Propionic acid has a ka of 1.3x10^(-5), 50.0ml of a buffer solution having a pH of 4.65 is to be prepared using a 0.20 M solutions of propionic acid and sodium propionate. How many ml of each of these solutions must be mixed?

Thanks in Advanced !

Use the HH equation. The way I read the problem you have a solution that is 0.2M in propionic acid and another solution 0.2M in sodium propionate.

pKa = -log Ka = about 4.89 but you should confirm that and adjust the figure to your liking.
4.65 = 4.89 + log (base/acid)
b/a = 0.575
Then base = 0.575*acid
Let x = mL acid and 50-x = mL base.
millimols acid = 0.2M*xmL acid
millimols base = 0.2M*(50-xmL acid) so
0.2*(50-xmLacid) = 0.575*0.2M*xmL acid
10-0.2xmL acid = 0.115x*mL acid
10 = 0.315x
x = 10/0.315 = 31.75 mL acid
50-x = 18.25 mL base

I always check these things to make sure they come out to the desired pH.
M base = 18.25*0.2/50 = 0.073
M acid = 31.75*0.2/50 = 0.127
pH = 4.89 + log(0.073/0.127) = 4.65. voila!
Note: I've carried this to too many significant figures and they should be adjusted to only two places so 32 and 18 would be appropriate I think.

Well, well, well! Mixing some solutions, are we? Let's get clowning!

To prepare this clown-worthy buffer solution, we need to make use of the Henderson-Hasselbalch equation. Are you ready? Here it goes:

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

We have the value of Ka, which is 1.3x10^(-5), and we want the pH to be 4.65. Since we have propionic acid and sodium propionate, we'll assign [HA] to the propionic acid and [A-] to the sodium propionate.

Now, let's plug in the values and solve for log([A-]/[HA]):

4.65 = -log(1.3x10^(-5)) + log([A-]/[HA])

Simplifying:

4.65 = 5.88 - log([A-]/[HA])

Rearranging:

log([A-]/[HA]) = 5.88 - 4.65

log([A-]/[HA]) = 1.23

Now, you'll need the antilog (also known as the "clown magic trick") to get rid of the logarithm and find [A-]/[HA]. Once you've got that ratio, you can do a simple calculation to determine the volume of each solution.

So, my dear friend, grab your calculator (and perhaps some clown makeup), follow the steps, and get mixing those solutions in the appropriate amounts. Happy buffering!

To calculate how many mL of each solution should be mixed, we can use the Henderson-Hasselbalch equation, which relates the ratio of the conjugate acid-base pair to the pH of the buffer solution:

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

Here, [A-] represents the sodium propionate concentration, and [HA] represents the propionic acid concentration.

Given:
pH = 4.65
pKa = -log(Ka) = -log(1.3x10^(-5))

Step 1: Calculate the ratio [A-]/[HA] from the Henderson-Hasselbalch equation.

4.65 = -log(1.3x10^(-5)) + log([A-]/[HA])
4.65 = 4.89 + log([A-]/[HA])
log([A-]/[HA]) = 4.65 - 4.89
log([A-]/[HA]) = -0.24

Step 2: Convert the ratio to a numerical value.

[A-]/[HA] = 10^(-0.24)

Step 3: Determine the required concentrations of propionic acid and sodium propionate.

Let x be the volume (in mL) of the 0.20 M propionic acid solution.
Therefore, the volume of the 0.20 M sodium propionate solution will be (50.0 - x) mL.

The moles of sodium propionate will be (0.20 M) * (50.0 - x) mL.
The moles of propionic acid will be (0.20 M) * x mL.

Step 4: Calculate the concentrations of propionic acid and sodium propionate.

The concentration of sodium propionate will be (moles of sodium propionate) / 50.0 mL.
The concentration of propionic acid will be (moles of propionic acid) / 50.0 mL.

Step 5: Substitute these concentrations into the [A-]/[HA] ratio.

[A-]/[HA] = concentration of sodium propionate / concentration of propionic acid

Step 6: Substitute the numerical value of [A-]/[HA] and solve for x.

10^(-0.24) = (concentration of sodium propionate) / (concentration of propionic acid)

Finally, you can solve for x to determine the volume of the 0.20 M propionic acid solution needed. Substituting this value back into (50.0 - x) will give you the volume of the 0.20 M sodium propionate solution needed.

To determine the volume of each solution needed to prepare the buffer, we need to use the Henderson-Hasselbalch equation:

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

In this case, propionic acid (HA) is being mixed with sodium propionate (A-) to form a buffer solution.

Given:
Ka = 1.3x10^(-5)
pH = 4.65
Volume of buffer = 50.0 mL
Concentration of propionic acid (HA) = 0.20 M

Step 1: Convert the given pH to [H+].
pH = -log[H+]
4.65 = -log[H+]
[H+] = 10^(-pH)
[H+] = 10^(-4.65)

Step 2: Calculate the ratio of [A-] to [HA].
[H+] = [A-]
[HA] = 0.20 M

Step 3: Use the Henderson-Hasselbalch equation to find the ratio of [A-] to [HA].
4.65 = -log(1.3x10^(-5)) + log([A-]/0.20)
4.65 + log(1.3x10^(-5)) = log([A-]/0.20)
log([A-]/0.20) = 4.65 + log(1.3x10^(-5))
log([A-]/0.20) = log(10^(4.65)) + log(1.3x10^(-5))
log([A-]/0.20) = log(10^(4.65) × 1.3x10^(-5))
[A-]/0.20 = 10^(4.65) × 1.3x10^(-5)
[A-]/0.20 = 48610 × 1.3x10^(-5)
[A-] = (48610 × 1.3x10^(-5)) × 0.20

Step 4: Calculate the volume of sodium propionate (A-) needed.
Volume of sodium propionate (A-) = ([A-] × volume of buffer)/concentration of sodium propionate (A-)
Volume of sodium propionate (A-) = (48610 × 1.3x10^(-5)) × 0.20 × (50.0 mL / 0.20 M)

Step 5: Calculate the volume of propionic acid (HA) needed.
Volume of propionic acid (HA) = (volume of buffer - volume of sodium propionate (A-))

Therefore, the final answer will be the volume of sodium propionate (A-) and the volume of propionic acid (HA) that make up the buffer solution.