Q1: You wish to prepare a buffer solution with pH = 11.10. What volume of 6.0 M HCl would you add to 500 mL of 0.10 M (C2H5)2NH to prepare the buffer? You may assume that the solution’s volume remains constant.

Q2: What is the resulting pH when 20 mL of 0.20 M NaOH is added to the buffer prepared in Q1?
Q3: Imagine diluting 250 mL of the buffer prepared in Q1 with distilled water to a final volume of 500 mL. What is the resulting pH when 20 mL of 0.20 M NaOH is added to the diluted buffer?

What pKa are you using for diethylamine. I found 11.02.

millimoles R2NH = mL x M = 500 x 0.1 = 50.

.........R2NH + H^+ ===> R2NH2^+
I.........50....0.........0
Add.............x...........
C.........-x...-x.........x
E.........50-x..0.........x
pH = pKa + log(base/acid)
11.10 = 11.02 + log(50-x/x)
Solve for x = millimols acid added.
M = mmols/mL or
mL = mmols/M = ?mmols/6 = mL 6M HCl.
I will leave the other parts for you. Post your work if get stuck.
)

Q1: To prepare a buffer solution with a specific pH, you need to choose an acid and its conjugate base whose pKa is close to the desired pH. In this case, the acid is (C2H5)2NH and its conjugate base is (C2H5)2NH2+.

To calculate the volume of 6.0 M HCl needed, we can use the Henderson-Hasselbalch equation:

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

First, let's convert the given concentrations to moles:

[HCl] = 6.0 M
[(C2H5)2NH] = 0.10 M

Next, we can assume that the volumes of the acid and base solutions are additive, so the final volume of the buffer solution will be 500 mL. Let's assign the volume of HCl as x mL. Therefore, the volume of (C2H5)2NH will be (500 - x) mL.

Now, let's set up the Henderson-Hasselbalch equation:

11.10 = pKa + log([(C2H5)2NH]/[(C2H5)2NH2+])

To find the pKa, we can use the relationship between pKa and Ka:

pKa = -log(Ka)

Assuming Ka for (C2H5)2NH is 1.8 x 10^-4, we get:

pKa = -log(1.8 x 10^-4) = 3.75

Now, we can substitute the values into the Henderson-Hasselbalch equation:

11.10 = 3.75 + log(0.10/(500 - x))

Simplifying the equation gives:

7.35 = log(0.10/(500 - x))

Next, we convert the equation to exponential form:

10^7.35 = 0.10/(500 - x)

Solving for x:

x = 500 - (0.10/10^7.35)

Now, you can calculate the value of x using a calculator.

Q2: To determine the resulting pH when 20 mL of 0.20 M NaOH is added to the buffer prepared in Q1, we can use the Henderson-Hasselbalch equation again.

Given the volume of the buffer solution is 500 mL, and we have already calculated the volume of HCl needed in Q1, which is x mL. Therefore, the volume of (C2H5)2NH is (500 - x) mL.

You can calculate the moles of NaOH added by multiplying the volume (20 mL) by the concentration (0.20 M). Let's call this molesNaOH.

To determine the new concentrations of (C2H5)2NH and (C2H5)2NH2+ after NaOH is added, we need to subtract molesNaOH from moles(C2H5)2NH and add molesNaOH to moles(C2H5)2NH2+. Then, divide the new concentrations by the total volume of the solution (500 mL) to obtain the new concentration of the buffer components.

Finally, you can use these new concentrations in the Henderson-Hasselbalch equation:

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

Q3: To find the resulting pH when 20 mL of 0.20 M NaOH is added to the diluted buffer, you can follow a similar procedure as in Q2.

First, you need to calculate the new concentrations of the buffer components after dilution. Since the final volume is 500 mL and you diluted 250 mL of the buffer prepared in Q1, the volume remaining from Q1 is 250 mL.

To calculate the moles of NaOH added, multiply the volume (20 mL) by the concentration (0.20 M). Let's call this molesNaOH.

Next, determine the remaining moles of the buffer components after dilution by subtracting molesNaOH from the initial moles of each component. Divide the remaining moles by the final volume (500 mL) to obtain the new concentration.

Finally, substitute these new concentrations into the Henderson-Hasselbalch equation to determine the resulting pH.