Question

Two students are preparing 850 mL of a buffer using trimethyl-acetic acid (CH3)3COOH, and its potassium salt, potassium trimethyl-acetate (CH3)3COOK. When they make their buffer, they used .36 moles of the acid, but forgot to rwrite down how much (CH3)3COOK they added. They know the Ka of the weak acid is 9.3x10^-6.

a.) Rather than start over, they know that if they measure the pH, they can figure out how much (CH3)COOK they used. THey find that the pH=5.25. How many moles of (CH3)3COOK did they add when they made their buffer?

b.) They know that a buffer resists pH changes because it cintains a weak acid and a weak base. One of these can neutralize added strong acif (H3O^+). The other can neutralize strong base (OH^-) if it is added. Complete the statements below...each answer should be a chemical formula.

The weak acid in the buffer is________________ and it would neutralize added _______________.

The weak base in the buffer is _______________ and it would neutralize added________________.

c.) They add 160 mL of 1.6M NaOH to their buffer. Write the reaction describing what happens. Do not include any spectator ions. Also, state (Yes or No) whether all of the strong base can be neutralized.

d.) Compute the concentrations of the (CH3)3COOK and (CH3)3COOH remaining after the strong base in part c is added and the pH of the buffer.

*Please show all work. Thank you!

Answer 1

a) Using Hendersen-Hasselbalck equation,

pH = pKa + log([base]/[acid])

Ka = 9.3 x 10^-6

pKa = 5.03

let x moles of base (CH3)3COOK is added then,

5.25 = 5.03 + log(x/0.36-x)

1.654(0.36-x) = x

x = 0.23 mols

So the student has added 0.23 mols of (CH3)3COOK

b) the weak acid in the buffer is (CH3)3COOH and it would neutralize added OH^-. the weak base in the buffer is (CH3)3COOK and it would neutralize added H3O^+.

c) moles of NaOH added = 1.6 M x 0.160 L = 0.256 mols

added base would neutralize weak acid and would increase the concentration of weak base in the solution.

the chemical equations would be thus,

(CH3)3COOH + NaOH ---> (CH3)3COONa + H2O

final concentration,

conc. of (CH3)3COOH = 0.36 - 0.256 = 0.104 mols

conc. of (CH3)3COOK = 0.23 + 0.256 = 0.486 mols

So all of the added base will be neutralized.

d) concentrations of the species in solution after the NaOH is added,

final concentration,

conc. of (CH3)3COOH = 0.36 - 0.256 = 0.104 mols

conc. of (CH3)3COOK = 0.23 + 0.256 = 0.486 mols

pH = 5.03 + log(0.486/0.104)

     = 5.70