Determine how many mL of solution A (acetic acid-indicator solution) must be added to solution B (sodium acetate-indicator solution) to obtain a buffer solution that is equimolar in acetate and acetic acid.

Question

Solution A:
10.0 mL 3.0e-4M bromescol green solution
25.0mL 1.60M acetic acid (HAc)
10.0mL .200M KCl solution

Solution B: 10.0mL 3.0e-4M bromescol green solution
10.0mL of .160M sodiuma cetate solution

a.Calculate moles Ac- in solution B
b. Calculate the molarity of HAc in solution A.
c. Calculate the volume of solution A needed when the moles of Ac- equals moles of HAc
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K HAc is approximately 2e-5 under your experimental conditions. Use pK HAc = pH + log(N HAc/ N Ac-)
a. Calculate pKa
b. calculate moles HAc in 5mL solution A added to solution B
c.Calculate moles Ac- in solution B
d. Rearrange equation to solve for pH

Answer 1

Determine how many mL of solution A (acetic acid-indicator solution) must be added to solution B (sodium acetate-indicator solution) to obtain a buffer solution that is equimolar in acetate and acetic acid.

Solution A:
10.0 mL 3.0e-4M bromescol green solution
25.0mL 1.60M acetic acid (HAc)
10.0mL .200M KCl solution

Solution B: 10.0mL 3.0e-4M bromescol green solution
10.0mL of .160M sodiuma cetate solution

The following three seem clear enough.
a.Calculate moles Ac- in solution B
(Note: mols = M x L = ? but I like to work in millimoles). So 10 mL x 0.160M = 1.6 mmols = 0.0016 mols.

b. Calculate the molarity of HAc in solution A.
1.6M x ((25 mL of HAc/45 mL total) = 0.888M HAc.

c. Calculate the volume of solution A needed when the moles of Ac- equals moles of HAc
mLA x MA = mmols B
mLA x 0.8889M = 1.6
mL A = 1.6/0.8889 = about 1.799 which I would round to 1.8 mL. You see why I like to work in mmols. mL x M = mmols AND M = mmols/mL.)

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K HAc is approximately 2e-5 under your experimental conditions. Use pK HAc = pH + log(N HAc/ N Ac-)
a. Calculate pKa
pKa = -log Ka = -log 2E-5 = about 4.70 but you need to confirm that and round to the right number of s.f.

b. calculate moles HAc in 5mL solution A added to solution B
mL A + M B = mmols HAc in B = 5 mL x 0.8889 = 4.45 millimols = 0.00445 mols HAc.

c.Calculate moles Ac- in solution B
Isn' that done above in the first part a? 0.0016 mols Ac^- (or 1.6 mmols).

d. Rearrange equation to solve for pH
The rearranged equation is pH = pKa + log(base)/(acid)
pH = 4.70 + log (1.60/4.45) = ? or if you want to use mols it is
pH = 4.80 + log(0.0016/0.00445) = ?
Check all of this carefully. Post any questions but show your work and explain exactly what your trouble is.

Answer 2

The molarity of HAc in solution A is wrong according to my TA.