Calculate the molar solubility of Ca(OH)_2 in 0.10 M NaOH solution at 25 degree
ID: 1060314 • Letter: C
Question
Calculate the molar solubility of Ca(OH)_2 in 0.10 M NaOH solution at 25 degree C. K_sp of Ca(OH)_2 1.8 times 10^-10 at 25 degree C. How would you prepare Ca(OH)_2 anyway? Use equations to show why silver chromate, Ag_2CrO.4, is soluble in solutions of i) sodium thiosulfate, Na_2S_2O_3 ii) nitric acid iii) ammonia. Calculate the molar concentration of Ag^+ ions in a solution containing 0.0200 M silver nitrate and 1.00 M ammonia solution. K_f = 1.7 times 10^7 Calculate the molar solubility of AgCI in 0.150 M ammonia. K_f for Ag(NH.3)_2^+ = 1.7 times 10^7 and K_sp for AgCI = 1.6 times 10^-10.Explanation / Answer
13) NaOH dissociates completely into its constituents ions and we have
NaOH (aq) -----> Na+ (aq) + OH- (aq)
[OH-] = 0.100 M
Consider the dissociation of Ca(OH)2:
Ca(OH)2 (s) <=====> Ca2+ (aq) + 2 OH- (aq); Ksp = [Ca2+][OH-]2 = 1.8*10-10
Look at the Ksp of Ca(OH)2. Since it is extremely small (of the order of 10-10), we can neglect the contribution of OH- from the dissociation of Ca(OH)2 and assume OH- to come entirely from NaOH. We are given that [OH-] = 0.100 M.
Let x be the solubility of Ca(OH)2 in 0.100 M NaOH. Therefore, [Ca2+] = x. Put in the Ksp expression to obtain
Ksp = [Ca2+][OH-]2 = (x).(0.100)2
===> 1.8*10-10 = x.(0.01)
===> x = 1.8*10-10/0.01 = 1.8*10-8
The solubility of Ca(OH)2 in 0.100 M NaOH is 1.8*10-8 M (ans).
14) Write down the dissociation of Ag2CrO4 as below:
Ag2CrO4 (s) <====> 2 Ag+ (aq) + CrO42- (aq)
Ksp = [Ag+]2[CrO42-]
(i) Na2S2O3 (aq) -----> 2 Na+ (aq) + S2O32- (aq)
Ag+ (aq) + 2 S2O32- (aq) ------> [Ag(S2O3)2]3- (aq)
Silver forms a soluble complex with thiosulfate ions as shown above. This reduces the effective concentration of Ag+ ions. Since Ksp is an equilibrium constant, the value of Ksp must remain unchanged at a particular temperature. As the temperature doesn’t change, more silver chromate must undergo dissociation (i.e., go into solution) to compensate for the loss of Ag+ ions. Therefore, Ag2CrO4 is soluble in sodium thiosulfate.
(ii) HNO3 is a strong electrolyte and undergoes complete dissociation to produce
HNO3 (aq) -----> H+ (aq) + NO3- (aq)
The proton combines with chromate ion to form soluble chromic acid.
2 H+ (aq) + CrO42- (aq) ----> H2CrO4 (aq)
Due to the formation of H2CrO4, the effective concentration of CrO42- reduces. To compensate for the reduced chromate ion concentration, more silver chromate must undergo dissociation producing more silver and chromate ions. Therefore, silver chromate is soluble in nitric acid.
(iii) Ammonia (NH3) combines with Ag+ to form a soluble silver amine complex as below:
Ag+ (aq) + 2 NH3 (aq) -----> [Ag(NH3)2]+ (aq)
The concentration of silver ions decreases due to the formation of the soluble silver amine complex. Since Ksp must remain constant at a particular temperature, hence, to make up for the formation of the above complex, more silver ions must be formed and hence silver chromate should undergo enhanced dissociation. Therefore, silver chromate is soluble in ammonia.
15) The given reaction is
Ag+ (aq) + 2 NH3 (aq) -----> [Ag(NH3)2]+ (aq); Kf = 1.7*107
Assume quantitative association of silver nitrate and ammonia so that [Ag(NH3)2]+ = 0.0200 M; therefore, in a 1.0 L vessel, we will have 0.0200 mole [Ag(NH3)2]+ and (1.00 – 0.0200) mol = 0.98 mole NH3. Now, write down the dissociation reaction as
[Ag(NH3)2]+ (aq) <======> Ag+ (aq) + 2 NH3 (aq)
Initial 0.0200 0 0.98
Change -x +x +2x
Equilibrium (0.0200 – x) x (0.98 + 2x)
Set up the expression for the new equilibrium constant Kd as
Kd = [Ag+][NH3]2/[Ag(NH3)2+] = 1/Kf = 1/(1.7*107) = 5.88*10-8; therefore,
5.88*10-8 = (x).(0.98 + 2x)2/(0.0200 – x)
Assume x << 0.0200 and write
5.88*10-8 = (x).(0.98)2/(0.0200)
====> x = 5.88*10-8*0.0200/(0.98)2 = 1.2244*10-9 1.22*10-9
The molar solubility of Ag+ ions in the solution of [Ag(NH3)2]+ is 1.22*10-9 M (ans).
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