Weak Acids & Weak Acids & Weak BasesWeak Bases

ReviewReview Try the next two questions to see what Try the next two questions to see what

you remember you remember

The pH of rainwater collected on a particular day was 4.82. What is the H+ ion concentration of the rainwater?

pH = -log [H+]

[H+] = 10-pH = 10-4.82 = 1.5 x 10-5 M

The OH- ion concentration of a blood sample is 2.5 x 10-7

M. What is the pH of the blood?

pH + pOH = 14.00

pOH = -log [OH-] = -log (2.5 x 10-7) = 6.60

pH = 14.00 – pOH = 14.00 – 6.60 = 7.40

New TopicNew Topic Weak Acids & Weak BasesWeak Acids & Weak Bases

Strong Electrolyte – 100% dissociation

NaCl (s) Na+ (aq) + Cl- (aq)

Weak Electrolyte – not completely dissociated

CH3COOH CH3COO- (aq) + H+ (aq)

Strong Acids are strong electrolytes

HCl (aq) + H2O (l) H3O+ (aq) + Cl- (aq)

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

HClO4 (aq) + H2O (l) H3O+ (aq) + ClO4- (aq)

H2SO4 (aq) + H2O (l) H3O+ (aq) + HSO4- (aq)

HF (aq) + H2O (l) H3O+ (aq) + F- (aq)

Weak Acids are weak electrolytes

HNO2 (aq) + H2O (l) H3O+ (aq) + NO2- (aq)

HSO4- (aq) + H2O (l) H3O+ (aq) + SO4

2- (aq)

H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

Strong Bases are strong electrolytes

NaOH (s) Na+ (aq) + OH- (aq)

KOH (s) K+ (aq) + OH- (aq)

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

F- (aq) + H2O (l) OH- (aq) + HF (aq)

Weak Bases are weak electrolytes

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

Conjugate acid-base pairs:• The conjugate base of a strong acid has no measurable

strength.

• H3O+ is the strongest acid that can exist in aqueous solution.

• The OH- ion is the strongest base that can exist in aqueous solution.

Strong Acid Weak Acid

What is the pH of a 2 x 10-3 M HNO3 solution?

HNO3 is a strong acid – 100% dissociation.

HNO3 (aq) + H2O (l) H3O+ (aq) + NO3- (aq)

pH = -log [H+] = -log [H3O+] = -log(0.002) = 2.7

Start

End

0.002 M

0.002 M 0.002 M0.0 M

0.0 M 0.0 M

What is the pH of a 1.8 x 10-2 M Ba(OH)2 solution?

Ba(OH)2 is a strong base – 100% dissociation.

Ba(OH)2 (s) Ba2+ (aq) + 2OH- (aq)

Start

End

0.018 M

0.018 M 0.036 M0.0 M

0.0 M 0.0 M

pH = 14.00 – pOH = 14.00 + log(0.036) = 12.6

HA (aq) + H2O (l) H3O+ (aq) + A- (aq)

Weak Acids (HA) and Acid Ionization Constants

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

Ka =[H+][A-][HA]

Ka is the acid ionization constant

Ka

weak acidstrength

Important PointsImportant Points When weak acids and bases are added to When weak acids and bases are added to

water the concentration of [Hwater the concentration of [H33OO++] and ] and [OH[OH--] is not the same as the concentration ] is not the same as the concentration of the dissolved acid. of the dissolved acid.

The concentration can be determined The concentration can be determined using the using the equilibrium constantequilibrium constant values values for acids and bases termed Kfor acids and bases termed Kaa and K and Kbb. .

Important PointsImportant Points KKaa values are the equilibrium constant values are the equilibrium constant

expressions for the dissolving of various acids expressions for the dissolving of various acids in water. in water.

The larger the value of KThe larger the value of Kaa the stronger the the stronger the acids are. acids are.

The same is true of bases and KThe same is true of bases and Kbb. The . The following link contains following link contains a table of acidsa table of acids and their and their KKaa values values

Calculating [HCalculating [H33OO++] and [OH] and [OH--] of Weak ] of Weak

Acids & BasesAcids & Bases We can calculate the concentration of the We can calculate the concentration of the

[H[H33OO++] or [OH] or [OH--] for a weak acid or base using K] for a weak acid or base using Kaa and Kand Kbb values and the following procedure. values and the following procedure.

The procedure assumes that the concentration The procedure assumes that the concentration of the dissolved acid or base does not change of the dissolved acid or base does not change significantly when the weak acid dissolves.significantly when the weak acid dissolves.

Procedure for Calculating [HProcedure for Calculating [H33OO++] and ] and [OH[OH--] of Weak Acids & Bases] of Weak Acids & Bases

1.1. Write the dissolving reaction for the weak acid. Write the dissolving reaction for the weak acid. 2.2. These are typically These are typically monoproticmonoprotic (acids or (acids or

bases) that produce only one Hbases) that produce only one H33OO++ ion or OH ion or OH-- ion. ion.

3.3. Write the KWrite the Kaa or K or Kbb expression for the acid or expression for the acid or base. base.

4.4. Calculate the concentration of the acid or base Calculate the concentration of the acid or base in moles per liter.in moles per liter.

