16.67The solubility product expression for magnesium hydroxide is Ksp=[Mg2+][OH−]2=1.2 ×10−11We find the hydroxide ion concentration when [Mg2+] is 1.0 ×10−10M1112101.2101.010−−⎛⎞×==⎜⎟⎜⎟×⎝⎠[OH ]0.35−MTherefore the concentration of OH−must be slightly greater than 0.35 M. 16.68 We first determine the effect of the added ammonia. Let's calculate the concentration of NH3. This is a dilution problem. MiVi=MfVf(0.60 M)(2.00 mL) =Mf(1002 mL) Mf=0.0012 MNH3Ammonia is a weak base (Kb=1.8 ×10−5). NH3+H2O UNH4++OH−Initial (M): 0.0012 0 0 Change (M): −x+x+xEquil. (M): 0.0012 −xxx4b3[NH ][OH ][NH ]+−=K
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 482 251.810(0.0012)−×=−xxSolving the resulting quadratic equation gives x=0.00014, or [OH−] =0.00014 MThis is a solution of iron(II) sulfate, which contains Fe2+ions. These Fe2+ions could combine with OH−to precipitate Fe(OH)2. Therefore, we must use Kspfor iron(II) hydroxide. We compute the value of Qcfor this solution. Fe(OH)2(s) UFe2+(aq) +2OH−(aq) Q=[Fe2+]0[OH−]02=(1.0 ×10−3)(0.00014)2=2.0 ×10−11Note that when adding 2.00 mL of NH3to 1.0 L of FeSO4, the concentration of FeSO4will decrease slightly. However, rounding off to 2 significant figures, the concentration of 1.0 × 10−3Mdoes not change. Qis larger than Ksp[Fe(OH)2] =1.6 ×10−14. The concentrations of the ions in solution are greater than the equilibrium concentrations; the solution is saturated. The system will shift left to reestablish equilibrium; therefore, a precipitate of Fe(OH)2will form. 16.71First find the molarity of the copper(II) ion 41 molMoles CuSO2.50 g0.0157 mol159.6 g=×=20.0157 mol[Cu]0.01740.90 L+==MAs in Example 16.15 of the text, the position of equilibrium will be far to the right. We assume essentially all the copper ion is complexed with NH3. The NH3consumed is 4 ×0.0174 M=0.0696 M. The uncombined NH3remaining is (0.30 −0.0696) M, or 0.23 M. The equilibrium concentrations of Cu(NH3)42+and NH3are therefore 0.0174 Mand 0.23 M, respectively. We find [Cu2+] from the formation constant expression. 2133 4f24243[Cu(NH )]0.01745.010[Cu][NH ][Cu](0.23)+++==×=K[Cu2+]=1.2 ×10−13M16.72Strategy:The addition of Cd(NO3)2to the NaCN solution results in complex ion formation. In solution, Cd2+ions will complex with CN−ions. The concentration of Cd2+will be determined by the following equilibrium Cd2+(aq) +4CN−(aq) UCd(CN)42−From Table 16.4 of the text, we see that the formation constant (Kf) for this reaction is very large (Kf=7.1 ×1016). Because Kfis so large, the reaction lies mostly to the right. At equilibrium, the concentration of Cd2+will be very small. As a good approximation, we can assume that essentially all the dissolved Cd2+ions end up as Cd(CN)42−ions. What is the initial concentration of Cd2+ions? A very small amount of Cd2+will be present at equilibrium. Set up the Kfexpression for the above equilibrium to solve for [Cd2+].
CHAPTER 16: ACID-BASE EQUILIBRIA AND SOLUBILITY EQUILIBRIA 483Solution:Calculate the initial concentration of Cd2+ions.