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 of0.50 M phosphate buffer, pH = 683, 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:0.304 M 4. Molar quantity of HP 18.2 mmol Molar quantity of H,PO,: 11.8 mmol Concentration of H,PO,:0.196 M You add 1.7 mL of 1.00 M HCI to the buffer. Calculate the molar quantity of H,O' added as HCl, and the final molar quantity of HPO a. and H,PO, at equilibrium? What is the new HPOMH,PO, ratio, and the new pH of the solution? The pK, of H,PO is 6.64. Use t molar ratio rather than the concentration ratio because both species are in the same volume.) b. he Henderson-Hasselbach equation to calculate the new pH. (Note: You can use the 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. c. 3.Aml Naot 100 mol NaoH 1.00 ml NaOH 7 6

Explanation / Answer

a)

Moles of acid added = Molarity X volume = 1 X 1.7 mL = 1.7 mmol

It will react with salt to give more acid (same moles)

Iniital moles of salt = 18.2 mmol

Moles of salt left = 18.2 - 1.7 = 16.5mmol

Iniital moles of acid = 11.8 mmole

Moles of acid [HPO4-] after addition of acid= 11.8 + 1.7 = 13.5 mmole

b) We will use Hendersen Hassalbalch's equation

pH = pKa + log [salt] / [acid]

pH = pKa + log [HPO4-2] / [H2PO4-]

pH = 6.64 + log [HPO4-2] / [H2PO4-]

pH = 6.64 + log [16.5] / [13.5]

pH = 6.64 + (0.0871)

pH = 6.73

Ratio of [HPO4-2] / [H2PO4-] = 1.22

c) moles of NaOH added = Molarity X volume = 1 X 3.7 = 3.7mmol

Moles of NaOH will react with equal moles of acid

Moles of acid left = 11.8 - 3.7 = 8.1mmole

Moles of salt after addition of NaOH = 18.2 + 3.7 = 21.9 mmole

pH = pKa + log [salt] / [acid]

pH = 6.64 + log [21.9] / [8.1] = 7.07

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