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

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

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 = 2.38. a. Calculate Ka for formic acid. b. What percentage of the acid ionized in this 0.10 M solution? 4/22/2017

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

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

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

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. 4/22/2017

And what about the percent ionization stuff? Formula to use: % = 4.2 x 10-3 x 100 = 4.2 % 0.10 4/22/2017

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? 4/22/2017

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 % 4/22/2017

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 4/22/2017

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

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

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 4/22/2017

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 4/22/2017

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

Solution HCN (aq)  H+ (aq) + CN-(aq) Ka = [H+ ] [CN-] = 4.9 x 10-10 I 0.20 M 0 0 C -x M +x M +x M E 0.20 – x x M x M 4/22/2017

(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 4/22/2017

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 Change -x +x Equilibrium 0.010-x x 4/22/2017

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 4/22/2017

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

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 0.020 M HCl with 125 mL of 0.010 M HI. pH=? 4/22/2017