A buffer containing 0.50 M of an acid (HA) and 0.10 M of its conjugate base (A–) has a pH of 10.30. Which of the following solutions will make a buffer with a pH of 11.30?
A. A solution with [HA] = 0.25 M and [A–] = 0.10 M.
B. A solution with [HA] = 0.50 M and [A–] = 0.50 M.
C. A solution with [HA] = 0.10 M and [A–] = 0.50 M.
D. A solution with [HA] = 0.25 M and [A–] = 0.50 M.
E. A solution with [HA] = 1.00 M and [A–] = 0.50 M.
we have
[HA] = 0.50 M
[A-] = 0.10 M
pH = 10.30
therefore using Henderson–Hasselbalch equation, we have
pH = pKa + log[A-]/[HA]
putting all the values , we get
10.30 = pKa + log(0.10/0.50)
pKa = 10.30 - (-0.70)
pKa = 10.30 + 0.70
pKa = 11.00
and
for the pH = 11.30 , we will have
pH = pKa + log[A-]/[HA]
11.30 = 11.00 + log[A-]/[HA]
or
log[A-]/[HA] = 11.30 - 11.00 = 0.30
or
[A-]/[HA] = 10^0.30 = 2.00
thus th correct answer is :
D. A solution with [HA] = 0.25 M and [A–] = 0.50 M.
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