posted by Caitlin
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.
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.