Calculating the pH of Acids and Bases Strong vs. Weak

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

Strong Acids & Bases n HCl n HNO 3 n HClO 4 n H 2 SO 4 n All alkali metal hydroxides

Weak Acids & Bases n Do not completely dissociate in water n The less dissociated they are…the weaker electrolytes they are n Any acid or base not on the aforementioned list is considered weak n Weak bases are often difficult to recognize…look for the presence of –NH 2, the amine group

pH of Strong Acids n Write the dissociation of HCl HCl(aq)  H + (aq) + Cl - (aq) Note the one- way arrow! n Let’s figure out the pH of a 12M solution of concentrated HCl.

pH of Strong Acids n Make a chart for the dissociation [HCl ] [H + ][Cl - ] i  f

pH of Strong Acids [HCl ] [H + ][Cl - ] i1200  f

pH of Strong Acids [HCl ] [H + ][Cl - ] i1200  f

pH of Strong Acids [HCl ] [H + ][Cl - ] i1200  f012

pH of Strong Acids n If the [H + ] is 12M, then we can determine the pH of the solution n pH = -log [H + ] n pH = -log [12] n pH = n Since it dissociates completely and an equilibrium is never reached, you really don’t need to make a chart.

pH of Strong Acids n What is the pH of concentrated sulfuric acid, 18M? n Only one H + dissociates at a time. n This is a diprotic acid…that is, it has two dissociable hydrogen ions n H 2 SO 4 (aq)  H + (aq) + HSO 4 1- (aq) n HSO 4 1- (aq)  H + (aq) + SO 4 2- (aq)

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

pH of Strong Bases n Write the dissociation of NaOH NaOH(aq)  Na + (aq) + OH - (aq) Note the one- way arrow! n Let’s figure out the pH of a 6M solution of NaOH.

pH of Strong Bases n Make a chart for the dissociation [NaOH][Na + ][OH - ] i  f

pH of Strong Bases [NaOH][Na + ][OH - ] i600  f

pH of Strong Bases [NaOH][Na + ][OH - ] i600  -6+6 f

pH of Strong Bases [NaOH][Na + ][OH - ] i600  -6+6 f066

pH of Strong Bases n If the [OH - ] is 6M, then we can determine the pH of the solution n pOH = -log [OH - ] n pOH = -log [6] n pOH = n pH + pOH = 14 n pH = 14 – (-0.778) n pH = 14.8

pH of Weak Acids n Write the dissociation of the weak acid HC 2 H 3 O 2 HC 2 H 3 O 2 (aq)  H + (aq) + C 2 H 3 O 2 - (aq) Note the two- way arrow! n Let’s figure out the pH of a 17.4M solution of concentrated HC 2 H 3 O 2.

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

pH of Weak Acids n K a = [H + ][C 2 H 3 O 2 - ] [HC 2 H 3 O 2 ] n The K a value for acetic acid is 1.76 x M n Make a chart for the dissociation

pH of Weak Acids [ HC 2 H 3 O 2 ][H + ][C2H3O2-][C2H3O2-] i  eq

pH of Weak Acids [ HC 2 H 3 O 2 ][H + ][C2H3O2-][C2H3O2-] i  -x+x eq

pH of Weak Acids [ HC 2 H 3 O 2 ][H + ][C2H3O2-][C2H3O2-] i  -x+x eq xxx

pH of Weak Acids n 1.76 x = [x][x] [ x] n Plug into the K a expression n Test the 5% rule n x =

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

pH of Weak Bases n Write the dissociation of the weak base NH 3 NH 3 (g) + H 2 O(l)  NH 4 + (aq) + OH - (aq) Note the two- way arrow! n When dissociating a weak base, react it with water to justify the acceptance of the H +

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

pH of Weak Bases n K b = [NH 4 + ][OH - ] [NH 3 ] n The K b value for ammonia is 1.75 x M n Make a chart for the dissociation of a 15.3M solution of NH 3 in water.

pH of Weak Bases [NH 3 ][NH 4 + ][OH - ] i  eq

pH of Weak Bases [NH 3 ][NH 4 + ][OH - ] i  -x+x eq

pH of Weak Bases [NH 3 ][NH 4 + ][OH - ] i  -x+x eq15.3 – xxx

pH of Weak Bases n 1.75 x = [x][x] [ x] n Plug into the K b expression n Test the 5% rule n x =

pH of Weak Bases n x = = [OH - ] n pOH = -log [0.0164] n pOH = 1.79 n pH = n pH = 12.2

pH of Multiprotic Weak Acids n Write the dissociations of tartaric acid, H 2 C 4 H 4 O 6 (found in cream of tartar) H 2 C 4 H 4 O 6 (aq)  H + (aq) + HC 4 H 4 O 6 1- (aq) HC 4 H 4 O 6 1- (aq)  H + (aq) + C 4 H 4 O 6 2- (aq)

pH of Multiprotic Weak Acids n Let’s figure out the pH of a 100-mL solution that contains 5.00g of H 2 C 4 H 4 O 6. K a1 is 9.20 x and K a2 is 4.31 x n First determine the initial molarity of the tartaric acid. 5g x 1mol x 1 = 0.333M 150.1g 0.1L n Make a chart for the 1 st dissociation.

pH of Multiprotic Weak Acids [H2C4H4O6][H2C4H4O6][H + ][ HC 4 H 4 O 6 - ] i  eq

pH of Multiprotic Weak Acids [H2C4H4O6][H2C4H4O6][H + ][ HC 4 H 4 O 6 - ] i  -x+x eq

pH of Multiprotic Weak Acids [H2C4H4O6][H2C4H4O6][H + ][ HC 4 H 4 O 6 - ] i  -x+x eq xxx

pH of Multiprotic Weak Acids n Write the K a1 expression K a1 = [H+][HC 4 H 4 O 6 1- ] [H 2 C 4 H 4 O 6 ] n Plug your values into the expression x = [x][x] [ x] n Test the 5% rule.

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

pH of Multiprotic Weak Acids [ HC 4 H 4 O 6 1- ][H + ][C4H4O62-][C4H4O62-] i  eq

pH of Multiprotic Weak Acids [ HC 4 H 4 O 6 1- ][H + ][C4H4O62-][C4H4O62-] i  eq

pH of Multiprotic Weak Acids [ HC 4 H 4 O 6 1- ][H + ][C4H4O62-][C4H4O62-] i  -x+x eq

pH of Multiprotic Weak Acids [ HC 4 H 4 O 6 1- ][H + ][C4H4O62-][C4H4O62-] i  -x+x eq – x x x

pH of Multiprotic Weak Acids n Write the K a2 expression K a2 = [H+][C 4 H 4 O 6 2- ] [HC 4 H 4 O 6 1- ] n Plug your values into the expression x = [ x][x] [ x] n Test the 5% rule.

pH of Multiprotic Weak Acids n 5% rule works! x = 4.31 x = [C 4 H 4 O 6 2- ] n So the total amount of hydrogen ion is represented by x … n [H+] = n pH = 1.77

pH of Multiprotic Weak Acids n What are the equilibrium concentrations of all of the species? [ HC 4 H 4 O 6 1- ] eq = – 4.31 x = [ C 4 H 4 O 6 2- ] eq = 4.31 x M [ H 2 C 4 H 4 O 6 ] eq = – =0.316 M M