Question

#16.110 A 130.0 −mL sample of a solution that is 2.9×10−3 M in AgNO3 is mixed with a 225.0 −mL sample of a solution that is 0.10 M in NaCN.

After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

Given that 130.0 −mL sample of a solution that is 2.9×10−3 M in AgNO3 is mixed with a 225.0 −mL sample of a solution that is 0.10 M in NaCN.

[AgNO3 ] = molarity x volume in litres = 2.9×10⁻³ M x 0.130 L = 0.000377 mol

   [NaCN ] = molarity x volume in litres = 0.1 M x 0.225 L = 0.0225  mol

When 0.000377 moles of Ag+ is mixed with 0.0225 moles of CN- ,

we say that essentially all of the 0.000377 moles of Ag+ is converted to [Ag(CN)2]^-1 .

The 0.000377 moles of [Ag(CN)2]^-1 has been diluted to a total volume of 355.0 ml.

Hence,

[Ag(CN)2]^-1 = (0.000377 moles of [Ag(CN)2]^-1) / (0.3550L) = 0.00106 M


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

Twice as much CN- is consumed, when it reacts with the 0.000377 moles of Ag+:

Hence,
(0.0225 moles of CN-) - (2) (0.000377 moles lost) = 0.021746 mol CN- remains

which has also been diluted to 355.0 ml .

Hence,

[CN-] reamains = (0.021746 mol CN- remains) / (0.355L) = 0.0612 M


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

1 X 10^21 = [0.00106] / [Ag+] [0.0612]^2

[Ag+] = [0.00106] / (1 X 10^21 ) [0.0612]^2

[Ag+] = 0.28 X 10^-21 Molar

Therefore,

concentration of Ag+(aq) remains =  0.28 X 10^-21 M