Solutions for Chemical Activities Exercises

Solutions for Chemical Activities Exercises 1. a) since this is a +1, -1 electrolyte, μ = [KNO3] = 0.20 mM b) 2. 3. (a) activity coefficients decrease with increasing ionic strength (b) activity coefficients decrease with increasing ionic charge (c) activity coefficients decrease with decreasing hydrated radius All values gotten from table of activity coefficients. For μ = 0.050M, γ(SO42-) = 0.445 μ = 0.10 M, γ(HS-) = 0.76 μ = 0.025M, 4. Determine the solubility of AgSCN in the presence of the common ion SCN-. AgSCN(s) X Ag+(aq) + SCN-(aq) init 0 0.060 equil x 0.060 + x where x equals the increase in the [Ag+] as a result of the dissolving process. at equil, Ksp = 1.1 x 10-12 = [Ag+]γAg+[SCN-]γSCNSince Ksp << 1 the contribution of AgSCN to the ionic strength is negligible. Since KSCN is a +1, -1 electrolyte, μ = [KSCN] = 0.060 M for μ = 0.060 M, Solutions for Chemical Activities Exercises for μ = 0.060 M, Substitute into the reaction quotient and solve for x. Ksp = 1.1 x 10-12 = (x)(0.790)(0.060 + x)(0.800) • (x)(0.790)(0.060)(0.800) assumes x << 0.060 Y 5. x = 2.90 x 10-11 M = [Ag+] Kw = 1.0 x 10-14 = [H3O+]γH3O+[OH-]γOH- at equilibrium [H3O+] = [OH-] = x Since LiBr is a +1, -1 electrolyte, μ = [LiBr] = 0.050 M for μ = 0.050 M γH+ = 0.86 γOH- = 0.81 Therefore, Kw = 1.0 x 10-14 = (x)(0.86)(x)(0.81) Y x = 1.20 x 10-7 = [H3O+] pH = -log (1.20 x 10-7 )(0.86) = 6.99 6. pH = -log aH+ = -log Kw/aOHaOH- = [OH-]γOHfor 0.010 M NaOH + 0.012 M LiNO3, μ = 0.022 M (since all electrolytes involve +1, -1 ions) for μ = 0.022 M, pH = -log (1.0 x 10-14)/(0.010)(0.873) = 11.94 Without using activities, pH = -log [H+] = -log Kw/[OH-] = -log (1.0 x 10-14)/(0.010) = 12.00 Solutions for Chemical Activities Exercises 7. Phenol (C6H5OH) is a weak acid and will not contribute significantly to the ionic strength. The ionic strength of the solution based on the LiBr electrolyte is 0.050 M. C6H5OH(aq) + H2O(l) init º H3O+(aq) + C6H5O-(aq) ~10-7 M 0.0500 M equil 0.0500 - x ~x for μ = 0.0500 M 0 γH+ = 0.86 x γC6H5O- = 0.835 γC6H5OH = 1 Y x = 2.65 x 10-6 M = [C6H5O-] = [H3O+] and pH = -log (2.65 x 10-6 )(0.86) = 5.64 8. The mixture is a HC2H3O2 /C2H3O2- buffer. HC2H3O2 + H2O init: 0.195 M equil: ~0.195 M X H3O+ + C2H3O2~10-7 M 0.040 M x ~0.040 M Solutions for Chemical Activities Exercises Since HC2H3O2 is a weak acid, the only important source of ions in the buffer solution is NaC2H3O2. [NaC2H3O2] = 0.040 M Y μ = 0.040 M Y γC2H3O2- = 0.902 - 0.062 = 0.840 γHC2H3O2 • 1 (since HC2H3O2 is uncharged) Y pH = 3.99