Question

1.) You have two buffered solutions. Buffered solution 1 consists of 5.0 M HOAc and 5.0 M NaOAc; buffered solution 2 is made of 0.050 M HOAc and 0.050 M NaOAc. How do the pHs of the buffered solutions compare? (Note: Ac^– = acetate ion, CH₃COO^–).

A. The pH of buffered solution 1 is greater than that of buffered solution 2.
B. None of the answers are correct
C. The pH of buffered solution 1 is equal to that of buffered solution 2.
D. The pH of buffered solution 2 is greater than that of buffered solution 1.
E. Cannot be determined without the Ka values.

2.) Calculate the pH of a solution that is 0.50 M in HF (K_a = 7.2 x10^–4) and 0.99 M in NaF.

A. 3.14
B. 3.44
C. 0.30
D. 10.56
E. 2.85

Answer 1

1)

we know that

for buffers

pH = pKa + log [salt / acid ]

pH = pKa + log [NaOAc / HOAc]

for solution 1

pH = pKa + log [ 5 / 5]

pH = pKa + log 1

pH = pKa + 0

pH = pKa

for solution 2

pH = pKa + log [ 0.05 / 0.05 ]

pH = pKa + log 1

pH = pKa + 0

pH = pKa

so

for both the solutions pH is same

so

the answer is C) the pH of the buffered solution1 is equal to that of buffered solution 2

2)

HF is a weak acid and NaF is its salt of its conjugate base

so

they form a buffer solution

we know that

for buffers

pH = pKa + log [salt / acid ]

also

pKa = -log Ka

so

pH = -log Ka + log [ NaF / HF]

given

Ka = 7.2 x 10-4

[NaF] = 0.99

[HF] = 0.5

so

using those values

we get

pH = -log ( 7.2 x 10-4) + log ( 0.99 / 0.5)

pH = 3.44

so

pH of the solution is 3.44

so

the answer is B) 3.44