Calculating the pH of Acids and Bases Pg. 127

Strong Acids & Bases Dissociate completely in water Also known as strong electrolytes Electrolytes conduct electricity in aqueous solutions The more ions dissociated…the more electricity conducted

Strong Acids and Strong Bases HCl HNO3 HClO4 H2SO4 All alkali metal hydroxides

Weak Acids & Bases Do not completely dissociate in water The less dissociated they are…the weaker electrolytes they are Weak bases are often difficult to recognize…look for the presence of –NH2, the amine group

HCl(aq)  H+(aq) + Cl-(aq) pH of Strong Acids Write the dissociation of HCl HCl(aq)  H+(aq) + Cl-(aq) Note the one-way arrow! Since there is complete dissociation, the concentration of the acid will also be the concentration of the H+ ion. Just plug into the pH formula (pH = -log[H+])

pH of Strong Acids example If the concentration of HCl is 12 M, what is the pH of the solution? pH = -log [H+] pH = -log [12] pH = -1.08

pH of Strong Acids (Multiprotic) What is the pH of concentrated sulfuric acid, 18M? This is a diprotic acid…that is, it has two dissociable hydrogen ions Only one H+ dissociates at a time. H2SO4(aq)  H+(aq) + HSO41-(aq) HSO41-(aq)  H+(aq) + SO42-(aq)

pH of Strong Acids (multiprotic) The first dissociation is complete, while the second is not. The second reaches equilibrium Thus, the first dissociation really determines the pH of a strong, multiprotic acid. pH = -log [18] pH = -1.255

NaOH(aq)  Na+(aq) + OH-(aq) pH of Strong Bases Write the dissociation of NaOH NaOH(aq)  Na+(aq) + OH-(aq) Note the one-way arrow! Again, since full dissociation, the concentration of the base is also the concentration of the [OH-] ion

pH of Strong Bases Determine the pH of a 6 M solution of NaOH (remember pH + pOH = 14) pOH = -log [OH-] pOH = -log [6] pOH = -0.778 pH + pOH = 14 pH = 14 – (-0.778) pH = 14.8

pH of Weak Acids Note the two-way arrow! Write the dissociation of the weak acid HC2H3O2 HC2H3O2(aq)  H+(aq) + C2H3O2-(aq) Note the two-way arrow! Problem: Determine the pH of a 17.4M solution of concentrated HC2H3O2.

pH of Weak Acids Since it’s weak, it will reach equilibrium. Since it will reach equilibrium, it has an equilibrium constant. The equilibrium constant of a weak acid is called a Ka. Write the Ka expression for the dissociation of acetic acid.

pH of Weak Acids Ka = [H+][C2H3O2-] [HC2H3O2] The Ka value for acetic acid is 1.76 x 10-5 M Make a chart for the dissociation

pH of Weak Acids [HC2H3O2] [H+] [C2H3O2-] i 17.4  eq

pH of Weak Acids [HC2H3O2] [H+] [C2H3O2-] i 17.4  -x +x eq

pH of Weak Acids [HC2H3O2] [H+] [C2H3O2-] i 17.4  -x +x eq 17.4 - x x

pH of Weak Acids Plug into the Ka expression 1.76 x 10-5 = [x][x] Neglect x x = 0.0175

pH of Weak Acids x = 0.0175 = [H+] pH = -log [0.0175] pH = 1.76

pH of Weak Bases Note the two-way arrow! Write the dissociation of the weak base NH3 NH3(g) + H2O(l)  NH4+(aq) + OH-(aq) Note the two-way arrow! When dissociating a weak base, react it with water to justify the acceptance of the H+

pH of Weak Bases Since it’s weak, it will reach equilibrium. Since it will reach equilibrium, it has an equilibrium constant. The equilibrium constant of a weak base is called a Kb. Write the Kb expression for the dissociation of ammonia.

pH of Weak Bases Kb = [NH4+][OH-] [NH3] The Kb value for ammonia is 1.75 x 10-5 M Make a chart for the dissociation of a 15.3M solution of NH3 in water.

pH of Weak Bases [NH3] [NH4+] [OH-] i 15.3  eq

pH of Weak Bases [NH3] [NH4+] [OH-] i 15.3  -x +x eq

pH of Weak Bases [NH3] [NH4+] [OH-] i 15.3  -x +x eq 15.3 – x x

pH of Weak Bases Plug into the Kb expression 1.75 x 10-5 = [x][x] Neglect x x = 0.0164

pH of Weak Bases x = 0.0164 = [OH-] pOH = -log [0.0164] pOH = 1.79

pH of Multiprotic Weak Acids Write the dissociations of tartaric acid, H2C4H4O6 (found in cream of tartar) H2C4H4O6(aq)  H+(aq) + HC4H4O61- (aq) HC4H4O61-(aq)  H+(aq) + C4H4O62- (aq)

pH of Multiprotic Weak Acids Determine the pH of a 100 mL solution that contains 5.00g of H2C4H4O6. Ka1 is 9.20 x 10-4 and Ka2 is 4.31 x 10-5. First determine the initial molarity of the tartaric acid. 5g x 1mol x 1 = 0.333M 150.1g 0.1L Make a chart for the 1st dissociation.

pH of Multiprotic Weak Acids [H2C4H4O6] [H+] [HC4H4O6-] i 0.333  eq

pH of Multiprotic Weak Acids [H2C4H4O6] [H+] [HC4H4O6-] i 0.333  -x +x eq

pH of Multiprotic Weak Acids [H2C4H4O6] [H+] [HC4H4O6-] i 0.333  -x +x eq 0.333 -x x

pH of Multiprotic Weak Acids Write the Ka1 expression Ka1 = [H+][HC4H4O61-] [H2C4H4O6] Plug your values into the expression. 9.20 x 10-4 = [x][x] [0.333 - x]

pH of Multiprotic Weak Acids x = 0.0175 = [H+] More hydrogen ion will dissociate in the next dissociation…this is just the amount from the first dissociation. Make a chart for the 2nd dissociation.

pH of Multiprotic Weak Acids [HC4H4O61-] [H+] [C4H4O62-] i  eq

pH of Multiprotic Weak Acids [HC4H4O61-] [H+] [C4H4O62-] i 0.0175  eq

pH of Multiprotic Weak Acids [HC4H4O61-] [H+] [C4H4O62-] i 0.0175  -x +x eq

pH of Multiprotic Weak Acids [HC4H4O61-] [H+] [C4H4O62-] i 0.0175  -x +x eq 0.0175 – x 0.0175 + x x

pH of Multiprotic Weak Acids Write the Ka2 expression Ka2 = [H+][C4H4O62-] [HC4H4O61-] Plug your values into the expression. Neglect x 4.31 x 10-5 = [0.0175 + x][x] [0.0175 - x] Neglect x

pH of Multiprotic Weak Acids x = 4.31 x 10-5 = [C4H4O62-] So the total amount of hydrogen ion is represented by 0.0175 + 4.31 x 10-5… [H+] = 0.0175431 pH = 1.76