Strong Acid calculations Weak Acid calculations Acids and Bases Strong Acid calculations Weak Acid calculations

Finding the pH of a Strong Acid Strong acids Ionize completely HA  H+ + A- Do it only one at a time So?? The concentration of the strong acid is EXACTLY the same as the [H+]

Finding the pH of a Strong Acid Example – Find the pH of 1.0 M HCl. What contributes to H+? HCl is an acid, water could be an acid [HCl] = 1.0 M  [H+] = 1.0 M [H+] in water = 1.0 x 10-7 M So HCl is the only significant acid pH = - log [H+] = - log [1.0] = 0.00

Finding the pH of a Strong Acid Example – Find the pH of 0.10 M HNO3. What contributes to H+? HNO3 is an acid, water could be an acid [HNO3] = 0.10 M  [H+] = 0.10 M [H+] in water = 1.0 x 10-7 M So HNO3 is the only significant acid pH = - log [H+] = - log [0.10] = 1.00

Finding the pH of a Strong Acid Example – Find the pH of 1.0 x 10-10 M HBr. What contributes to H+? 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] = 7.00

Finding the pH of a Weak Acid Don’t ionize completely (only partially) So they make an EQUILIBRIUM Write the equation with water using the Bronsted-Lowry definition They’ll have a Ka value 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 Weak acids have known Ka values (look them up) 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 Example – find the pH of 1.0 M HF. Look up the Ka for a weak acid. HF (aq) + H2O (l)  F- (aq) + H3O+ (aq) I 1.0 0 0 C - x + x + x E 1.0 – x x x

Finding the pH of a Weak Acid 7.4 x 10-4 = [x][x] [1.0-x] very small x = 2.7 x 10-2  5% rule? What’s x? So [H3O+] = 2.7 x 10-2 pH = - log [0.027] = 1.57

A Little Different Weak Acid calculation Find the molarity of acetic acid is 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

A Little Different Weak Acid calculation pH = 4.15 So [H3O+] = 10-4.15 = 7.1 x 10-5 M So x = 7.1 x 10-5 M 1.8 x 10-5 = [7.1 x 10-5] [7.1 x 10-5] [??? – x] Solve for ??? 2.8 x 10-4 M

Last thing they could do Give you the initial concentration of an unknown acid and its pH, find Ka 0.0500 M HQ, pH of 5.31

Mixtures of Acids When two or more acids are mixed together, you must calculate them in order from best “strongest” to weakest Based on Ka

Mixtures of Acids If both are strong…. Basically finding the new molarity of H+ and taking the pH Ex – calculate the pH of a mixture if 50.0 mL of 1.00 M HCl and 100.0 mL of 3.00 M HNO3 are mixed.

Mixtures of Acids Calculate the pH of a solution that contains 1.00 M HCN (Ka=6.2 x 10-10) and 5.00 M HNO2 (Ka=4.0 x 10-4). Also calculate the [CN-] in this solution.

Mixtures of Acids Start with the better acid HNO2 (aq) + H2O (l)  NO2- (aq) + H3O+ (aq) I 5.00 0 0 C - x + x + x E 5.00 – x x x 4.0 x 10-4 = [x][x]  x = 0.045 = [H3O+] [5.00-x]

Mixtures of Acids Then do the next acid (BUT YOU ALREADY HAVE A CONCENTRATION OF H3O+ FROM THE FIRST ACID) HCN (aq) + H2O (l)  CN- (aq) + H3O+ (aq) I 1.00 0 0.045 C - x + x + x E 1.00 – x x 0.045 + x 6.2 x 10-10 = [0.045 +x][x]  x = 1.4 x 10-8 = [CN-] [1.00-x] [H3O+] = 0.045  pH = - log [0.045] = 1.35

Polyprotic Acids Acids that furnish more than one H+ Ex: H2SO4, H2CO3, H3PO4 *****It is always one proton at a time**** H2CO3  H+ + HCO3- Ka1 HCO3-  H+ + CO32- Ka2

Polyprotic Acids Typically: Ka1 > Ka2 > Ka3 Usually by at least a factor of 100 x Why? The conjugate base of a weak acid is a pretty “good” base – so it would rather act like a base than an acid in most cases

Polyprotic Acid Calculation Find the pH of 5.0 M H3PO4 and the equilibrium [ ] of each conjugate base. What are we trying to find? [H2PO4-] [HPO42-] [PO43-] [H3O+]  pH

Polyprotic Acid Calculation Start with the first H+ (Ka1) H3PO4 (aq) + H2O (l)  H2PO4- (aq) + H3O+ (aq) I 5.0 0 0 C - x + x + x E 5.0 – x x x 7.5 x 10-3 = [x][x]  x = 0.038 = [H3O+] = [H2PO4-] [5.0-x]

Polyprotic Acid Calculation Then do the next hydrogen (Ka2)– use the values for the things you just calculated in this one too!! H2PO4- (aq) + H2O (l)  HPO42- (aq) + H3O+ (aq) I 0.038 0 0.038 C - x + x + x E 0.038 – x x 0.038 + x 6.2 x 10-8 = [0.038 +x][x]  x = 6.2 x 10-8 = [HPO42-] [0.038-x] [H3O+] now = 0.038 + 6.2 x 10-8 = 0.038

Polyprotic Acid Calculation Then do the last hydrogen (Ka3) – use the values for the things you just calculated in this one too!! HPO42- (aq) + H2O (l)  PO43- (aq) + H3O+ (aq) I 6.2 x 10-8 0 0.038 C - x + x + x E 6.2 x 10-8 – x x 0.038 + x 4.8 x 10-13 = [0.038 +x][x]  x = 7.8 x 10-19 = [PO43-] [6.2 x 10-8 -x] [H3O+] still 0.038 M  pH = 1.42