Ka Acid Ionization by Abhishek Jaguessar

8/4/2019 Ka Acid Ionization by Abhishek Jaguessar

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Ka, Kb

BYAbhishek Jaguessar



Comparing 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)=1.31.3 2.3 ?

10 –pH = 0.05 10 –pH = 0.005



QuestionsRead15.3. (pg. 607+)1. Based on your results, which acid ionizes

(forms ions) to a greater degree?

2. Which two measurements taken in the labsupport 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 lab

6. Solve for PE 8, 9 (use this equilibrium for

butyric acid: HBu H+ + Bu –)

7. For HF(aq) set up a RICE chart, then solvefor Ka. How does your value for Kacompare to the accepted value (pg. 608)?

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



Answers1. HCl ionizes more than HF

2. HCl has a lower pH (indicating more H+), & ahigher conductivity (indicating more ions)

3. Ka: acid ionization constant

4. HNO2 H+ + NO2 –, Ka=[H+][NO2 –]/[HNO2]HPO4

2 – H+

+ PO43 –,Ka=[H+][PO4

3 –]/[HPO42 –

]

5. HCl H+ + Cl –, Ka=[H+][Cl –]/[HCl]HF H+

+ F –, Ka=[H+][F –]/[HF]



PE 8 - pg. 610 HBu H+ + Bu –

R

I

CE

HBu

H+ Bu –

1 1 1

0.0100 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

I

CE

HBu

H+ Bu –

1 1 1

0.0100 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 –

RI

CE

HF H+ F –

1 1 10.05 0 0

-0.005 +0.005 +0.005

0.045 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 –

RI

CE

HC2H4NO2 H+ C2H4NO2 –

1 1 10.010 0 0

-x +x +x

0.010 - 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 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 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. 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 is similar to Ka except b stands for base

• The general reaction involving a base can bewritten 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 notneed to know the chemical formula ofmorphine. Symbolize it with M)



PE 12

a) 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 10.010 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 10.020 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-] = 3.22

pH = 14 - pOH = 10.78

[0.020]

x2

Ka Acid Ionization by Abhishek Jaguessar

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