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 fore given pH Chang 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: Concentration of HPO_4: 0.304 M Concentration of H_2P0_4: 0.196 M Molar quantity of HP0_4: 18.2 moll Molar quantity of H_2P0_4: 11.8 moll You add 1.7 mL of 1.00 M HCI to the buffer. Calculate the molar quantity of H_3O added as HCI, and the final molar quantity of HP0_4^2- and H_2P0_4- at equilibrium? What is the new HP0427H, P04' ratio, and the new pH of the solution? The pK of h, po4 is 6.64. Use the Henderson-Hassel Bach equation to calculate the new pH. (You can use the molar ratio rather than the concentration ratio because both species are in the same volume) Now take a fresh 60 mL of the 0.50 M pH 6.83 buffer and add 3.7 mL of 1.00 M NaOH. Using steps similar to those above, calculate the new pH of the solution.

Explanation / Answer

a. moles of H3O+ added = 1 M x 1.7 ml = 1.7 mmol

new moles of HPO4^2- = 18.2 - 1.7 = 16.5 mmol

new moles of H2PO4- = 11.8 + 1.7 = 13.5 mmol

b. ratio (HPO4^2-/H2PO4-) = (16.5/13.5) = 1.222 mmol

new pH = 6.64 + log(base/acid)

             = 6.64 + log(1.222)

             = 6.73  

c. when moles of NaOH added = 1.0 M x 3.7 ml = 3.7 mmol

new moles of HPO4^2- = 18.2 - 3.7 = 21.9 mmol

new moles of H2PO4- = 11.8 - 3.7 = 8.1 mmol

new ratio (HPO4^2-/H2PO4-) = (21.9/8.1) = 2.704 mmol

New pH = 6.64 + log(2.704) = 7.072

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