I would do this. First, note that the concns of acid and base are equal which means we can dispense with the molarity and just use volume of 150 mL.
..........HCOOH + OH^- ==>HCOO^- + H2O
We must take some quantity of HCOOH and add NaOH to it so volume NaOH = X.
Since the total volume is to be 150 mL, then we know HCOOH must be 150-x to start. What will it be at the end
.........HCOOH + OH^- ==> HCOO^- + H2O
begin....150-x....0.........0.......0
add...............x..................
react.....-x......-x........+x....+x
final...150-2x....0..........x.....x
3.5 = 3.75 + log[x/(150-2x)]
Solve for x (I get about 40 mL)
Then 150-x is about 110 mL HCOOH to begin and it will be 110-x = about 70 after reaction with the NaOH. So in the end we will have
pH = 3.75 + log(40/70) = close to 3.5
You need to do it more accurately than I've done.
I like to substitute these amounts, multiplied by the molarity or course, react them, and substitute the final mmoles into the HH equation and see if it gives me 3.5. If not we have trouble, if so we're ok. I tried it and it looks ok to me.
How would you prepare a 150.0mL of a 3.50 pH buffer solution using .500M NaOH and .500M HCOOH. I've gotten the base/acid ratio for the Henderson-H. equation. But now I'm not sure how to figure out the balance of solutions of HCOOH and NaOH to get both 150mL total and still have the ratio between COOH- and HCOOH
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