Question

need help with all parts of question 2. thank you :)

equivalence points and the pH at the half equivalence points for the three titration regions. Question 2 Part A Given the Ksp = 2.0 x 10-29 for the sparingly soluble salt, calcium phosphate (Ca3(PO4)2), determine the molar solubility of the salt in a solution that is 0.20 M in calcium perchlorate, Ca(C104)2. Part B The cation, Co2+, is able to form a complex ion, (Co(SCN)4] 2-, with the thiocyanate anion, SCN- (a) Write the chemical equation for the (a) Write the chemical equation for the formation of this complex ion with the corresponding expression for the formation constant, kf, and (b) Given Kf for (Co(SCN)4] - = 1 x103, explain how having SCN-in solution would affect the solubility of Co(OH)2 (Ksp for Co(OH)2 = 1.3 x 10-15). Part C When the sparingly soluble salt lead fluoride, PbF2, is added to an acidic solution with (H+) = 1.6 x 10-4 M, the resulting concentration of [Pb2+) after the salt dissolves is determined to be 7.75 x 10-5 M. (a) Determine the equilibrium constant for the dissolving of PbF2 in an acidic solution of H3O+ and (b) given Ka = 6.8 x 10-4, use the value determined in part a to calculate the Ksp for PbF2 through manipulation of the equilibrium equations. Show work.

Answer 1

Note that all the parts are seperate questions and are not related. Here Part A) and Part B) with their sub-parts have been answered. Kindly post Part C) separately

Part A ) is an example of common ion effect. If a common ion is present, in this case Co2+, the solubility of the sparingly soluble salt decreases further, according to Le-Chatelier's principle. And the reduced solubility (s) is now determined by the concentration of the ion(s) which is not common, here PO43- .

Part B) is an example of increasing the solubility of a sparing soluble salt , Co(OH)2, by removing one of the ions, here Co2+ by a parallel reaction, here reaction with SCN-, to form the complex, which is much faster than the rate of dissolution of Co(OH)2 . Again due to Le-Chatelier's principle since one of 'products' , here Co2+ , is removed, the reaction will move further in the forward direction to keep its Ksp constant. Thus greater dissolution of Co(OH)2 implies its solubility will increase, in the presence of SCN-

See below for calculations of both parts:

$1585048748943_img295.png$