pH and K a values of Weak Acids

1. pH and Ka values of Weak Acids AP Chem April 24, 2012

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.

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?

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

5. HCHO2(aq) H+(aq) + CHO2-(aq)

6. Assumed from the pH  [H+]

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.

8. And what about the percent ionization stuff? • Formula to use: % = 4.2 x 10-3 x 100 = 4.2 % 0.10

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?

10. Niacin Problem #1 • pH = 3.26 [H+] = ? • [H+] = 10-3.26 = 5.50 x 10-4M • Percent Ionization = [H+]equilibriumx 100 [Acid]Initial = 5.50 x 10-4M / 0.02 M x 100 = 2.7 %

11. 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 10-45.50 x 10-4 • Ka = (5.50 x 10-4)2 = 1.55 x 10-5 0.019

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.

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]

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

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+]equilibriumx 100 [Acid]Initial % ionization = 0.0023 M x 100 = 0.77% 0.30 M

16. Calculate the pH of a 0.20 M solution of HCN (Refer to table 16.2 or Appendix D for the Ka value.)

17. Solution • HCN (aq)  H+ (aq) + CN-(aq) • Ka = [H+ ] [CN-] = 4.9 x 10-10 [HCN] I 0.20 M 0 0 C -x M +x M +x M E 0.20 – x x M x M

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

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

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 = 0 x = 3.92 x 10 -4 = [H+] pH = -log(3.92 x 10 –4) pH = 3.41

21. 5.00 mL of 0.250 M HClO3 diluted to 50.0 mL; pH =?

22. A solution formed by mixing 50.0 mL of 0.020 M HCl with 125 mL of 0.010 M HI. pH=?