2. a. Experimental Procedure. Cite the reason for each of the five cautions in t
ID: 1027415 • Letter: 2
Question
2. a. Experimental Procedure. Cite the reason for each of the five cautions in the experiment. 130 b. Experimental Procedure, Part E. How is "bumping" avoided in the preparation of a hot water bath? The following chemical equilibria are studied in this experiment. To become familiar with their behavior, indicate the direction, left or right, of the equilibrium shift when the accompanying stress is applied to the system. 3. a. NH3(aq) is added to Ag+(aq) + Cl-(aq) AgCl(s) b. HNO3(ag) is added to Ag.co3(s) Ag+(aq) + CO32-(aq) c. KI(ag) is added to Ag+(aq) + 2 NH,(aq) Ag(NH3)2rap d. Na, S(ag) is added to Agl(s) Ag (aa) +I(a) e. KOH(a) is added to CH COOH(a) + H2O(l)HO (a) +CH,CO2 (aq) f. HCa) is added to 4 CI(a) +Co(H,0)(aa)CoCL, (a) +6 H,00) Experiment 16 21Explanation / Answer
a) Ag+ (aq) + Cl- (aq) <======> AgCl (s)
K = 1/[Ag+][Cl-] (the concentration of a solid species is taken as unity)
NH3 combines with Ag+ to form silver diamine complex as below.
Ag+ (aq) + 2 NH3 (aq) -------> [Ag(NH3)2]+ (aq)
This reduces [Ag+]. In order to counteract the reduced [Ag+], the solid AgCl dissolves, furnishing more Ag+. Thus, the formation of solid AgCl is suppressed by the addition of NH3 and the equilibrium shifts to the left.
b) Ag2CO3 (s) <=======> 2 Ag+ (aq) + CO32- (aq)
K = [Ag+]2[CO32-]
HNO3 dissociates completely producing protons as
HNO3 (aq) --------> H+ (aq) + NO3- (aq)
H+ combines with CO32- to form carbonic acid (H2CO3) which decomposes to gaseous carbon dioxide (CO2) as below.
2 H+ (aq) + CO32- (aq) --------> H2CO3 (aq) ---------> CO2 (g) + H2O (l)
Due to the formation of gaseous CO2, [CO32-] is reduced. To counteract the reduced [CO32-], more Ag2CO3 dissolves in solution, producing CO32-. Thus, the solubility of Ag2CO3 in water increases with the addition of HNO3. Consequently, the equilibrium shifts to the right.
c) Ag+ (aq) + 2 NH3 (aq) <======> [Ag(NH3)2]+ (aq)
K = [Ag(NH3)2+]/[Ag+][NH3]2
KI ionizes completely to produce I- which combines with Ag+ to form insoluble AgI as below.
Ag+ (aq) + I- (aq) --------> AgI (s)
Therefore, [Ag+] reduces due to the formation of AgI. In order to combat the reduced [Ag+], the concentration of the diamine complex, i.e, [Ag(NH3)2+] must reduce proportionately. This is possible when the reverse reaction is favored, i.e, the complex formation reaction is suppressed by the addition of KI. Thus, the equilibrium shifts to the left.
d) AgI (s) <======> Ag+ (aq) + I- (aq)
K = [Ag+][I-]
Na2S ionizes in solution to produce S2-. The produced S2- combines with Ag+ to form insoluble Ag2S as
2 Ag+ (aq) + S2- (aq) --------> Ag2S (s)
Due to the formation of solid Ag2S, [Ag+] reduces and hence K tends to decrease. However, K is an equilibrium constant and therefore, must remain constant at the temperature of the reaction. In order to keep K constant, more AgI must solubilize, producing Ag+. Therefore, the equilibrium shifts to the right.
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