In: Chemistry
A. Calculate the solubility of silver chloride in a solution that is 0.150.
B. 120.0 −mL sample of a solution that is 2.8×10−3 M in AgNO3 is mixed with a 225.0 −mL sample of a solution that is 0.11 M in NaCN.
A) AgCl(S) = Ag+(aq) + Cl-(aq)
Ag+(aq) + 2NH3(aq) = Ag(NH3)2+(aq)
adding the 2 equations we get:
. .AgCl + 2NH3 <---> Ag(NH3)2+ + Cl-
I. . . . . . . 0.150. . . . . . . .0. . . . . . . .0
E. . . . . . 0.150 -2s. . . . .s . . . . . . . .s
K = [Ag(NH3)2+][Cl-] / [NH3]^2
0.00272 = s^2 / (0.150-2s)^2
s = [AgCl] = 0.00472 mol/lit.
B) 125.0 mL sample of a solution that is 2.7×10−3 M in AgNO3
is
(0.125 L ) (2.7×10−3 mol / L) = 0.0003375 moles of Ag+
225.0 mL sample of a solution that is 0.11 M in NaCN is
(0.225 L) (0.11 M in NaCN) = 0.02475 moles of CN-
when 0.0003375 moles of Ag+ is mixed with 0.02475 moles of
CN-
we say that all of the 0.0003375 moles of Ag+ is converted to
[Ag(CN)2]^-1
the 0.0003375 moles of [Ag(CN)2]^-1 has been diluted to a total
volume of 345.0 ml:
(0.0003375 moles of [Ag(CN)2]^-1) / (0.3450L) = 0.000978 Molar
[Ag(CN)2]^-
======================================...
by the equation:
1 Ag+ & 2 CN-1 --> [Ag+(CN-)2]^-1
twice as much CN- is consumed, when it reacts with the 0.0003375
moles of Ag+:
(0.02475 moles of CN-) - (2) (0.0003375 moles lost) = 0.0240 mol
CN- remains
which has also been diluted to 345.0 ml
(0.0240 mol CN- remains) / (0.345L) = 0.0697 Molar CN-
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Kf = [Ag+(CN-)2] / [Ag+] [CN-]^2
1*10^21 = [0.000978] / [Ag+] [0.0697]^2
[Ag+] = [0.000978] / (1*10^21 ) [0.0697]^2
[Ag+] = [0.000978] / (1*10^21 ) (0.00485)
Ag = 2.0 *10^-22 Molar