Ka, KbKa, Kb

Comparing the pH of two acidsComparing the pH of two acids1. Predict the pH of HCl and HF (below)2. Calibrate a pH meter3. Measure the pH of HCl(aq) and HF(aq)4. Complete the chart below

HCl (aq) HF (aq)

[ ] in mol/L (on label)

Net ionic equation

Predicted [H+]

Predicted pH

pH measured

Actual [H+]

Conductivity (demo) Higher / stronger Lower / weaker

0.05 0.05

HCl H+ + Cl– HF H+ + F–

0.05 0.05-log(0.05)=1.3 -log(0.05)=1.3

1.3 2.3 ?

10–pH = 0.05 10–pH = 0.005

QuestionsQuestionsRead15.3. (pg. 607+)1. Based on your results, which acid ionizes (forms

ions) to a greater degree? 2. Which two measurements taken in the lab

support your answer to 1?3. What is another name for Ka?4. Solve PE 5, 65. Write the Ka equation for HCl (aq) and HF (aq)

from today’s lab6. Solve for PE 8, 9 (use this equilibrium for butyric

acid: HBu H+ + Bu–)7. For HF(aq) set up a RICE chart, then solve for

Ka. How does your value for Ka compare to the accepted value (pg. 608)?

8. Try PE 10 (follow example 15.7 on pg. 610)

AnswersAnswers1. HCl ionizes more than HF 2. HCl has a lower pH (indicating more H+), & a

higher conductivity (indicating more ions)3. Ka: acid ionization constant4. HNO2 H+

+ NO2–, Ka=[H+][NO2

–]/[HNO2]

HPO42–

H+ + PO4

3–,Ka=[H+][PO43–]/[HPO4

2–]

5. HCl H+ + Cl–, Ka=[H+][Cl–]/[HCl]

HF H+ + F–, Ka=[H+][F–]/[HF]

PE 8 - pg. 610 HBu H+ + Bu–

R

ICE

HBu H+ Bu–

1 1 10.0100 0 0-0.0004 +0.0004 +0.00040.0096 0.0004 0.0004

[HBu]Ka =

[H+][Bu–]=

[.0096]

[0.0004]2= 1.67 x 10

– 5

[H+] = 10– pH = 10–

3.40 = 3.98 x 10–

4

PE 9 - pg. 610 HBu H+ + Bu–

R

ICE

HBu H+ Bu–

1 1 10.0100 0 0-0.001 +0.001 +0.0010.009 0.001 0.001

[HBu]Ka =

[H+][Bu–]=

[.009]

[0.001]2= 1.1 x 10

– 4

[H+] = 10– pH = 10–

2.98 = 1.05 x 10–

3

Question 7: HF H+ + F–

R

ICE

HF H+ F–

1 1 10.05 0 0

-0.005 +0.005 +0.0050.045 0.005 0.005

[HF]Ka =

[H+][F–]=

[.045]

[0.005]2= 5.6 x 10

– 4

[H+] = 10– pH = 10–

2.3 = 0.005

Accepted value of Ka for HF is 6.4 x 10 –

4

10: HC2H4NO2 H+ + C2H4NO2–

R

ICE

HC2H4NO2 H+ C2H4NO2–

1 1 10.010 0 0

-x +x +x0.010 - x x x

[HF]Ka =

[H+][C2H4NO2–]

=[0.010 - x]

[x]2= 1.4 x 10

– 5

Since x is small 0.010 – x = 0.010

x= 3.74 x 10–5 M, pH = 3.43[0.010]

[x]2=1.4 x 10 – 5

Ka summaryKa summary• Ka follows the pattern of other “K” equations• I.e. for HA(aq) + H2O(l) H3O+(aq) + A–(aq)

• Ka = [H3O+][A–] / [HA]

• Notice that H2O is ignored because it is liquid

• HA cannot be ignored because it is aqueous• This is different than with Ksp. In Ksp, solids

could only be in solution as ions• Acids can be in solution whether ionized or not• The solubility of acids makes sense if you

think back to the partial charges in HCl for ex.

