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04/19/23 2
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.
04/19/23 3
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?
04/19/23 4
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
04/19/23 6
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+]
04/19/23 7
So, now for the Ka calculation:
= 1.8 x 10-4
Is our answer reasonable? Yes, Ka values for weak acids are usually between 10-3
and 10-10.
2
2
HCHO
CHOHKa
0958.0
)102.4)(102.4( 33 aK
04/19/23 8
And what about the percent ionization stuff?
• Formula to use:
% = 4.2 x 10-3 x 100 = 4.2 % 0.10
100IonizationPercent
initial
equil
Acid
H
04/19/23 9
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 acid-dissociation constant, Ka, for niacin?
04/19/23 10
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 %
04/19/23 11
Solution to Niacin Problem
• Ka = [H+] [ niacin ion-] [niacin] Niacin H+ niacin ionI 0.02 0 0C - 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
04/19/23 12
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.
04/19/23 13
First step: Write the ionization equilibrium for acetic acid:
• HC2H3O2(aq) H+ (aq) + C2H3O2- (aq)
Second Step: Write the equilibrium-constant expression
• Ka = [H+ ] [C2H3O2- ] = 1.8 x 10-5
[HC2H3O2]
04/19/23 14
Step 3: Set up an ICE calculation
HC2H3O2(aq) H+ (aq) + C2H3O2- (aq)
I 0.30 M 0 0
C -x M +x M +x M
E (0.30 – x) M x M x M
04/19/23 15
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
04/19/23 16
• Calculate the pH of a 0.20 M solution of HCN (Refer to table 16.2 or Appendix D for the Ka value.)
04/19/23 17
Solution
• HCN (aq) H+ (aq) + CN-(aq)• Ka = [H+ ] [CN-] = 4.9 x 10-10
[HCN]
I 0.20 M 0 0C -x M +x M +x ME 0.20 – x x M x M
04/19/23 18
• (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
04/19/23 19
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
Niacin H+ niacin ion
Initial 0.010 0 0
Change -x +x +x
Equilibrium 0.010-x x x
04/19/23 20
• 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 = 0x = 3.92 x 10 -4 = [H+]pH = -log(3.92 x 10 –4) pH = 3.41
04/19/23 21
5.00 mL of 0.250 M HClO3 diluted to 50.0 mL; pH =?
6.1)025.0log(
H M025.0soln L 0.050
mol 00125.0
H 00125.0soln L 1
mol 250.0
mL 1000
L 1 mL 00.5
pH
mol