5.5. Using a table of acids and bases locate the Using a table of acids and bases locate the value of Kvalue of Kaa or K or Kbb for the acid and base and for the acid and base and then use the following formulas to calculate the then use the following formulas to calculate the [H[H33OO++] or [OH] or [OH--]]

[H[H33OO++]]22 = [Acid]* K = [Acid]* Kaa

[OH[OH--]]2 2 = [Base] *K= [Base] *Kbb

6.6. Using the KUsing the Kww formula calculate the other formula calculate the other concentration or [OHconcentration or [OH--] or [H] or [H33OO++]]

Problem 1:Problem 1: Calculate the concentration of [HCalculate the concentration of [H33OO++] and [OH] and [OH--] ]

ions in a 0.0056 mol/l solution of HCN . ions in a 0.0056 mol/l solution of HCN . (K(Kaa value = 6.2 X10 value = 6.2 X10-10-10) )

Strategy:Strategy:1.1. Write the dissolving reaction: Write the dissolving reaction:

HCNHCN(aq) (aq) + H+ H22O O (l)(l) ↔ H ↔ H33OO++(aq)(aq) + CN + CN-- (aq)(aq)

2.2. Write the equilibrium expression KWrite the equilibrium expression Kaa : : KKaa = [H = [H33OO++]*[CN]*[CN--] / [HCN]] / [HCN]

3.3. Determine the concentration of the weak acid , Determine the concentration of the weak acid , [HCN] = 0.0056 mol/l [HCN] = 0.0056 mol/l

3.3. Using Using a table of acids a table of acids determine that the Kdetermine that the Ka a

value is = 6.2 X10value is = 6.2 X10-10-10

4.4. Calculate the [HCalculate the [H33OO++] by substituting information ] by substituting information into the formula to get into the formula to get [H[H33OO++]]22 = (0.0056 mol/L)(6.2 X10 = (0.0056 mol/L)(6.2 X10-10-10))[H[H33OO++] = 1.86 X10] = 1.86 X10-6-6 mol/L mol/L

5.5. Using the KUsing the Kww formula calculate the [OH formula calculate the [OH--] ; ] ; remember Kremember Kww = 1.0 X10 = 1.0 X10-14-14

[OH[OH--] = K] = Kww / [H / [H33OO++]][OH[OH--] = 1.0 X10] = 1.0 X10-14-14/ 1.86 X 10/ 1.86 X 10-6-6 mol/L mol/L[OH[OH--] = 5.37 X10] = 5.37 X10-9-9 mol/L mol/L

Important Note:Important Note: The procedure and formula used above only The procedure and formula used above only

work if the concentration of the weak acids and work if the concentration of the weak acids and bases is not significantly changed when it bases is not significantly changed when it dissolves. dissolves.

This is true when the [Acid] ÷ KThis is true when the [Acid] ÷ Kaa > 500 > 500

Work through the following questions by Work through the following questions by creating ICE tables that support this same creating ICE tables that support this same logic or strategy!logic or strategy!

What is the pH of a 0.5 M HF solution (at 250C)?

HF (aq) H+ (aq) + F- (aq) Ka =[H+][F-][HF]

= 7.1 x 10-4

HF (aq) H+ (aq) + F- (aq)

Initial (M)

Change (M)

Equilibrium (M)

0.50 0.00

-x +x

0.50 - x

0.00

+x

x x

Ka =x2

0.50 - x= 7.1 x 10-4

Ka x2

0.50= 7.1 x 10-4

0.50 – x 0.50The ratio is > 500,

x2 = 3.55 x 10-4 x = 0.019 M

[H+] = [F-] = 0.019 M pH = -log [H+] = 1.72

[HF] = 0.50 – x = 0.48 M

When can I use the approximation?

0.50 – x 0.50Ka << 1

When the ratio of initial concentration/Ka is greater than 500.

x = 0.5 0.50 M7.1 x 10-4

= 714 which is > 500

What is the pH of a 0.05 M HF solution (at 250C)?

0.05

7.1 X 10-4 = 71.4 which is < 500

We can not neglect the “x” value. We must solve for x exactly using quadratic equation or method of successive approximation.

Solving weak acid ionization problems:

1. Identify the major species that can affect the pH.

• In most cases, you can ignore the auto-ionization of water.

• Ignore [OH-] because it is determined by [H+].

2. Use ICE to express the equilibrium concentrations in terms of single unknown x.

3. Write Ka in terms of equilibrium concentrations. Solve for x by the approximation method. If approximation is not valid, solve for x exactly.

4. Calculate concentrations of all species and/or pH of the solution.

What is the pH of a 0.122 M monoprotic acid whose Ka is 5.7 x 10-4?

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

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

Ka =x2

0.122 - x= 5.7 x 10-4

Ka x2

0.122= 5.7 x 10-4

0.122 – x 0.122Ka << 1

0.1225.7 X 10-4

= 214.04This value is < 500

Approximation not valid.

Ka =x2

0.122 - x= 5.7 x 10-4 x2 + 0.00057x – 6.95 x 10-5 = 0

ax2 + bx + c =0-b ± b2 – 4ac

2ax =

x = 0.0081 x = - 0.0081

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

Initial (M)

Change (M)

Equilibrium (M)

0.122 0.00

-x +x

0.122 - x

0.00

+x

x x

[H+] = x = 0.0081 M pH = -log[H+] = 2.09

percent ionization = Ionized acid concentration at equilibrium

Initial concentration of acidx 100%

For a monoprotic acid HA

Percent ionization = [H+]

[HA]0

x 100% [HA]0 = initial concentration

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

Weak Bases and Base Ionization Constants

Kb =[NH4

+][OH-][NH3]

Kb is the base ionization constant

Kb

weak basestrength

Solve weak base problems like weak acids except solve for [OH-] instead of [H+].

Chemistry In Action: Antacids and the Stomach pH Balance

NaHCO3 (aq) + HCl (aq)

NaCl (aq) + H2O (l) + CO2 (g)

Mg(OH)2 (s) + 2HCl (aq)

MgCl2 (aq) + 2H2O (l)