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
(Ka = 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 |
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
1.) You have two buffered solutions. Buffered solution 1 consists of 5.0 M HOAc and 5.0 M NaOAc; buffered solution 2 is...
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