Acids Lesson 8 Weak Acids pH, Ka Calculations

Weak Acids calculations Helpful Hints: Weak acids do NOT ionize 100% There is an equilibrium state NEED….ICE tables! Ka can be used directly from table!

Weak Acids calculations Two types of questions -calculate pH, or [H3O+] -calculate Ka

Calculating pH or [H3O+]

1. Calculate the pH of 0.45 M HCN. You need an ICE chart for weak acids or bases! HCN ⇄ H+ + CN- I 0.45 M 0 0 C -x x x E 0.45 - x x x Ka = [H+][CN-] = 4.9 x 10-10 [HCN] Ka = x2 = 4.9 x 10-10 0.45 - x

The Ka is small, so x is small We will find that x = 0.000015 0.45 - 0.000015 = 0.45 This means we can make the approximation that 0.45 - x = 0.45 We can do this anytime the Ka has an exponent of 10-4 or less Ka = x2 = 4.9 x 10-10 0.45 - x 0 small ka

x2 = 4.9 x 10-10 0.45 x = [H+] = 0.000014849 M pH = -Log[0.000014849] pH = 4.83 2 sig figs due to molarity and Ka

“Percentage dissociation” is not more than 5% what you just learnt: You need to ALWAYS state this assumption of x being insignificant, otherwise it is considered a chemical error not to state it. You can use it anytime Ka has an exponent of 10-4 or less In this course, we will always make that assumption just be sure to state it! There is another way you can also justify it: “Percentage dissociation” is not more than 5%

“Percentage dissociation” is not more than 5% Refers to the amount of [H3O+] produced as a percentage of the original amount of acid present. % dissociation = [H3O+] found at equilibrium x 100% [HA] at start

2. Calculate the pH of 0.60 M H3BO3 H3BO3 ⇄ H+ + H2BO3- I 0.60 M 0 0 C -x x x E 0.60 - x x x Anytime there is a polyprotic acid, only use the first proton for the ionization. 0 small ka x2 = 7.3 x 10-10 0.60

x2 = 7.3 x 10-10 0.60 x = [H+] = 2.09 x 10-5 M pH = -Log[2.09 x 10-5] pH = 4.68 2 sig figs due to molarity and Ka

3. Calculate the pH of a 0.20 M diprotic acid with a Ka = 4.7 x 10-7 H2X ⇄ H+ + HX- I 0.20 M 0 0 C -x x x E 0.20 - x x x 0 small ka x2 = 4.7 x 10-7 0.20

x = [H+] = 3.066 x 10-4 M pH = -Log[3.066 x 10-4] pH = 3.51 2 sig figs due to molarity and Ka

What if asked to calculate Ka?

Ex: The pH of 0.100 M H2C2O4 is 1.28. Calculate the Ka for the weak acid. [H+] = 0.05248 M H2C2O4 ⇄ H+ + HC2O4- I 0.100 0 0 C - 0.05248 0.05248 0.05248 E 0.04752 0.05248 0.05248 [H+][HC2O4-] Ka = (0.05248)2 = = 5.8 x 10-2 [H2C2O4] 0.04752

TRY: The pH of a 1. 0 M triprotic weak acid is 4. 568 TRY: The pH of a 1.0 M triprotic weak acid is 4.568. Calculate the Ka and identify the acid.

TRY: The pH of a 1. 0 M triprotic weak acid is 4. 568 TRY: The pH of a 1.0 M triprotic weak acid is 4.568. Calculate the Ka and identify the acid. Ka = 7.3 x 10-10 Hint- it’s not very exciting Boric acid

TRY: The pH of a 0. 010 M solution of HBr is 2. 00 TRY: The pH of a 0.010 M solution of HBr is 2.00. What is the Ka for HBr?

Another example Calculate the pH of 0.30 M solution of NH4NO3 79 p152

Homework Read Pages 148-151 Go through examples Pg 152 #74-81