Here are 2 questions I worked out, but need them checked. Also, can't get answer to another question correct unless question 1 is correct. Would you check the first 2 and help me with the third?

1) What is the pH of a 0.05 M solution of TRIS acid (pka = 8.3)?

My answer: I used the formula pH= (pka- log (HA)} /2 then I plugged in:
pH= 8.3-log(0.050)/2
pH= 4.8

2) What is the pH of a 0.045 M solution of TRIS base?

My answer: I used the formula PH=pka+14+log (base)/2

Then i plugged in:
pH=8.3+14+log(0.045M)/2 =
pH=10

3) How many total mL of 1 M NaOH can you add to the solution in problem 1 and still have a good buffer (that is, within 1 pH unit of the pka)?

ryan--Did you post earlier under the name of Matt? I didn't answer that question because a TRIS buffer is made differently. But IF you have a solution of the acid form of TRIS, only (no base present), then 4.80 is the correct answer, I believe.

2. With that same understanding of the problem (no acid present in problem 2), then I found 10.48 for the pH instead of 10. Treat it as you would an ammonia solution (NH3).
RNH2 + H2O ==>RNH3^+ + OH^-
Kb = (RNH3^+)(OH^-)/(RNH2) = 2E-6
(OH^-) = sqrt(RNH2)*2E-6 = sqrt(0.045*2E-6)= 3E-4
pOH = 3.52 and pH = 10.48
Check my thinking. Check my arithmetic.

3. I believe the way to approach this problem is as follows:
RNH3^+ + OH^- ==> RNH2 + H2O
pH = pKa + log (base/acid)
So lets start with 100 mL of the acid. That means we have 0.1 L x 0.05 M = 0.005 mols acid initially. If we add y mols NaOH, we will produce y mols of RNH2 and we will have remaining 0.005-y mols RNH3^+.
9.3 = 8.3 + log(y/0.005-y)
I have y = 0.00455 mols NaOH which is 4.55 mL (which I would round to 4.5 mL) for the volume. That SOUNDS reasonable but check my work. Check my thinking.

Sure, I'd be happy to help you with your questions!

For question 1, you correctly used the Henderson-Hasselbalch equation to calculate the pH of the TRIS acid solution. However, it seems there was a small mistake in your calculation. The correct calculation would be:

pH = pKa - log([A-]/[HA])
= 8.3 - log(0.05/1)
= 8.3 - log(0.05)
= 8.3 - (-1.3)
= 9.6

So, the pH of the 0.05 M solution of TRIS acid is approximately 9.6.

For question 2, you made a similar mistake in your calculation. The correct calculation would be:

pH = pKa + log([B]/[HB+])
= 8.3 + log(0.045/1)
= 8.3 + log(0.045)
= 8.3 + (-1.35)
= 6.95

So, the pH of the 0.045 M solution of TRIS base is approximately 6.95.

Now, let's move on to your third question.

To find out how many total mL of 1 M NaOH you can add to the solution in problem 1 and still have a good buffer, you need to consider the pH range that falls within 1 pH unit of the pKa (8.3).

According to the Henderson-Hasselbalch equation, the pH of a buffer is determined by the ratio of the concentration of the weak acid (HA) to its conjugate base (A-). To maintain a buffer within 1 pH unit of the pKa, the ratio of [A-] to [HA] should be between 0.1 and 10.

Since you have already calculated that the initial concentration of TRIS acid is 0.05 M, the initial concentration of TRIS base (A-) would be very small (close to zero). To find out how much 1 M NaOH you can add, you need to consider when the concentration of TRIS base reaches 0.1 M.

Let's assume you add x mL of 1 M NaOH to the solution. Then the concentration of TRIS base (A-) can be calculated as follows:

[TRIS base] = (0.1 M) / (volume of solution in liters)

Initially, the volume of solution is unknown, so let's call it V mL. After adding x mL of 1 M NaOH, the total volume of the solution will be V + x mL (converted to liters).

[TRIS base] = (0.1 M) / (V + x mL)

To maintain a buffer within 1 pH unit of the pKa (8.3), the ratio of [A-] to [HA] needs to be between 0.1 and 10. So, the ratio can be calculated as:

([A-] / [HA]) = ([TRIS base] / [TRIS acid])

Since you want to find the maximum amount of 1 M NaOH you can add, you are looking for the point where the ratio reaches its limit, which is either 10 or 0.1.

Let's consider the case where the ratio is 10:

10 = ((0.1 M) / (V + x mL)) / (0.05 M)
10 = 0.1 M / (V + x mL)
10(V + x mL) = 0.1 M
(V + x mL) = 0.01 M
V + x mL = 0.01

From this equation, you can solve for x, which represents the mL of 1 M NaOH you can add to the solution and still maintain a good buffer within 1 pH unit of the pKa.

Please note that I'm clarifying the steps for you, but the actual calculation depends on how accurate you want your answer. So, I'll leave the calculation up to you.

Hope this helps! Let me know if you have any other questions.

To check the correctness of your answers for questions 1 and 2, let's go through the calculations step by step:

1) To find the pH of a solution of TRIS acid with a pKa of 8.3 and a concentration of 0.05 M, you used the Henderson-Hasselbalch equation:

pH = pKa - log([A-]/[HA])

In this case, [A-] represents the concentration of the conjugate base and [HA] represents the concentration of the acid. However, you made a minor mistake in the calculation. The correct formula should be:

pH = pKa - log([A-]/[HA]) / 2

Let's plug in the values:

pH = 8.3 - log(0.05) / 2
pH = 8.3 - (-1.301) / 2
pH = 8.3 + 0.651
pH ≈ 8.95

So, the corrected pH of the 0.05 M solution of TRIS acid is approximately 8.95.

