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

Answer:

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

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

Answer 2

Answer:

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

Explanation:

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

$n_{Ag+} =110mL*\frac{1L}{1000mL} *\frac{2.7x10^{-3}moles }{1L} =2.97x10^{-4} moles$

The number of moles of CN⁻ is:

$n_{CN-} =220mL*\frac{1L}{1000mL} *\frac{0.11moles}{1L} =0.0242moles$

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

$M=\frac{2.97x10^{-4} }{0.33} =9x10^{-4} mol/L$

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:

$M=\frac{0.0236}{0.33} =0.0715 mol/L$

The Kf of the reaction is:

$Kf=\frac{[Ag(CN)_{2}] }{[Ag][CN]^{2} }$

Replacing:

$1x10^{21} =\frac{9x10^{-4} }{[Ag][0.0715]^{2} }$

Solving for [Ag]:

$[Ag]=\frac{9x10^{-4} }{1x10^{21}*(0.0715^{2}) } =1.76x10^{-22} mol/L$