Buffer capacity refers to the amount of acid or base a buffer can absorb without

Question

Buffer capacity refers to the amount of acid or base a buffer can absorb without a significant pH change. It is governed by the concentrations of the conjugate acid and base components of the buffer. A 0.5 M buffer can 'absorb" five times as much acid or base as a 0.1 M buffer for a given pH change. in this problem, you begin with a buffer of known pH and concentration and calculate the new pH after a particular quantity of acid or base is added. You are given 60 ml. of 0.50 M phosphate buffer. pH = 6.83. to test. The starting composition of the buffer, both in terms of the concentration and the molar quantity of the two major phosphate species, is: You add 1.7 mL of 1.00 M HCl to the buffer. Calculate the molar quantity of H_3 O^+ added as HCl, and the final molar quantity of HPO^2-_4 and H_2 PO^-_4 at equilibrium? What is the new HPO^2-_4/H_2 PO_4^- ratio, and the new pH of the solution? the pKa of H_2 PO_4^- is 6.64. Use the Henderson - Hassel Bach equation to calculate the new pH.

Explanation / Answer

a)

mol of H3O+ = MV = 1.7*1 = 1.7 mmol = 1.7*10^-3 mol of H3O+

then

HPO4-2 +H+ = H2PO4-

therefore

HPO4-2 = initial - reacted = left

H2PO4- = initial + reacted = formed

substitute

HPO4-2 = 18.2- 1.7 = 16.5

H2PO4- = 11.8+ 1.7 = 13.5

b)

finr ratio:

HPO4- / H2PO4-2 = 16.5 /13.5 = 1.222

pH new:

pH = pKa2 + log(HPO4- / H2PO4-2)

pH = 7.21 + log(1.222) = 7.29707

c)

V = 60 mL of buffer

V = 3.7 mL of NaOH...

mmol of NAOH = MV = 3.7*1 = 3.7 mmol

therefore

HPO4-2 = initial + reacted = left

H2PO4- = initial - reacted = formed

substitute

HPO4-2 = 18.2 + 3.7 = 21.9

H2PO4- = 11.8 - 3.7 = 8.1

then

ratio = 21.9/8.1 = 2.703

pH = 7.21 + log(2.703)

pH = 7.6418

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