AP Chem Today: Strong vs. Weak Acid Calculations Acid/Base Equilibrium

Finding the pH of a Strong Acid Strong acids Ionize completely: HA  H+ + A- So, the [strong acid] is the same as the [H+]

Finding the pH of a Strong Acid Example – Find the pH of 0.10 M HNO3. HNO3 is a strong acid, complete dissociates HNO3  H+ + NO3- [HNO3] = 0.10 M, therefore [H+] = 0.10 M pH = - log [H+] = - log [0.10] = 1.00

Finding the pH of a Strong Acid What if you have a very dilute amount of acid in water? Example – Find the pH of 1.0 x 10-10 M HBr. pH = - log [H+] = - log [1.0 x10-10] = 10…. But an acid can’t have a pH greater than 7!

Find the pH of 1.0 x 10-10 M HBr HBr is an acid, water could be an acid [HBr] = 1.0 x 10-10 M, [H+] = 1.0 x 10-10 M [H+] in water = 1.0 x 10-7 M So WATER is actually a better acid pH = - log [H+] = - log [1.0 x 10-7 + 1.0 x10-10] = 6.9995

Finding the pH of a Weak Acid Don’t ionize completely (only partially), so they make an EQUILIBRIUM HA ↔ H+ + A They’ll have an equilibrium acid dissociation constant value, Ka, associated with them

A Couple Things about Ka The relative strength of acids compared to each other can be found by comparing the Ka values for the acids Strong acids have a Ka of infinity (very large) Weak acids have known Ka values (will be given or can calculate) You can also find the Kb of its conjugate base Ka x Kb = 1.00 x 10-14 (same as [H+][OH-] = 1.00 x 10-14 )

Finding the pH of a Weak Acid Find the pH of 1.0 M HF. (Ka = 7.2 x 10-4) HF (aq) + H2O (l)  F- (aq) + H3O+ (aq) I 1.0 0 0 C - x + x + x E 1.0 – x x x 7.2 x 10-4 = [x][x] [1.0-x] x = 2.7 x 10-2 = [H3O+] pH = - log [0.027] = 1.57

A Little Different Weak Acid calculation Find the molarity of acetic acid if the pH of the solution is 4.15. HC2H3O2 (aq) + H2O (l)  C2H3O2- (aq) + H3O+ (aq) I ??? 0 0 C - x + x + x E ??? – x x x 1.8 x 10-5 =[7.1 x 10-5][7.1 x 10-5] [??? – 7.1 x 10-5] Solve for ??? =3.49 x 10-4 M pH = 4.15 So [H3O+] = 10-4.15 = 7.1 x 10-5 M = x