Question

± pH Changes in Buffers When a solution contains a weak acid and its conjugate base...

± pH Changes in Buffers

When a solution contains a weak acid and its conjugate base or a weak base and its conjugate acid, it will be a buffer solution. Buffers resist change in pH following the addition of acid or base. A buffer solution prepared from a weak acid (HA) and its conjugate base (A−) is represented as

HA(aq)⇌H+(aq)+A−(aq)

The buffer will follow Le Châtelier's principle. If acid is added, the reaction shifts to consume the addedH+, 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 0.809 mol of the weak acid HA to 0.507 mol of NaA in 2.00 Lof solution? The dissociation constant Ka of HA is 5.66×10−7.

Express the pH numerically to three decimal places.

pH =

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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 three decimal places.

pH =

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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 three decimal places.

pH =

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Homework Answers

Answer #1

A)

pKa = -log Ka

   = -log (5.66*10^-7)

   = 6.247

use:

pH = pKa + log {[A-]/[HA]}

we can use number of moles instead of concentration, since volume occurring in both numerator and denominator will cancel each other

pH = pKa + log {nA-/nHA}

= 6.247 + log (0.507/0.809)

   = 6.044

B)

acid added = 0.150 mol

so,

0.150 mol of A- will react to form additional 0.150 mol of HA

pH = pKa + log {nA-/nHA}

= 6.247 + log ((0.507 – 0.150)/(0.809+0.150))

   = 5.818

c)

BASE ADDED = 0.195 mol

0195 mol of HA will react to form additional 0.195 mol of A-

pH = pKa + log {nA-/nHA}

= 6.247 + log ((0.507+0.195)/(0.809 – 0.195))

   = 6.305

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