Question

A 130.0 −mL sample of a solution that is 3.0×10−3 M in AgNO3 is mixed with a 225.0 −mL sample of a solution that is 0.12 M in NaCN. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

ANSWER:-
130.0 mL sample of a solution that is 3.0×10−3 M in AgNO3 is
(0.130 L ) (3.0×10−3 mol / L) = 0.00039 moles of Ag+

225.0 mL sample of a solution that is 0.12 M in NaCN is
(0.225L) (0.12M in NaCN) = 0.027 moles of CN-

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when 0.00039 moles of Ag+ is mixed with 0.027 moles of CN-
we say that essentially all of the 0.00039moles of Ag+ is converted to [Ag(CN)2]^-1

the 0.00039 moles of [Ag(CN)2]^-1 has been diluted to a total volume of 355.0 ml:
(0.00039moles of [Ag(CN)2]^-1) / (0.3550L) = 0.0010986 Molar [Ag(CN)2]^-

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by the equation:
1 Ag+ & 2 CN-1 --> [Ag+(CN-)2]^-1
twice as much CN- is consumed, when it reacts with the 0.00039 moles of Ag+:
(0.027 moles of CN-) - (2) (0.00039 moles lost) = 0.02622 mol CN- remains

which has also been diluted to 355.0 ml
(0.02622 mol CN- remains) / (0.355L) = 0.0738 Molar CN-

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Kf = [Ag+(CN-)2] / [Ag+] [CN-]^2

1 X 10^21 = [0.0010986] / [Ag+] [0.07386]^2

[Ag+] = [0.0010986] / (1 X 10^21 ) [0.07386]^2

[Ag+] = [0.0010986] / (1 X 10^21 ) (0.0054553)

Ag = 2.0138X 10^-22 Molar