1. pH and Ka values of Weak Acids
AP Chem April 24, 2012
4/22/2017

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.
4/22/2017

3. A student prepared a 0. 10 M solution of formic acid (HCHO2)
A student prepared a 0.10 M solution of formic acid (HCHO2). A pH meter shows the pH = a. Calculate Ka for formic acid. b. What percentage of the acid ionized in this 0.10 M solution?
4/22/2017

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
4/22/2017

5. HCHO2 (aq)  H+ (aq) + CHO2- (aq)I
0.10 M C E
4/22/2017

6. HCHO2 H+ CHO2- I 0.10 M C -4.2 x 10-3 M +4.2 x 10-3 M EC
-4.2 x 10-3 M
+4.2 x 10-3 M E
x 10-3 M =
4.2 x 10-3 M
Assumed from the pH  [H+]
4/22/2017

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
4/22/2017

8. And what about the percent ionization stuff?
Formula to use:
% = 4.2 x x = 4.2 %
0.10
4/22/2017

9. Niacin, one of the B vitamins, has the
following molecular
structure:
                   
A M solution of niacin has a pH of (a) What percentage of the acid is ionized in this solution? (b) What is the acid-dissociation constant, Ka, for niacin?
4/22/2017

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 = 2.7 %
4/22/2017

11. Solution to Niacin Problem
Ka = [H+] [ niacin ion-]
[niacin]
Niacin H niacin ion I
C x x x 10-4
E x x x 10-4
Ka = (5.50 x 10-4)2 = x 10-5
0.019
4/22/2017

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.
4/22/2017

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]
4/22/2017

14. Step 3: Set up an ICE calculation
HC2H3O2(aq)  H+ (aq) + C2H3O2- (aq)
I M
C x M x M x M
E (0.30 – x) M x M x M
4/22/2017

15. Fourth Step: Substitute the equilibrium conc into expression.
Ka = [H+ ] [C2H3O2- ] = 1.8 x 10-5
[HC2H3O2]
= (x) (x) = 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 = M x 100 = 0.77%
0.30 M
4/22/2017

16. Calculate the pH of a 0. 20 M solution of HCN (Refer to table 16
Calculate the pH of a 0.20 M solution of HCN (Refer to table 16.2 or Appendix D for the Ka value.)
4/22/2017

17. Solution HCN (aq)  H+ (aq) + CN-(aq) Ka = [H+ ] [CN-] = 4.9 x 10-10
I M
C x M x M x M
E – x x M x M
4/22/2017

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)
x x 10-10x – 9.8 x = 0
x = 9.9 x 10-6 = [H+]
pH = -log(9.9 x 10-6)
pH = 5.00
4/22/2017

19. Second Niacin problem
The Ka for niacin is 1.6 x What is the pH of a M solution of niacin?
1st find the [H+] at equilibrium
Niacin H+
niacin ion
Initial
0.010
Change -x +x
Equilibrium
0.010-x x
4/22/2017

20. Ka = [H+] [niacin ion] = 1.6 x 10-5 [niacin]
1.6 x = x2 / (0.010-x)
x x 10-5 x x = 0
x = 3.92 x = [H+]
pH = -log(3.92 x 10 –4)
pH = 3.41
4/22/2017

21. 5.00 mL of 0.250 M HClO3 diluted to 50.0 mL; pH =?
4/22/2017

22. A solution formed by mixing 50. 0 mL of 0. 020 M HCl with 125 mL of 0
A solution formed by mixing 50.0 mL of M HCl with 125 mL of M HI. pH=?
4/22/2017