For #4 I am very confuse as to why there is two extra volumes. If I use that I w
ID: 505504 • Letter: F
Question
For #4 I am very confuse as to why there is two extra volumes. If I use that I will be canceling the volumes of the concentrations and then I am only left with moles which I am getting confuse at and I am probably doig it wrong. 3. Assuming that no equilibria other than dissolution are involved, calculate the concentration of all solute species in each of the following solutions of salts in contact with a solution containing a common ion. Show that changes in the initial concentrations of the common ions can be neglected. (a) AgCl(s) in 0.025 M NaCl (b) CaF2(s) in 0.00133 M KF (c) Ag2SO4(s) in 0.500 L of a solution containing 19.50 g of K2SO4 (d) Zn(OH02(s in a solution buffered at a pH of 11.45 4. Calculate the cadmium ion concentration, [Cd2 in a solution prepared by mixing 0.100 L of 0.0100 M Cd(NO3 2 with 1.150 L of 0.100 NH3(aq). Hint: Cadmium ions associate with ammonia molecules in solution to form the complex ion [CdCNH which is defined by the following equilibrium K, 4.0 x 10
Explanation / Answer
Take the example of AgCl in 0.025 M NaCl. two equilibrium will occur in water :
AgCl (s) <==> Ag+ + Cl-
NaCl (s) <==> Na+ + Cl-
Here common ion is Cl-.. We will consider concentration of Cl- to be 0.025 M. But stricly speaking, Cl- concentration will be slightly less than 0.025 M. As the NaCL dissolution reaction is in equilibrium, so, some og the Cl- will react with Na+ to go back to NaCl. This problem is asking you to consider than this amount is very small and can be neglected.
Ksp values of AgCl is 1.77*10^-10 (Refer to your textbook for Ksp values)
AgCl <==> Ag+ + Cl-
s s (s+ 0.025)
s * (s+0.025) = 1.77*10^-10
or, s^2 + 0.025s - 1.77*10^-10 = 0
s = 7.08*10^-9 M
[Ag+] = 7.08*10^-9 M
[Cl-] = 0.025 + 7.08*10^-9 M = 0.025 M
[Na+] = 0.025 M
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In the second problem you have to find out the concentration of free Cd ^2+ ion in equilibrium with [Cd(NH3)4]^2+
Cd^2+ + 4NH3 (aq) --------> Cd(NH3)4^2+
Kf = [Cd(NH3)4^2+]/[Cd^2+][NH3]^4
[Cd(NH3)2^2+] = 0.1L * 0.01 M = 0.001 moles
[Nh3] = 1.15 L * 0.1M = 0.15 moles
4*10^6 = 0.001 /[Cd^2+][0.15]^4
or, [Cd^2+] = 0.001/4*10^6* (0.15)^4 = 4.94*10^-7 moles
Molarity of Ca^2+ = 4.94*10^-7 moles/total volume = 4.94*10^-7 moles/1.25 = 3.95*10^-7 M