Ka summaryKa summary• Generally Ka tells you about acid strength• Strong acids have high Ka values• A “strong” acid is an acid that completely

ionizes. E.g. HCl + H2O H3O+ + Cl–

• A “weak” acid is an acid that doesn’t ionize completely. E.g. HF + H2O H3O+ + F–

• Note: don’t get confused between strength and concentration. 1 M HCN has a smaller [H+], thus a higher pH, than 0.001 M HCl

• In general: Ka < 10 –

3 Weak acid

10 –

3 < Ka < 1 Moderate acid

Ka > 1 Strong acid

Dissociation vs. IonizationDissociation vs. Ionization• Ionization and dissociation indicate ions form• Dissociation: ions form when a chemical comes

apart. E.g. NaCl melts to form Na+, Cl–

• Ionization: ions form when two chemicals react. E.g. HCl(aq) + H2O H3O+(aq) + Cl–(aq)

• Even though we write HCl H+ + Cl– , this is just an abbreviation. In reality HCl reacts with H2O,

thus it is an ionization not a dissociation• Note that NaCl can also dissociate in water. This

is not an ionization, since water is only required to stabilize ions (it is not needed as a reactant involved in forming ions)

Kb – the last K (I promise)Kb – the last K (I promise)• Kb is similar to Ka except b stands for base• The general reaction involving a base can be

written as B(aq) + H2O BH+(aq) + OH–(aq)

• Thus Kb = [BH+] [OH–] / [B]• Recall: shorthand for Ka is HA H+ + A–

• Kb has no shorthand form• Read pg. 614 - 617• Try PE 12 (a-c), 13, 14 (for 13, you do not

need to know the chemical formula of morphine. Symbolize it with M)

PE 12PE 12a) CN–(aq) + H2O HCN(aq) + OH–(aq)

Kb = [HCN][OH–] / [CN–]

b) C2H3O2–(aq) + H2O HC2H3O2(aq) + OH–(aq)

Kb = [HC2H3O2][OH–] / [C2H3O2–]

c) C6H5NH2(aq) + H2O C6H5NH3+(aq) + OH–(aq)

Kb = [C6H5NH3+][OH–] / [C6H5NH2]

PE 13 - pg. 617 M + H2O MH+ + OH–

R

I

CE

M MH+ OH–

1 1 1

0.010 0 0-0.00013 +0.00013 +0.00013

0.000130.000130.00987

[M]Kb =

[MH+] [OH–]= =1.7 x 10-6

pOH = 14 - pH = 14 - 10.10 = 3.90[OH-] = 10-pOH = 10-3.90 = 1.26 x 10-4

[0.00987]

[0.00013] [0.00013]

PE 14 - pg. 617 M + H2O MH+ + OH–

R

I

CE

NH3 NH4+ OH–

1 1 1

0.020 0 0-x +x +x

xx0.020 - x

[0.020]Kb =

[x] [x]= = 1.8 x 10-5

x= 6.0 x 10-4

pOH = -log[OH-] = 3.22pH = 14 - pOH = 10.78

[0.020]

x2

Strength of conjugatesConsider HCl(l) + H2O Cl–(aq) + H3O+(aq)

The Ka for HCl is [Cl–(aq)] [H3O+(aq)] / [HCl(aq)]

Also, Cl–(aq) + H2O(aq) HCl(l) + OH–

The Kb for Cl– is [HCl(aq)] / [Cl–(aq)] [H3O+(aq)]

Relative values of KaRecall for HX H+ + X–, Ka = [H+][X–] / [HX]Q - what does a large Ka indicate?A - equilibrium is far to the right (all dissociates)Thus a large Ka = strong acidLook at Table 15.4 on page 608The text uses this definition:

Ka < 10–3 is a weak acid10–3 < Ka < 1 is a moderate acid 1 < Ka is a strong acid

These definitions are somewhat arbitrary, we will not focus on this. Just remember a high Ka means the acid is strong.

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