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Acids & Bases. Calculating Ka from % Dissociation Polyprotic Acids Properties of Salts. Example. Lactic acid (HC 3 H 5 O 3 ) is a waste product that accumulates in the muscle during exertion, leading to pain & a feeling of fatigue. A 0.100 M solution is 3.7% dissociated. Calculate K a. - PowerPoint PPT Presentation
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Acids & Bases
Calculating Ka from % DissociationPolyprotic AcidsProperties of Salts
Example
Lactic acid (HC3H5O3) is a waste product that accumulates in the muscle during exertion, leading to pain & a feeling of fatigue. A 0.100 M solution is 3.7% dissociated. Calculate Ka.
Example
I
C
E
HC3H5O3 H+ + C3H5O3 -
Example
I 0.1 0 0
C -0.0037 +0.0037 +0.0037
E 0.9963 0.0037 0.0037
HC3H5O3 H+ + C3H5O3 -
0.100 x 3.7% = 0.0037
Example
Ka = [H+][C3H5O3-] / [HC3H5O3]
Ka = [0.0037][0.0037] / [0.9963] Ka = 1.37 x 10-5
Polyprotic Acids
More that one ionizable H+ H+ come off one at a timeEach one has a specific Ka valueKa1>>>>Ka2>>Ka3
When calculating the pH of a polyprotic acid, you only use the 1st Ka because all of the others are so small
Complete Ionization
Write the steps for the complete ionization of H2Se
H2Se
H2Se + H20 H30+ + HSe- Ka1 = #
HSe- + H20 H30+ + Se-2 Ka2 = #
Complete Ionization
Write the steps for the complete ionization of H3AsO
H3AsO + H2O H2AsO - + H3O + Ka1 = #
H2AsO - + H2O HAsO -2 + H3O + Ka2 = #
HAsO -2 + H2O AsO -3 + H3O + Ka3 = #
Example
Calculate the pH & the [ ] of all species of a 3.0 M solution of H3PO4.
Ka1 = 7.5 x 10-3, Ka2 = 6.2 x 10-8, Ka3 = 4.8 x 10 -13
Example
H3PO4 H+ + H2PO4- Ka1 = 7.5 x 10-3
I
C
E
Example
H3PO4 H+ + H2PO4- Ka1 = 7.5 x 10-3
I 3 0 0
C -x +x +x
E 3-x x x
2.85 0.146 o.146
7.5 x 10-3 = x2 / (3-x)
7.5 x 10-3 (3-x) = x2
x = 0.146
Example
H2PO4 - H+ + HPO4
-2 Ka1 = 6.2 x 10-8
I
C
E
Example
H2PO4 - H+ + HPO4
-2 Ka2 = 6.2 x 10-8
I 0.146 0.146 0
C -x +x +x
E 0.146-x 0.146+x x
0.146 0.146 6.2x10-8
6.2 x 10-8 =(0.146+x)x / (0.146-x)
x = 6.2 x 10-8
Example
HPO4 -2
H+ + PO4-3 Ka3 = 4.8 x 10-13
I
C
E
Example
HPO4 -2 H+ + PO4-3 Ka3 = 4.8 x 10-13
I 6.2x10-8 0.146 0
C -x +x +x
E 6.2x10-8-x 0.146+x x
6.2x10-8 0.146 2.0x10-19
4.8x 10-8 =(0.146+x)x/(6.2x10-8-x)
x = 2.0x10-19
Example
pH = 0.836 [H3PO4] = 2.85M
[H2PO4-] = 0.146M
[HPO4-2] = 6.2x10-8
[PO4-3] = 2.0x10-19M
[H+] = 0.146M
Acid Base Properties of Salts
Salts of weak acids produce basic solutions
NaC2H3O2 Na+ + C2H3O2-
NaOH HC2H3O2
SB WAC2H3O2
- + H2O HC2H3O2 + OH-
Resulting - basic solution