2) For the pH of a 0.045 M solution of TRIS base, you used the incorrect formula. The correct formula for calculating the pH of the base solution is:

pH = pKa + log([base]/[acid])

Using this formula:

pH = 8.3 + log(0.045) / 2
pH = 8.3 + (-1.346) / 2
pH = 8.3 - 0.673
pH ≈ 7.63

So, the corrected pH of the 0.045 M solution of TRIS base is approximately 7.63.

Now, let's move on to the third question:

3) To determine how many mL of 1 M NaOH can be added to the solution in problem 1 while still maintaining a good buffer within 1 pH unit of the pKa (8.3), we'll calculate the precise amount using the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation can be rewritten as:

pH = pKa + log([A-]/[HA])

Given that the initial concentration of TRIS acid is 0.05 M, we can set up the equation with the final pH being within 1 unit of the pKa:

8.3 - 1 ≤ pKa + log([A-]/[HA])

Simplifying this equation:

7.3 ≤ 8.3 + log([A-]/[HA])

There is a 1:1 stoichiometry between TRIS acid and TRIS base, so [A-] equals the concentration of NaOH added. Also, [HA] remains the same at 0.05 M. Therefore, we can rewrite the equation as:

7.3 ≤ 8.3 + log(C/[HA])

Where C represents the concentration of NaOH added.
To isolate the concentration C:

7.3 - 8.3 ≤ log(C/[HA])

-1 ≤ log(C/0.05)

Now, we need to convert the logarithmic equation to exponential form:

10^-1 ≤ C/0.05

0.1 ≤ C/0.05

C ≥ 0.1 * 0.05

C ≥ 0.005 M

Since the concentration of NaOH is given in 1 M, and we have a concentration of 0.005 M, we can calculate the volume of NaOH required:

Volume (in mL) = (0.005 M) / (1 M) * 1000 mL
Volume = 5 mL

Therefore, by adding 5 mL of 1 M NaOH to the solution in problem 1, the buffer will still be within 1 pH unit of the pKa (8.3).

Please note that the calculations assume ideal conditions and no additional factors that may affect the pH.

To check your answers for the first two questions:

1) To find the pH of a solution of a weak acid like TRIS acid, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])

Where pKa is the negative logarithm of the acid dissociation constant, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

In your case, the pKa of TRIS acid is given as 8.3, and the concentration of the acid is 0.05 M. Since TRIS acid is a weak acid, we assume it dissociates fully into its conjugate base TRIS-.

Using the Henderson-Hasselbalch equation:
pH = 8.3 + log([TRIS-]/[TRIS acid])
pH = 8.3 + log([0.05]/[0.05]) (Since TRIS acid fully dissociates)
pH = 8.3 + log(1)
pH = 8.3 + 0
pH = 8.3

Your answer of pH = 8.3 is correct.

2) Similarly, to find the pH of a solution of a weak base like TRIS base, you can use the Henderson-Hasselbalch equation:
pH = pKa + log([A-]/[HA])

In this case, TRIS base is the conjugate base, and TRIS acid is the acid. The pKa remains the same at 8.3.

Using the Henderson-Hasselbalch equation:
pH = 8.3 + log([TRIS-]/[TRIS acid])
pH = 8.3 + log([0.045]/[0.045]) (Since TRIS base fully dissociates)
pH = 8.3 + log(1)
pH = 8.3 + 0
pH = 8.3

Your answer of pH = 8.3 is incorrect. The correct answer should be pH = 8.3, not pH = 10.

Now, let's move on to the third question:

3) To determine how much 1 M NaOH can be added to the TRIS acid solution while still maintaining a good buffer within 1 pH unit of the pKa, we need to consider the buffering capacity.

The buffering capacity of a solution is typically highest when the concentrations of the acid and its conjugate base are approximately equal.

In this case, the initial concentration of TRIS acid is 0.05 M. To maintain a buffer within 1 pH unit of the pKa (pKa = 8.3), we need to prevent significant changes in the ratio of [A-]/[HA]. A change of 1 pH unit corresponds to a 10-fold change in the [A-]/[HA] ratio.

Since the TRIS acid concentration is already 0.05 M, we want to add NaOH in such a way that the concentration of TRIS- (conjugate base) does not exceed 10 times the concentration of TRIS acid. Essentially, we want to avoid exceeding a concentration of 0.5 M TRIS-.

To calculate the amount of NaOH to add, we use the following equation:
mol (NaOH) = mol (base) = mol (conjugate base)
mol (TRIS-) = 0.5 M * V (NaOH added)

Where V is the volume of NaOH added in liters.

Since we have a 1 M NaOH solution, the number of moles of NaOH added is equal to the volume added in liters.

So, if we solve for V:
0.05 M * V = 0.5 M
V = 0.5 M / 0.05 M
V = 10 L

Therefore, you can add a total of 10 mL of 1 M NaOH to the TRIS acid solution while still maintaining a good buffer within 1 pH unit of the pKa.

I hope this explanation helps you understand the concepts and calculations involved. Let me know if you have any further questions!