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Calculating the pH of Acids and Bases. Strong vs. Weak. 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 & Bases. HCl HNO 3
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Calculating the pH of Acids and Bases Strong vs. Weak
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 & 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 • Any acid or base not on the aforementioned list is considered weak • Weak bases are often difficult to recognize…look for the presence of –NH2, the amine group
Note the one-way arrow! pH of Strong Acids • Write the dissociation of HCl HCl(aq) H+(aq) + Cl-(aq) • Let’s figure out the pH of a 12M solution of concentrated HCl.
pH of Strong Acids • Make a chart for the dissociation
pH of Strong Acids • If the [H+] is 12M, then we can determine the pH of the solution • pH = -log [H+] • pH = -log [12] • pH = -1.08 • Since it dissociates completely and an equilibrium is never reached, you really don’t need to make a chart.
pH of Strong Acids • 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 • 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
Note the one-way arrow! pH of Strong Bases • Write the dissociation of NaOH NaOH(aq) Na+(aq) + OH-(aq) • Let’s figure out the pH of a 6M solution of NaOH.
pH of Strong Bases • Make a chart for the dissociation
pH of Strong Bases • If the [OH-] is 6M, then we can determine the pH of the solution • pOH = -log [OH-] • pOH = -log [6] • pOH = -0.778 • pH + pOH = 14 • pH = 14.8 • pH = 14 – (-0.778)
Note the two-way arrow! pH of Weak Acids • Write the dissociation of the weak acid HC2H3O2 HC2H3O2(aq) H+(aq) + C2H3O2-(aq) • Let’s figure out 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 • Plug into the Ka expression • 1.76 x 10-5 = [x][x] [17.4 - x] • Test the 5% rule • x = 0.0175
pH of Weak Acids • x = 0.0175 = [H+] • pH = -log [0.0175] • pH = 1.76
Note the two-way arrow! pH of Weak Bases • Write the dissociation of the weak base NH3 NH3(g) + H2O(l) NH4+(aq) + OH-(aq) • 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 • Plug into the Kb expression • 1.75 x 10-5 = [x][x] [15.3 - x] • Test the 5% rule • x = 0.0164
pH of Weak Bases • x = 0.0164 = [OH-] • pOH = -log [0.0164] • pOH = 1.79 • pH = 14 - 1.79 • pH = 12.2
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)
5g x 1mol x 1 = 0.333M 150.1g 0.1L pH of Multiprotic Weak Acids • Let’s figure out 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. • Make a chart for the 1st dissociation.
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] • Test the 5% rule.
pH of Multiprotic Weak Acids • Use the quadratic to solve for x. x = 0.0170 = [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 • Write the Ka2 expression Ka2 = [H+][C4H4O62-] [HC4H4O61-] • Plug your values into the expression. 4.31 x 10-5 = [0.0170 + x][x] [0.0170 - x] • Test the 5% rule.
pH of Multiprotic Weak Acids • 5% rule works! x = 4.31 x 10-5 = [C4H4O62-] • So the total amount of hydrogen ion is represented by 0.0170 + 4.31 x 10-5… • [H+] = 0.0170431 • pH = 1.77
pH of Multiprotic Weak Acids • What are the equilibrium concentrations of all of the species? [H2C4H4O6]eq = 0.333 – 0.017 = 0.316 M [HC4H4O61-]eq = 0.017 – 4.31 x 10-5 = 0.01696 M [C4H4O62-]eq = 4.31 x 10-5M