A solution of a weak acid (HA) and its conjugate base (A), HA(aq)H+(aq)+A(aq) is

Question

A solution of a weak acid (HA) and its conjugate base (A),

HA(aq)H+(aq)+A(aq)

is called a buffer and will resist a change in pH. If acid is added, the reaction shifts to consume the added H+, forming more HA. When base is added, the base will react with H+, reducing its concentration. The reaction then shifts to replace H+ through the dissociation of HA into H+ and A. In both instances, [H+] tends to remain constant.

The pH of a buffer is calculated by using the Henderson-Hasselbalch equation:

pH=pKa+log[A][HA]

Part A

What is the pH of a buffer prepared by adding 0.506 mol of the weak acid HA to 0.406 mol of NaAin 2.00 L of solution? The dissociation constant Ka of HA is 5.66×107.

Express the pH numerically to two decimal places.

Part B

What is the pH after 0.150 mol of HCl is added to the buffer from Part A? Assume no volume change on the addition of the acid.

Express the pH numerically to two decimal places.

Part C

What is the pH after 0.195 mol of NaOH is added to the buffer from Part A? Assume no volume change on the addition of the base.

Express the pH numerically to two decimal places.

Explanation / Answer

moles of acid = 0.506

moles of salt = 0.406

Ka =5.66×107.

pKa = -logKa = -log (5.66×107.) = 6.25

Part A

pH = pKa + log[salt/acid]

pH = 6.25 + log[0.406/0.506]

pH = 6.15

part B)

no of moles of acid HCl = C =0.150

on addition of ’ C’ moles of acid to acidic buffer salt moles decreases and acid moles increases

so

pH = pKa + log [Salt –C/acid + C]

pH = 6.25 + log [0.406 –0.150 / 0.506 + 0.150 ]

pH = 5.84

part C )

no of moles of acid NaOH = C =0.195

on addition of ’ C’ moles of base to acidic buffer salt moles increases and acid moles decreases

so

pH = pKa + log [Salt + C / acid - C]

pH = 6.25 + log [0.406 + 0.195 / 0.506 - 0.195 ]

pH = 6.54

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