Updates

• Assignment 06 is due Mon., March 12 (in class)

• Midterm 2 is Thurs., March 15 and will cover Chapters 16 & 17– Huggins 10, 7-8pm– For conflicts: ELL 221, 6-7pm (must arrange at

least one week in advance)

Acid-Base Equilibria andSolubility Equilibria

Chapter 17

Calculating pH Changes in Buffers

A buffer is made by adding 0.300 mol CH3CO2H and 0.300 mol CH3CO2Na to enough water to make 1.00 L of solution. The pH of the buffer is 4.74. Calculate the pH of this solution after 0.020 mol of NaOH is added.

Calculating pH Changes in Buffers

Before the reaction, since

mol CH3CO2H = mol CH3CO2−

pH = pKa = −log (1.8 10−5) = 4.74

Calculating pH Changes in Buffers

The 0.020 mol NaOH will react with 0.020 mol of the acetic acid:

CH3CO2H (aq) + OH−(aq) CH3CO2−(aq) + H2O(l)

CH3CO2H CH3CO2− OH−

Before reaction 0.300 mol 0.300 mol 0.020 mol

After reaction 0.280 mol 0.320 mol 0.000 mol

New concentrations: 0.280 mol/ 1 L = 0.280 M CH3CO2H 0.320 mol/ 1 L = 0.320 M CH3CO2

-

Calculating pH Changes in Buffers

Now use the Henderson–Hasselbalch equation to calculate the new pH:

pH = 4.74 + log(0.320)(0. 280)

pH = 4.74 + 0.06

pH = 4.80

Which of the following are buffer systems? (a) KF/HF (b) KBr/HBr, (c) Na2CO3/NaHCO3

(a) KF is a weak acid and F- is its conjugate basebuffer solution

(b) HBr is a strong acidnot a buffer solution

(c) CO32- is a weak base and HCO3

- is it conjugate acidbuffer solution

17.3

= 9.20

Calculate the pH of the 0.30 M NH3/0.36 M NH4Cl buffer system. What is the pH after the addition of 20.0 mL of 0.050 M NaOH to 80.0 mL of the buffer solution?

NH4+ (aq) H+ (aq) + NH3 (aq)

pH = pKa + log[NH3][NH4

+]pKa = 9.25 pH = 9.25 + log

[0.30][0.36]

= 9.17

NH4+ (aq) + OH- (aq) H2O (l) + NH3 (aq)

start (moles)

end (moles)

0.029 0.001 0.024

0.028 0.0 0.025

pH = 9.25 + log[0.25][0.28]

[NH4+] =

0.0280.10

final volume = 80.0 mL + 20.0 mL = 100 mL

[NH3] = 0.0250.10

17.3

Titration

In a titration a solution of accurately known concentration is added gradually to another solution of unknown concentration until the chemical reaction between the two solutions is complete.

Titration

A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

TitrationsEquivalence point – the point at which the reaction is complete

Indicator – substance that changes color at (or near) the equivalence point

Slowly add baseto unknown acid

UNTIL

The indicatorchanges color

(pink)

17.3

Change in indicator color occurs at the point when all ofthe acid or base being titrated has reacted. At this pointany additional titrant reacts with the indicator.

Titration of a Strong Acid with a Strong Base

From the start of the titration to near the equivalence point, the pH goes up slowly.

Just before and after the equivalence point, the pH increases rapidly.

Titration of a Strong Acid with a Strong Base

At the equivalence point, moles acid = moles base, and the solution contains only water and the salt from the cation of the base and the anion of the acid.

Titration of a Strong Acid with a Strong Base

As more base is added, the increase in pH again levels off.

Titration of a Strong Acid with a Strong Base

• Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.

• The pH at the equivalence point will be >7.

• Phenolphthalein is commonly used as an indicator in these titrations.

Titration of a Strong Acid with a Strong BaseWeak acid/strong base or strong

acid/weak base

Strong Acid-Strong Base Titrations

NaOH (aq) + HCl (aq) H2O (l) + NaCl (aq)

OH- (aq) + H+ (aq) H2O (l)

17.4

Weak Acid-Strong Base Titrations

CH3COOH (aq) + NaOH (aq) CH3COONa (aq) + H2O (l)

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

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

At equivalence point (pH > 7):

17.4

Strong Acid-Weak Base Titrations

HCl (aq) + NH3 (aq) NH4Cl (aq)

NH4+ (aq) + H2O (l) NH3 (aq) + H+ (aq)

At equivalence point (pH < 7):

17.4

H+ (aq) + NH3 (aq) NH4Cl (aq)

Exactly 100 mL of 0.10 M HNO2 are titrated with a 0.10 M NaOH solution. What is the pH at the equivalence point ?

HNO2 (aq) + OH- (aq) NO2- (aq) + H2O (l)

start (moles)

end (moles)

0.01 0.01

0.0 0.0 0.01

NO2- (aq) + H2O (l) OH- (aq) + HNO2 (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.05 0.00

-x +x

0.05 - x

0.00

+x

x x

[NO2-] =

0.010.200 = 0.05 MFinal volume = 200 mL

Kb =[OH-][HNO2]

[NO2-]

=x2

0.05-x= 2.2 x 10-11

0.05 – x 0.05 x 1.05 x 10-6 = [OH-]

pOH = 5.98

pH = 14 – pOH = 8.02

Acid-Base Indicators

HIn (aq) H+ (aq) + In- (aq)

10[HIn]

[In-]Color of acid (HIn) predominates

10[HIn]

[In-]Color of conjugate base (In-) predominates

17.5

The titration curve of a strong acid with a strong base.

17.5

Which indicator(s) would you use for a titration of HNO2 with KOH ?

Weak acid titrated with strong base.

At equivalence point, will have conjugate base of weak acid.

At equivalence point, pH > 7

Use cresol red or phenolphthalein

17.5