Question

A 110.0-mL sample of a solution that is 2.7×10−3 M in AgNO3 is mixed with a 220.0-mL sample of a solution that is 0.11 M in NaCN. For Ag(CN)2−, Kf=1.0×1021. After the solution reaches equilibrium, what concentration of Ag+(aq) remains?

Answer 1

The concentration of Ag+(aq) remains is 1.76x10⁻²²mol/L

Step-by-step explanation:

The number of moles of Ag⁺ in the solution is:

`n_(Ag+) =110mL*(1L)/(1000mL) *(2.7x10^(-3)moles )/(1L) =2.97x10^(-4) moles`

The number of moles of CN⁻ is:

`n_(CN-) =220mL*(1L)/(1000mL) *(0.11moles)/(1L) =0.0242moles`

At a total volume of 330 mL = 0.33L, the molarity is:

`M=(2.97x10^(-4) )/(0.33) =9x10^(-4) mol/L`

The reaction is:

Ag + 2CN → Ag(CN)₂

You can see that 1 mol of Ag requires 2 moles of CN, so:

`moles-of-CN-remain=0.0242-(2*2.97x10^(-4) )=0.0236moles`

The molarity is:

`M=(0.0236)/(0.33) =0.0715 mol/L`

The Kf of the reaction is:

`Kf=([Ag(CN)_(2)] )/([Ag][CN]^(2) )`

Replacing:

`1x10^(21) =(9x10^(-4) )/([Ag][0.0715]^(2) )`

Solving for [Ag]:

`[Ag]=(9x10^(-4) )/(1x10^(21)*(0.0715^(2)) ) =1.76x10^(-22) mol/L`

Answer 2

The concentration of Ag+ that remains is 1.76 *10^-22 M

Step-by-step explanation:

Step 1: Data given

Volume of AgNO3 = 110.0 mL = 0.110 L

Molarity of AgNO3 = 2.7 * 10^-3 M

Volume of NaCN = 220.0 mL = 0.220 L

Molarity of NaCN = 0.11 M

Kf of Ag(CN)2− = 1.0 * 10^21

Step 2: Calculate moles AgNO3

Moles AgNO3 = molarity AgNO3 * volume

Moles AgNO3 = 2.7 * 10^-3 M * 0.110 L

Moles AgNO3 = 0.000297 moles

Step 3: Calculate moles Ag+

For 1 mol AgNO3 we have 1 mol Ag+ and 1 mol NO3-

For 0.000297 moles AgNO3 we have 0.000297 moles Ag+

Step 4: Calculate moles NaCN

Moles NaCN = 0.11 M * 0.220 L

Moles NaCN = 0.0242 moles

Step 5: Calculate moles CN-

For 1 mol NaCN we have 1 mol CN-

For 0.0242 moles NaCN we have 0.0242 moles CN-

Step 6: The balanced equation

Ag+ + 2CN- → Ag(CN)2-

Step 7: Calculate the limiting reactant

For 1 mol Ag+ we need 2 moles CN- to produce 1 mol of Ag(CN)2-

Ag+ is the limiting reactant, it will completely be consumed (0.000297 moles). CN- is in excess. There will react 2*0.000297 moles = 0.000594 moles. There will remain 0.0242 - 0.000594 = 0.023606 moles CN-

There will be formed 0.000297 moles Ag(CN)2-

Step 8: Define Kf

Kf = 1.0 * 10^21

Kf = [Ag(CN)2-] / [Ag+][CN-]²

Step 9: Calculate concentration of Ag+ that remains

Final volume of solution = 110 mL + 220 mL = 330 mL = 0.330 L

[Ag(CN)2-] = 0.000297 moles/0.330 L

[Ag(CN)2-] = 0.0009 M

[CN-] = 0.023606 moles / 0.330 L

[CN-] = 0.0715 M

[Ag+] = TO BE DETERMINED

1.0 * 10^21 = 0.0009 M / ([Ag+](0.0715²)

[Ag+] = 0.0009 / (1.0 * 10^21 * 0.0715²)

[Ag+]= 1.76 *10^-22 M

The concentration of Ag+ that remains is 1.76 *10^-22 M