Silver ion reacts with excess CN− to form a colorless complex ion, Ag(CN)−2, which has a formation constant Kf=3.0×1020.
Calculate the concentration of Ag+ in a solution prepared by mixing equal volumes of 2.3×10−3M AgNO3 and 0.21M NaCN.
After mixing but before reacting the concns are (Ag^+) = 0.0023/2 = 0.00115M
(CN^-) = 0.21/2 = 0.105M
I've divided by 2 since equal volumes of each have been added which means both dilute each other by a factor of 2. Then we do an ICE chart.
..........Ag^+ + 2CN^- --> Ag(CN)2^-
I....0.00115....0.105........0
C...-0.00115...-0.00115....+0.0115
E......0........0.10385....+0.0115
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We can make this assumption (that with such a huge Kf that the reaction goes to completion and there is no free Ag^+ floating around. This isn't EXACTLY right but with K of 10^20 it isn't far from right. THEN we start with another ICE chart but this time work backwards,(from right to left) like this.
......Ag^+ + 2CN^- <-- Ag(CN)2^-
......0.....0.10385....0.00115.....I
......x......+x.........-x.........C
......x.....0.10385+x...0.00115-x..E
Now substitute the E line of this last ICE chart into Kf expression and solve for x = (Ag^+). x should be a very small number.
To calculate the concentration of Ag+ in the solution, we will use the concept of the formation constant and the common ion effect.
Step 1: Write the balanced equation for the reaction:
Ag+ + 2CN- -> Ag(CN)2-
Step 2: Determine the initial concentrations of Ag+ and CN-:
The initial concentration of Ag+ is given as 2.3×10^-3 M.
The initial concentration of CN- is given as 0.21 M.
Step 3: Find the final concentration of CN- after the reaction:
Since we have equal volumes of AgNO3 and NaCN, the final concentration of CN- would be half of the initial concentration, i.e., 0.21 M / 2 = 0.105 M.
Step 4: Set up an equilibrium expression using the formation constant:
Kf = [Ag(CN)2-] / ([Ag+] * [CN-]^2)
Step 5: Substitute the known values into the equilibrium expression:
3.0×10^20 = [Ag(CN)2-] / ([Ag+] * (0.105 M)^2)
Step 6: Solve the equation for [Ag+] by rearranging the equation:
[Ag+] = [Ag(CN)2-] / (3.0×10^20 * (0.105 M)^2)
Step 7: Calculate [Ag+] using the formation constant and the concentration of CN-:
[Ag+] = [Ag(CN)2-] / (3.0×10^20 * 0.105 M * 0.105 M)
[Ag+] = [Ag(CN)2-] / 3.32625×10^20
Step 8: Determine the concentration of [Ag(CN)2-]:
Since equal volumes of AgNO3 and NaCN are mixed, the concentration of [Ag(CN)2-] will be determined by the limiting reagent. In this case, AgNO3 will be the limiting reagent.
Using the balanced equation, we can calculate the moles of Ag(CN)2- produced from the moles of AgNO3:
1 mole of AgNO3 produces 1 mole of Ag(CN)2-
Moles of AgNO3 = concentration of AgNO3 * volume of solution (since volume = concentration)
Moles of AgNO3 = 2.3×10^-3 M * V
(where V is the volume of the solution)
Moles of Ag(CN)2- = Moles of AgNO3
Now we can substitute the value of [Ag(CN)2-] into the equation for [Ag+] calculated in Step 7 to determine the concentration of Ag+.
Step 9: Calculate the concentration of Ag+:
[Ag+] = [Ag(CN)2-] / 3.32625×10^20
[Ag+] = (Moles of Ag(CN)2-) / (3.32625×10^20 * V)
Substitute the value for the moles of Ag(CN)2-, which is equal to the moles of AgNO3, into the equation:
[Ag+] = (2.3×10^-3 M * V) / (3.32625×10^20 * V)
Step 10: Calculate the concentration of Ag+:
[Ag+] = (2.3×10^-3) / 3.32625×10^20
(assuming V cancels out in the equation)
[Ag+] ≈ 6.91×10^-24 M (rounded to 3 significant figures)
Therefore, the concentration of Ag+ in the solution is approximately 6.91×10^-24 M.
To solve this problem, we need to consider the reaction between silver ion (Ag+) and cyanide ion (CN-) to form the complex ion Ag(CN)−2:
Ag+ + 2CN- --> Ag(CN)−2
First, we need to determine the initial concentration of Ag+ in the solution. Since we are mixing equal volumes of 2.3×10−3 M AgNO3 and 0.21 M NaCN, the initial concentration of Ag+ ions will be half of the AgNO3 concentration.
Initial concentration of Ag+ = (2.3×10−3 M) / 2 = 1.15×10−3 M
Next, we need to determine the concentration of CN- ions in the solution. Since we are mixing equal volumes of AgNO3 and NaCN, the concentration of CN- ions will be the same as the NaCN concentration.
Concentration of CN- = 0.21 M
Now we can use the formation constant (Kf) for Ag(CN)−2 to calculate the concentration of Ag+ ions in the solution.
Using the formation constant expression:
Kf = [Ag(CN)-2] / ([Ag+] * [CN-]^2)
We can rearrange the equation to solve for [Ag+]:
[Ag+] = [Ag(CN)-2] / (Kf * [CN-]^2)
Substituting the known values:
[Ag+] = [Ag(CN)-2] / (Kf * [CN-]^2)
= (1.15×10−3 M) / (3.0×10^20 * (0.21 M)^2)
= 24.7 x 10^-24 M
Therefore, the concentration of Ag+ in the solution is approximately 24.7 x 10^-24 M.