Question

Calculate the pH of (a) a solution that is 0.060M in potassium propionate (C2H5COOK or KC3H5O2) and 0.085M in proprionic acid (C2H5COOH or HC3H5O2); (b) a solution that is 0.075M in trimethylamine, (CH3)3N, and 0.10M in trimethylammonium chloride, (CH3)3NHCl; (c) a solution that is made by mixing 50.0 mL of 0.15M acetic acid and 50.0 mL of 0.20M sodium acetate.

Answer 1

Question a) You have produced a propionic acid. potassium propionate buffer soloution. The pH of a buffer solution is determined using the Henderson - Hasselbalch equation:
You need:
pKa propionic acid = 4.87
Molarity of CH3CH2COOH = 0.085M
Molarity of CH3CH2COOK = 0.06M
Equation:
pH = pKa + log ([salt]/[acid] )
pH = 4.87 + log (0.06/0.085)
pH = 4.87 + log 0.706
pH = 4.87 + (-0.15)
pH = 4.72

Question b) you have a basic buffer. Easier to solve for pOH and convert to pH
The Kb value for (CH3)3N is 6.5 x 10-5.
pKb = -log (6.5*10^-5) = 4.19

pOH = pKa + log (salt]/[base])
pOH = 4.19 + log ( 0.10/0.075)
pOH = 4.19 + log 1.33
pOH = 4.19 + 0.12
pOH = 4.31
pH = 14.00- pOH
pH = 9.69


Question c: Once again you hace an acetic acid/ sodium acetate buffer
pKa for CH3COOH = 4.74
Molarity of Acetic acid - the final volume is 100mL
M1V1 = M2V2
M1*100 = 0.15* 50
M1 = 0.15*50/100 = 0.075M
Molarity of CH3COONa
M1*100 = 0.20/50 = 0.10M

Now substitute into the H-H equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log( 0.10/0.075)
pH = 4.74 + log 1.33
pH = 4.74 + 0.12
pH = 4.86