Question C Due to the formation of a complex, increased [NH3] results in a large

Question

Question C Due to the formation of a complex, increased [NH3] results in a larger Agcl solu O True O False QUESTION 2 1 points Save Given that the ksp of cacog is 1 x 10-12 andthe formation constamt of CarNta)k2 s 10x 107 wha does the ammonia concentration of a solution need to be in order for the solubility of Caco3 to be 0.1 M? QUESTION 3 points Save The closer a process is to being reversible, the more efficient the process can convert heat to work O True O False 1 pointsSave QUESTION 4 What is the change in entropy when 30.0 grams of ethanol (CH3CH20H) are vaporized at its boiling point of 78.0oc? AH 39.3 k/mol. Answer in units of /K but do not include units.

Explanation / Answer

1) The solubility of AgCl in water can be denoted as

AgCl (s) --------> Ag+ (aq) + Cl- (aq)

Ksp = [Ag+][Cl-]

The formation of silver-amine complex can be denoted as

Ag+ (aq) + 2 NH3 (aq) ---------> [Ag(NH3)2]+ (aq)

Increased [NH3] results in more complex formation, thereby reducing [Ag+]. Therefore, in order to keep Ksp constant, more AgCl must solubilize in water forming Ag+. Therefore, the given statement is true.

2) Write down the dissociation of CdCO3 in water as

CdCO3 (aq) --------> Cd2+ (aq) + CO32- (aq)

Ksp = [Cd2+][CO32-] = 1*10-12

Given that the solubility of CdCO3 is 0.1 M, we must have [Cd2+] = 0.1 M.

The complex formation equation is given as

Cd2+ (aq) + 4 NH3 (aq) ---------> [Cd(NH3)4]2+ (aq)

Kf = [Cd(NH3)4]2+/[Cd2+][NH3]4 = 1.0*107

Since Kf is high, the entire Cd2+ obtained from the dissociation of CdCO3 must complex, i.e, [Cd(NH3)4]2+ = 0.1 M.

Again we have

1 mole Cd2+ = 4 moles NH3.

Therefore, 0.1 M Cd2+ = (4*0.1 M) = 0.4 M [Cd(NH3)4]2+.

Let the initial concentration of NH3 be x M; therefore, [NH3]e = (x – 0.4) M.

Set up the Kf expression as

Kf = (0.1)/(0.1)(x – 0.4)4

=====> 1.0*107 = 1/(x – 0.4)4

=====> (x – 0.4)4 = 1/(1.0*107) = 1.0*10-7

=====> log [(x – 0.4)4] = log (1.0*10-7)

=====> 4*log (x – 0.4) = -7

=====> log (x – 0.4) = -7/4 = -1.75

=====> (x – 0.4) = antilog(-1.75) = 0.01778

=====> x = 0.4 + 0.01778 = 0.41778 ? 0.42

The concentration of NH3 must be 0.42 M (ans).

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