Question

A 120.0 −mL sample of a solution that is 2.6×10⁻³ M in AgNO3 is mixed with a 230.0 −mL sample of a solution that is 0.11 M in NaCN. a.) After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

Number of moles = molaritry * volume in L

1 Mole AgNO3 = 1 Mole Ag+

And moles of AgNO3: 2.6* 10^-3 * 120/1000

= 0.000312 moles AgNO3 or Ag+

And,

Moles of NaCN or CN-:

0.11*230/1000= 0.0253 moles CN-

when 0.000312 moles of Ag+ is mixed with 0.0253 moles of CN-
Then; 0.000312 moles of Ag+ is converted to [Ag(CN)2]^-1

total volume = 230+ 120 = 350 ml

=350/1000= 0.350 L

Molarity = number of moles / volume in L

= 0.000312 moles [Ag(CN)2]^-1 /0.350L

= 0.000891 M [Ag(CN)2]^-1

And;
by the equation:
1 Ag+ & 2 CN-1 --> [Ag+(CN-)2]^-1

twice as much CN- is consumed, when it reacts with the 0.000312moles of Ag+:
(0.0253 moles of CN-) - (2) (0.000312 moles lost) = 0.024676mol CN- remains

molarity of CN- = 0.024676mol CN- /0.350 L

= 0.071M CN-

We know that Kf of [Ag+(CN-)2] =5.3*10^18
formation constant of [Ag+(CN-)2] is given as follows:

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

5.3 X 10^18 = [0.000891] / [Ag+] [0.071]^2

[Ag+] = [0.000891] / (5.3 X 10^18 ) [0.071]^2

[Ag+] = [0.000951] / (1 X 10^21 ) (0.007885)

Ag = 3.33 X 10^-21 Molar