Question

In: Chemistry

A 130.0 mL sample of a solution that is 2.7×10−3 M in AgNO3 is mixed with...

A 130.0 mL sample of a solution that is 2.7×10−3 M in AgNO3 is mixed with a 230.0 mL sample of a solution that is 0.14 M in NaCN.

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

Solutions

Expert Solution

130.0 mL sample of a solution that is 2.7 x 10-3 M in AgNO3 is---( we will find the moles of Ag+) ---
(0.130 L )* (2.7x 10-3 mol / L) = 0.000351 moles of Ag+

and 230.0 mL sample of a solution that is 0.14 M in NaCN is--( we will find the moles of CN- )---
(0.230 L) * (0.14 mol /L ) = 0.0322 moles of CN-

When 0.000351 moles of Ag+ is mixed with 0.0322 moles of CN- ,then we can assume that necessarily all of the 0.000351 moles of Ag+ has been converted to [Ag(CN)2]-1 .

Again, 0.000351 moles of [Ag(CN)2]-1 has been diluted to a total volume of (130.0 mL + 230.0 mL) = 360.0 mL
Now,

(0.000351 moles of [Ag(CN)2]-1) / (0.3600L) = 0.000975 M [Ag(CN)2]-1

From the equation ,
1 Ag+ + 2 CN-1 --> [Ag+(CN-)2]-1
This shows that twice as much CN- is consumed when it reacts with the 0.000351 moles of Ag+
(0.0322 moles of CN-) - [2 x (0.000351 moles lost) ]

= (0.0322 - 0.000702) mol CN- remains

=0.031498 mol CN- remains.

And which has also been diluted to (130.0 + 230.0) mL =360.0 mL = 0.3600 L
(0.031498 mol CN- remains) / (0.3600 L) = 0.0875 M CN-

Finally,
Kf = [Ag(CN)2]-1 / [Ag+] [CN-]2

=> 1 X 1021 = [0.000975] / [Ag+] [0.0875]2

=> [Ag+] = [0.000975] / (1 X 1021 ) [0.0875]2

=> [Ag+] = [0.000975] / (1 X 1021 ) (0.007656)

=> [Ag+] = (0.000975 / 0.007656) x 10-21

=> [Ag+] = 1.27 x 10-1 x 10-21

=> [Ag+] = 1.27 x 10-22 M


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