chapter 17 part 1 acid-base equilibria - miami dade college · acid-base equilibria • common-ion...

Acid-Base Equilibria

Chapter 17 – Part 1

Acid-Base Equilibria

•  Common-Ion Effect •  Buffers

•  Acid-Base Titration

•  Titration Curves

The Common Ion Effect

A system at equilibrium will shift in response to being stressed.

The addition of a reactant or a product can be an applied stress.

[H+] = [CH3COO–] = 1.34 x 10–3 M; pH = 2.87

Initial concentration (M) 0.10 0 0

Change in concentration (M) –x +x +x

Equilibrium concentration (M) 0.10 – x x x

CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)

CH3COONa(aq) Na+(aq) + CH3COO–(aq) H2O

CH3COOH(aq) H+(aq) + CH3COO–(aq)

Equilibrium is driven toward reactant

addition

The Common Ion Effect-Practice Problem

Determine the pH at 25°C of a solution prepared by adding 0.050 mol of sodium acetate to 1.0 L of 0.10 M acetic acid solution, assume constant volume.

Buffer Solutions

A solution that contains a weak acid and its conjugate base (or a weak base and its conjugate acid) is a buffer. Buffer solutions resist changes in pH.

CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq) acid conjugate base

reacts with added base

reacts with added acid

CH3COOH(aq) + OH–(aq) CH3COO– (aq) + H2O(l)

CH3COOH(aq) CH3COO– (aq) + H+(aq)

Buffer Solutions-Practice Problems

Calculate the pH of a buffer that contains 1.0 M acetic acid (Ka = 1.8 x 10-5) and 1.0 M sodium acetate.

Buffer Solutions

Now because this is a buffer, its resists a change in the pH upon the addition of acid (H+) or base (OH-) Ex: Calculate the change in pH that occurs when 0.10 mol HCl is added to 1.0 L of the previous buffer.

Buffer Solutions

In a buffered solution, the pH is governed by the ration of [HA] (weak acid) to the common ion [A-] The pH of a buffer solution can often be calculated with the Henderson-Hasselbalch equation.

[ ][ ]a

conjugate basepH p log

weak acidK= +

Henderson-Hasselbalch Equation-Practice

Starting with 1.00 L of a buffer that is 1.00 M in acetic acid and 1.00 M in sodium acetate, calculate the pH after the addition of 0.100 moles of NaOH. (assume the addition does not change the volume of the solution)

Buffer Solutions

A solution is only a buffer if it has the capacity to resist pH change either when an acid or a base is added. Buffers must have conjugate acid and base concentrations within a factor of 10. The pH of a buffer cannot be more than one pH unit different than the pKa of the weak acid it contains.

[ ][ ]

conjugate baseweak acid

10 0.1≥ ≥

Buffer Solutions

To make a buffer with a specific pH:

1) Pick a weak acid whose pKa is close to the desired pH.

2) Substitute the pH and pKa into the equation below to obtain the necessary [conjugate base]/[weak acid] ratio.

[ ]a

ApH = p log

HAK

−⎡ ⎤⎣ ⎦+

Buffer Solutions

Select an appropriate acid from the table below to describe how you would prepare a buffer with pH = 4.5.

Weak Acid Ka pKa

HF 7.1 x 10–4 3.15

HNO2 4.5 x 10–4 3.35

HCOOH 1.7 x 10–4 3.77

C6H5COOH 6.5 x 10–5 4.19

CH3COOH 1.8 x 10–5 4.74

HCN 4.9 x 10–10 9.31

C6H5OH 1.3 x 10–10 9.89

Solution:

Step 1: Pick the acid with pH ≈ pKa. CH3COOH

Step 2: Substitute into the equation below:

[ ]a

ApH = p log

HAK

−⎡ ⎤⎣ ⎦+

[ ]A

4.5 = 4.74 logHA

−⎡ ⎤⎣ ⎦+

Buffer Solutions

Select an appropriate acid from the table below to describe how you would prepare a buffer with pH = 4.5.

Solution:

Step 2 (cont):

[ ]A

4.5 = 4.74 logHA

−⎡ ⎤⎣ ⎦+[ ]

0.24A

10 0.6HA

−

−⎡ ⎤⎣ ⎦ = =

Dissolve 0.6 mol of CH3COONa and 1 mol of CH3COOH in enough water to make 1 L of solution.

