Question

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.

Answer 1

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.