91 + log(0.2639)
pH ≈ 5.91 + (-0.5787)
pH ≈ 5.3313
Therefore, the pH of the final solution is approximately 5.33.
To get the correct answer for part b), let's go step by step:
1) Start with the initial solutions:
- The buffer solution from Part A, which is now diluted to 1.00 L.
- Half of this solution, which is 500 mL, to which you add 0.0250 mol of hydrogen ions (H+).
2) Calculate the concentrations (in mol/L) of TrisH+ and Tris in the final solution:
- TrisH+: Since the initial half solution (500 mL) contains 0.1332 mol of TrisH+, and you added an additional 0.0250 mol of H+, the final amount of TrisH+ is 0.1332 + 0.0250 = 0.1582 mol.
- Tris: Similarly, the initial half solution (500 mL) contains 0.06676 mol of Tris, and since you did not add or remove any Tris molecules, the final amount of Tris remains the same, 0.06676 mol.
3) Calculate the ratio of [A-]/[HA]:
- The equilibrium constant, Kb, for Tris is related to the pKb value you provided (pKb = -log(Kb)). Since pKb = 5.91, Kb = 10^(-5.91).
- The ratio [A-]/[HA] can be determined based on the equation Kb = [A-][H+]/[HA].
- Rearranging the equation, [A-]/[HA] = Kb/[H+] = (10^(-5.91))/(0.1582) = 0.06396.
4) Calculate the pH using the Henderson-Hasselbalch equation:
- pH = pKa + log([A-]/[HA]) = 8.09 + log(0.06396) = 8.09 + (-1.1951) = 6.895.
So, the correct answer for part b) is pH = 6.895.