Acid-Base Titrations

1. Strong Acid-Strong Base Titrations The reaction between the strong acid HCl and the strong base NaOH can be represented by:

NaOH(aq) + HCl(aq) → NaCl(aq) + H2O(l) or by the net ionic equation,

OH–(aq) + H+(aq) → H2O(l)


1. Strong Acid-Strong Base Titrations

Titration of 25.0 mL of 0.100 M HCl with 0.100 M NaOH

Practice – Strong Acid-Strong Base Titration

Calculate the pH when (a) 49.0 mL and (b) 51.0 mL of 0.100 M NaOH solution have been added to a 50.0 mL of 0.100 M HCl.


2.   Weak Acid-Strong Base Titrations: Major difference: to calculate [H+], you much deal with the the weak acids’ equilibrium expression. Consider the neutralization between acetic acid and sodium hydroxide:

CH3COO– (aq) + H2O(l)(aq) ⇌ CH3COOH(aq) + OH–(aq)

CH3COOH(aq) + OH–(aq) → CH3COO– (aq) + H2O(l)

The acetate ion that results from this neutralization undergoes hydrolysis:


2. Weak Acid-Strong Base Titrations

Titration of 25.0 mL of 0.100 M acetic acid with 0.100 M NaOH


The initial pH is determined by the ionization of acetic acid.

x = [H+] = 1.34 x 10–3 M; pH = 2.87


Change in concentration (M)

Equilibrium concentration (M)

CH3COOH(aq) ⇌ H+(aq) + CH3COO–(aq)

–x +x +x

0.10 – x x x

25

a 1.8 100.10xK

x−= = ×

−



After the addition of base, some of the acetic acid has been converted to acetate ion: The solution is a buffer and the Henderson-Hasselbalch equation can be used to calculate pH. After 10.0 mL of base has been added:

CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)

[ ]a

ApH = p log

HAK

−⎡ ⎤⎣ ⎦+[ ][ ]1.0 mmol

pH = 4.74 log 4.561.5 mmol

+ =

Volume of OH– added (mL)

OH– added (mol)

CH3COOH remaining

CH3COO– produced pH

10.0 1.0 1.5 1.0 4.56



At the equivalence point, all the acetic acid has been neutralized.

CH3COOH(aq) + OH–(aq) → CH3COO–(aq) + H2O(l)


OH– added (mol)

CH3COOH remaining


25.0 2.5 0.0 2.5 8.72




CH3COO–(aq) + H2O(l) ⇌ OH–(aq) + CH3COOH (aq)

Must use TOTAL VOLUME to calculate concentration. 3

2.5 mmolCH COO 0.050 50.0 mL

M−⎡ ⎤ = =⎣ ⎦


Use Ka x Kb = Kw to get Kb for the acetate ion; Kb = 5.6 x 10–10

x = [OH–] = 5.3 x 10–6 ; pOH = 5.28; pH = 8.72





CH3COO–(aq) + H2O(l) ⇌ OH–(aq) + CH3COOH (aq)

[ ] 23 10

b3

OH CH COOH5.6 10

0.050CH COOxK

x

−

−

−

⎡ ⎤⎣ ⎦= = = ×−⎡ ⎤⎣ ⎦



After the equivalence point, all the acetic acid has been neutralized, nothing is left to neutralize the added strong base.

Volume of OH– added

(mL)

OH– added (mmol)

Excess OH–

(mmol)

Total Volume

(mL)

[OH– ] (mol/L)

pOH pH

30.0 3.0 0.5 55.0 0.0091 2.04 11.96

35.0 3.5 1.0 60.0 0.017 1.78 12.22


OH– added (mol)

CH3COOH remaining


25.0 2.5 0.0 2.5 8.72


Practice – Weak Acid-Strong Base Titration

Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M acetic acid, CH3COOH. (Ka = 1.8 x 10-5).

Practice – Weak Acid-Strong Base Titration

Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.


3. Strong Acid-Weak Base Titrations

Titration of 25.0 mL of 0.100 M NH3 with 0.100 M HCl


The equivalence point in a titration can be visualized with the use of an acid-base indicator. The endpoint of a titration is the point at which the color of the indicator changes.

HIn(aq) ⇌ H+(aq) + In–(aq)

Objectives •  Know what the common ion effect for is for weak acids and

weak bases. •  Know what a buffer is and how to determine the pH of a

buffered solution. •  Be able to solve problems using the Henderson-Hasselbalch

equation. •  Know what titration curves are (know the information that it tells

you about the buffer). Be able to distinguish the titration curve of a strong acid-strong base, strong acid-weak base, and weak acid-strong base.

chapter 17 part 1 acid-base equilibria - miami dade college · acid-base equilibria • common-ion...

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