Ion Concentration and pH in Detail Strong Acids      HClO4, HI, HBr, HCl, HNO3 & H2SO4 If we use the general symbol, X, to represent the anion of these acids, then the general equation becomes: HX + H2O ---> H3O+(aq) + X-(aq) Dissociation is 100% complete. Strong Bases Any alkali metal with hydroxide is considered to be a strong base and capable of undergoing complete dissociation. O2- and NH2- also react with water to give OH- to completion. It is easy to calculate the [H+] and the pH of a strong acid or base. Since a strong acid is completely dissociated in dilute solution the [H3O+] essentially equals the original concentration of the solute. eg.         a 1.0 x 10-3 mol/L (0.0010 mol/L) HCl solution HCl      +    H2O --------> H3O+(aq) + Cl-(aq)   start  0.0010 M                               -------         -------   after        0.0 M                            0.0010 M     0.0010 M                                                    (1.0 x 10-3 M) The pH will be 3, the pOH will be 11. Weak Monoprotic Acids The general equation for a weak acid is: HX + H2O ------> H3O+ + X- Applying the equilibrium law to this equation we get: In dilute solutions of acids and bases the [H2O] is essentially a constant at 55.4 mol/L. (1000 mL of H2O = 1000 g of  H2O/18.02 g/mol = 55.4 moles). This, when multiplied with Keq gives us a new constant value which we call Ka. OR The Ka values for acids and bases stronger than H3O+ are not listed because of their high degree of dissociation, the denominator would approach zero and result in an infinitely large Ka. Calculate (a) the [H3O+], (b) the pH and (c) the % dissociation for a 0.100 mol/L solution of acetic acid at 25oC, Ka for CH3COOH is 1.8 x 10-5. The equation for the dissociation is:                  CH3COOH + H2O -------> H3O+(aq) + CH3 COO-(aq) Let 'x' be the concentration of acetic acid that dissociates:                 CH3COOH + H2O -------> H3O+(aq) + CH3 COO-(aq)    start          0.100 M                            -------            -------   finish          0.100 - x                               x                      x 1.8 x 10-3 =    (x) (x)                     0.100 - x 1.8 x 10-4 - 1.8 x 10-3 x = x2 Upon rearrangement x2 +1.8 x 10-3x -1.8 x 10-4 = 0 Use the quadratic equation to solve for 'x' a = 1    b = 1.8 x 10-3    c = -1.8 x 10-4 Therefore the [H3O+] is 1.25 x 10-2 mol/L b) The pH is 1.9 based upon -log[H3O+] c) The % dissociation is 1.25 x 10-2 M * 100 = 12.5%                                       0.100 M
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