Acid Base Properties of Salts
Salts of weak bases produce acidic solutions
NH4Cl NH4+ + Cl-
NH3 HCl
WB SANH4
+ + H2O NH3 + H3O+
Resulting – acidic solution
Acid Base Properties of Salts
Salts of strong acids & Bases produce neutral solutions
NaCl Na+ + Cl-
NaOH HCl SB SAResulting solution is neutral
Calculations with salts
KaxKb = Kw
Can find Ka or Kb with this equationOn AP formula sheet
Example
What is the pH if a 0.140M solution of NaC2H3O2 (Ka HC2H3O2 = 1.8x10-5)
NaC2H3O2 Na+ + C2H3O2-
C2H3O2- + H2O HC2H3O2 + OH-
I
C
E
Example
What is the pH if a 0.140M solution of NaC2H3O2 (Ka HC2H3O2 = 1.8x10-5)
NaC2H3O2 Na+ + C2H3O2-
C2H3O2- + H2O HC2H3O2 + OH-
I 0.140 0 0
C -x +x +x
E 0.14-x x x
Example
C2H3O2- + H2O HC2H3O2 + OH-
Ka HC2H3O2 = 1.8x10-5 but we are starting with a base C2H3O2 –
Ka x Kb = Kw(1.8x10-5)(Kb) = 1 x 10-14
Kb = 5.56x10-10
Example
Kb = [HC2H3O2][OH-] / [C2H3O2-]
5.56x10-10 = [x2]/ [0.14-x]X=8.82x10-6 = [OH-]pOH = 5.05pH = 8.95
Another example
What is the pH of a 0.140M solution of diethylaminochloride (C2H5)2NH3Cl.
Kb (C2H5)2NH3+ = 1.3x10-3
(C2H5)2NH3Cl (C2H5)2NH3+ + Cl-
(C2H5)2NH3+ + H2O (C2H5)2NH2 + H3O +
I
C
E
Another example
What is the pH of a 0.140M solution of diethylaminochloride (C2H5)2NH3Cl.
Kb (C2H5)2NH2 = 1.3x10-3
(C2H5)2NH3Cl (C2H5)2NH3+ + Cl-
(C2H5)2NH3+ + H2O (C2H5)2NH2 + H3O +
I 0.140 0 0
C -x +x +x
E 0.14-x x x
Another Example
(C2H5)2NH3+ + H2O (C2H5)2NH2 + H3O+
Kb (C2H5)2NH2 = 1.3x10-3 but we are starting with an acid (C2H5)2NH3
+
Ka x Kb = Kw(1.3x10-3)(Kb) = 1 x 10-14
Kb = 7.7x10-12
Example
Kb = [(C2H5)2NH2][H3O+] / [(C2H5)2NH3+]
7.7x10-12 = [x2]/ [0.14-x]X=1.4x10-6 = [H3O+]
pH = 5.98
Qualitative Predictions of pH
Ka>Kb acidicKb>Ka basicKa = Kb neutral
pH Predictions
Predict if an aqueous solution of the following are acidic, basic, or neutral
NH4C2H3O2
Kb NH4+ = 1.8x10-5
Ka HC2H3O2 = 1.8x10-5
NH4C2H3O2 NH4+ + C2H3O2
- CA (Ka) CB (Kb) 5.56x10-10 = 5.56x10-10
neutral
pH Predictions
Predict if an aqueous solution of the following are acidic, basic, or neutral
NH4CN Kb NH4
+ = 1.8x10-5
Ka HCN= 6.2x10-10
NH4CN NH4+ + CN-
CA (Ka) CB (Kb) 5.56x10-10 < 1.6x10-5
Basic
Effect of Structure of Acid Base Properties
HF weakest because the H – X bond the strongest, so it won’t let the H pop off easily
HI strongest because the H – X bond the weakest, so the H pop off easily
Effect of Structure of Acid Base Properties
HClO4
The O’s are pulling the e- (they are e- hogs) leaving the H to easily pop off
HClO The O is not pulling
as much as HClO4, so it is harder for the H to pop off
