Transcript pH and Ka values of Weak Acids

pH and Ka values of Weak Acids
AP Chem April 24, 2012
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Weak Acids: Calculation of Ka
from pH
• Will need to use ICE skills for solving
equilibrium problems.
• Because the concentration of the acid
(reactant side) does NOT equal the
concentration of the H+ ion (product side)
• There is far less than 100% ionization taking
place.
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A student prepared a 0.10 M solution
of formic acid (HCHO2).
A pH meter shows the pH = 2.38.
a. Calculate Ka for formic acid.
b. What percentage of the acid
ionized in this 0.10 M solution?
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HCHO2 (aq) 
+
H
-
(aq) + CHO2 (aq)
• First, let’s find the [H+] from the
pH
• [H+] = 10(-2.38)
•
= 4.2 x 10-3 M
• Great, Now for some ICE
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HCHO2 (aq)  H+ (aq) + CHO2- (aq)
I
HCHO2
H+
CHO2-
0.10 M
0
0
C
E
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HCHO2
H+
CHO2-
I
0.10 M
0
0
C
-4.2 x 10-3 M
+4.2 x 10-3 M +4.2 x 10-3 M
E 0.10 -4.2 x 10-3 M
= 0.0958
4.2 x 10-3 M
4.2 x 10-3 M
Assumed from the pH  [H+]
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So, now for the Ka calculation:

H  CHO 



Ka
2
HCHO2 
(4.2 103 )(4.2 103 )
Ka 
= 1.8 x 10-4 0.0958
Is our answer reasonable?
Yes, Ka values for weak acids are usually between 10-3
and 10-10.
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And what about the percent ionization
stuff?
• Formula to use:
PercentIonization
% = 4.2 x 10-3 x 100
0.10
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H

equil
 100
Acidinitial 
= 4.2 %
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Niacin, one of the B
vitamins, has the
following molecular
structure:
A 0.020 M solution of niacin has a
pH of 3.26. (a) What percentage
of the acid is ionized in this
solution? (b) What is the aciddissociation constant, Ka, for
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niacin?
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Niacin Problem #1
• pH = 3.26 [H+] = ?
• [H+] = 10-3.26 = 5.50 x 10-4 M
• Percent Ionization = [H+]equilibrium x 100
[Acid]Initial
= 5.50 x 10-4 M / 0.02 M x 100 = 2.7 %
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Solution to Niacin Problem
• Ka = [H+] [ niacin ion-]
[niacin]
Niacin
H+
niacin ion
I 0.02
0
0
C - 5.50 x 10-4
+ 5.50 x 10-4 + 5.50 x 10-4
E 0.02 - 5.50 x 10-4 5.50 x 10-4
5.50 x 10-4
• Ka = (5.50 x 10-4)2 = 1.55 x 10-5
0.019
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Using Ka to Calculate pH
• Similar to the approach we used in Chapter 15,
sometimes using the quadratic equation to solve for the
equilibrium concentrations. Once you know the
equilibrium concentration of [H+], you can calculate the
pH.
• Need to have Ka value and the initial concentration of the
weak acid
• Start by writing equation and equilibrium-constant
expression for the reaction.
• Let’s calculate the pH of a 0.30 M solution of acetic acid
(HC2H3O2) at 250C.
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First step: Write the ionization
equilibrium for acetic acid:
• HC2H3O2(aq)  H+ (aq) + C2H3O2- (aq)
Second Step: Write the equilibriumconstant expression
• Ka = [H+ ] [C2H3O2- ] = 1.8 x 10-5
[HC2H3O2]
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Step 3: Set up an ICE calculation
I
HC2H3O2(aq)  H+ (aq) + C2H3O2- (aq)
0.30 M
0
0
C
-x M
E (0.30 – x) M
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+x M
+x M
xM
xM
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Fourth Step: Substitute the
equilibrium conc into expression.
• Ka = [H+ ] [C2H3O2- ] = 1.8 x 10-5
[HC2H3O2]
•
= (x) (x) = 1.8 x 10-5
(0.30 –x)
Solve using quadratic equation: x = 2.3 x 10-3 M
• Percent Ionization = [H+]equilibrium x 100
[Acid]Initial
% ionization = 0.0023 M x 100 = 0.77%
0.30 M
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• Calculate the pH of a 0.20 M solution of
HCN (Refer to table 16.2 or Appendix D
for the Ka value.)
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Solution
• HCN (aq)  H+ (aq) + CN-(aq)
• Ka = [H+ ] [CN-] = 4.9 x 10-10
[HCN]
I
C
E
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0.20 M
-x M
0.20 – x
0
+x M
xM
0
+x M
xM
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• (x) (x) = 4.9 x 10-10
(0.20 –x)
Use quadratic equation to solve for x:
x2 = 4.9 x 10-10(0.20 – x)
x2 + 4.9 x 10-10x – 9.8 x 10-11 = 0
x = 9.9 x 10-6 = [H+]
pH = -log(9.9 x 10-6)
pH = 5.00
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Second Niacin problem
The Ka for niacin is 1.6 x 10-5. What is the pH of
a 0.010 M solution of niacin?
1st find the [H+] at equilibrium
Initial
Change
Equilibrium
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Niacin
H+
niacin ion
0.010
0
0
-x
+x
+x
0.010-x
x
x
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• Ka = [H+] [niacin ion] = 1.6 x 10-5
[niacin]
1.6 x 10-5 = x2 / (0.010-x)
x2 + 1.6 x 10-5 x - 1.6 x 10-7 = 0
x = 3.92 x 10 -4 = [H+]
pH = -log(3.92 x 10 –4)
pH = 3.41
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5.00 mL of 0.250 M HClO3
diluted to 50.0 mL; pH =?
1L
0.250 mol

5.00 mL 

 0.00125m ol H
1000mL 1 L soln
0.00125mol

 0.025M H
0.050L soln
pH   log(0.025)  1.6
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A solution formed by mixing 50.0 mL of 0.020 M
HCl with 125 mL of 0.010 M HI.
pH=?
1 L soln 0.020 mol
50.0 mL

 0.001mol H 
1000mL 1 L soln
1 L soln 0.010 mol
125 mL

 0.00125mol H 
1000mL 1 L soln


0.00225mol H
0.00225mol H

 0.0129M H 
(0.050L 0.125L)
0.175L
pH   log(0.0129)  1.89
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