You have to prepare a pH 3.50 buffer, and you have the following 0.10M solutions available HCOOH, CH3COOH, H3PO4,CHOONa, CH3COONa, and NaH2PO4.

a) What solutions would you use?
b) How many milliliters of each solution would you use to make approximately 1 liter of the buffer?

I figured out the solutions you would use based on the pKa values. The solutions you would use would be CHOOH and CHOONa.

Im stuck on how to find the milliliters of each solution. I used the Henderson-Hasselbalch equation to find the ratio between CHOO- and CHOOH and that is .575 but I'm confused on where to go from there.

It would help if you gave the values of Ka when problems like this is posted. Since Ka values differ from text to text (and from website to website) we can't get the same answer you get if we don't use the same numbers you have. Anyway, I took your 0.575 and calculated that you used 3.74 for pKa formic acid.

Don't you have a strength of the buffer. 1L of 0.1M buffer; 1 L of 0.3M buffer, or what.

Well, it seems you've got the right solutions for your buffer, which is a good start! Now, let's move on to finding the milliliters of each solution.

To determine the exact amounts of each solution needed, you'll need to consider the Henderson-Hasselbalch equation and the desired pH of the buffer. The Henderson-Hasselbalch equation for a weak acid buffer is given by:

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

In this case, the weak acid is HCOOH (formic acid) and its conjugate base is CHOONa (sodium formate). We know that the desired pH is 3.50, so let's substitute the values into the equation:

3.50 = pKa + log([CHOONa]/[HCOOH])

As you mentioned, the ratio of [CHOONa]/[HCOOH] is 0.575. Now, we can rearrange the equation to solve for the ratio [CHOONa]/[HCOOH]:

log([CHOONa]/[HCOOH]) = 3.50 - pKa

Using the pKa value for formic acid (HCOOH), which is 3.75, we can calculate the log term:

log([CHOONa]/[HCOOH]) = 3.50 - 3.75
log([CHOONa]/[HCOOH]) = -0.25

Now, we can convert the log back into a ratio:

[CHOONa]/[HCOOH] = 10^(-0.25)
[CHOONa]/[HCOOH] ≈ 0.562

Since we want the total volume of the buffer to be approximately 1 liter, we can choose any values we want for the volume of HCOOH and CHOONa, as long as they add up to 1 liter.

Let's say we choose to have 0.600 liters (600 mL) of HCOOH. This means that the remaining 0.400 liters (400 mL) will be CHOONa. Now, we can calculate the actual milliliters needed for each solution based on their concentrations:

For HCOOH:
Volume (HCOOH) = (0.600 L) x (0.10 M) = 60 mL

For CHOONa:
Volume (CHOONa) = (0.400 L) x (0.10 M) = 40 mL

So, approximately 60 mL of HCOOH and 40 mL of CHOONa would be needed to make a 1 liter pH 3.50 buffer.

Remember, these calculations assume ideal behavior and that the pKa values and concentrations remain constant. I hope this helps and adds a little "pHunny" to your chemistry!

To find the milliliters of each solution needed to prepare the pH 3.50 buffer, you need to consider the Henderson-Hasselbalch equation and use the information you have.

The Henderson-Hasselbalch equation for a buffer is:

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

Where pH is the desired pH of the buffer, pKa is the acid dissociation constant of the acid component of the buffer, [A-] is the concentration of the conjugate base, and [HA] is the concentration of the acid.

For a pH 3.50 buffer, you are given two possible solutions: HCOOH and CHOONa. Since you've determined that CHOOH and CHOONa are the solutions to use based on the pKa values, we can proceed with those.

Now, let's calculate the ratio of the concentrations of CHOO- and CHOOH using the Henderson-Hasselbalch equation:

3.50 = pKa + log([CHOONa]/[CHOOH])

Since CHOOH is the acid and CHOONa is its conjugate base, we can rewrite this as:

3.50 = pKa + log([CHOONa]/[CHOO-])

Now, substitute the pKa value for CHOOH into the equation. The pKa of CHOOH is typically provided in the question. Let's assume it is 3.75 for this explanation:

3.50 = 3.75 + log([CHOONa]/[CHOO-])

Now, let's rearrange the equation to solve for the ratio ([CHOONa]/[CHOO-]):

log([CHOONa]/[CHOO-]) = 3.50 - 3.75
log([CHOONa]/[CHOO-]) = -0.25

Next, take the antilog (inverse log) of both sides to find the actual ratio:

[CHOONa]/[CHOO-] = 10^(-0.25)
[CHOONa]/[CHOO-] = 0.5623

Now, since the ratio of [CHOONa] to [CHOO-] is 0.5623, you can determine the relative amounts of CHOO- and CHOOH needed to prepare the buffer.

If you want to prepare approximately 1 liter of buffer, you can choose a total volume for the buffer, such as 1000 ml. You can then assign a particular volume to one of the solutions, and adjust the volume of the other solution based on the desired ratio.

Let's assign a volume of V1 ml to the CHOONa solution:

Volume of CHOOH solution = V2 ml (since it is the other solution)

Since the ratio of [CHOONa] to [CHOO-] is 0.5623, you can express this as:

V1/V2 = 0.5623

You can choose any value for V1 as long as it is reasonable and convenient to work with. For example, let's say you choose V1 = 500 ml.

Now, you can substitute the value of V1 and solve for V2:

500/V2 = 0.5623

Cross-multiplying gives:

V2 = 500/0.5623
V2 ≈ 888.38 ml (rounded to two decimal places)

So, approximately 500 ml of CHOONa solution and 888.38 ml of CHOOH solution should be mixed to prepare approximately 1 liter of the pH 3.50 buffer.

Please note that this calculation assumes ideal behavior, and in practice, you may need to adjust the volumes slightly to account for practical considerations and molarities of the solutions.

If you don't care what the strength of the buffer is you can do it this way. Then when you finish you can calculate the strength of the solution you have prepared.

let x = mL base; then 1000-x = mL acid.
3.5 = 3.74 + log (0.1x)/[(0.1)(1000-x)]
Then solve for x and 1000-x.
If I punched in the right numbers I get approx 400 mL base and 600 mL acid. You can get a better number when you work the problem. I always like to check thse things.
pH = 3.74 + log (400/600)
pH = about 3.56 which is close. Remember that 400 and 600 are just estimates. I think you will come out with 3.5 when you do it right.
Then you can calculate the M of the final solution.
You have 0.1 x apprx 400 = approx 40 millimols base and approx 60 mmols acid which makes approx 100 mmols in 1000 mL or 0.1M for the mixture. Again, that isn't exact but you see how to do it from this. In fact, whatever numbers you get the concn will be 0.1M doing it this way.