Silver ions will react with cyanide ions to form the complex ion Ag(CN)2 a) 0.20
ID: 506108 • Letter: S
Question
Silver ions will react with cyanide ions to form the complex ion Ag(CN)2 a) 0.200 moles of AgNO3 and 0.800 moles of NaCN are mixed together in 1.00 liter of solution. Determine the equilibrium concentrations of the Agt (aq), CN (aq), and Ag(CN)2 (aq) ions in the resulting solution. b) Silver chloride is almost insoluble in pure water, but it is much more soluble in a solution containing sodium cyanide due to the formation of the Ag(CN)2 complex ion. (i) Determine the molar solubility of silver chloride in pure water. (i) Write a balanced chemical reaction for the dissolving of AgCl (s) in the presence of excess CN (aa), and determine the numerical value of the equilibrium constant for this reaction. Please include state symbols such as (s) and (aq). (iii) 0.200 moles of AgCl (s) is added to 1.00 L of 0.500 MNaCN. Using a numerical calculation, show that all of the AgCl (s) will dissolve in this solution. c) Using any data from the Useful Information given above, determine the numerical value of the equilibrium constant for the following reaction: Ag(CN)2- (aa) 2 H+ (aq) Ag+ (aq) 2 HCN (aq)Explanation / Answer
Answer for question (a):
Silver nitrate and sodium cyanide dissociate in aqueous solutions as given below
AgNO3 (aq) <============> Ag+ (aq) + NO3- (aq)
NaCN (aq) <=============> Na+ (aq) + CN- (aq)
Initially as sodium cyanide is added it will form insoluble AgCN precipitate as shown below.
AgNO3 (aq) + NaCN (aq) <=============> AgCN (s) + NaNO3 (aq)
Upon further addition of excess sodium cyanide then AgCN forms a dicyanide complex
AgCN (s) + NaCN <===================> Na[Ag(CN)2]
It's assumed that AgNO3 completly reacts with excess NaCN to form the equivalent amount of silver cyanide complex. So overall stoichiometric equation can be writtten as follows.
AgNO3 + 2NaCN <=============> Na[Ag(CN)2] + NaNO3
0.2 M 0.8 M 0.2 M 0.2 M
If 0.4 M NaCN is consumed as per the stoichiometric equation then 0.4 M solution of NaCN will remain
Equilibrium constant, Kc = [Na[Ag(CN)2]] [NaNO3] / [AgNO3] + [NaCN]2
= [0.2] [0.2] / [0.2]0.4]2
= 0.04 / 0.032
Kc = 1.25
xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
Answer for question (b. i): Molar solubility of AgCl
AgCl (s) dissociates into the corresonding ions as shown below:
AgCl (s) =============> Ag+ + Cl-
Solubility product, Ksp = [Ag+][Cl-]
Let the molar solubility of AgCl be x
So Ksp = [x][x] = x2
x = square root of Ksp of AgCl
Ksp of AgCl = 1.8 x 10-10
x = Squar root of 1.8 x 10-10
x = 1.34 x 10-5 M
Therefore molar solubility of AgCl = 1.34 x 10-5 M
(b. ii): Balanced chemical reaction:
AgCl (s) + 2 NaCN (aq) <===================> Na[Ag(CN)2] (aq) + NaCl (aq)
(b. iii): Following the balanced chemical reaction
AgCl (s) + 2 NaCN (aq) <===================> Na[Ag(CN)2] (aq) + NaCl (aq)
0.2 M 0.5 M 0.2 M 0.2 M
Since 1 mole of AgCl reacts with 2 moles of NaCN and forms 1 mole of sodium siver cyanide complex apart from the formation of 1 mole of NaCl.
So as shown 0.2 M AgCl upon reaction with excess NaCN forms 0.2 M sodium siver cyanide complex apart from the formation of 0.2 M of NaCl.
Answer for question (c):
Ag(CN)2- (aq) + 2H+ (aq) <===============> Ag+ (aq) + 2 HCN (aq)
0.2 M 0.5 M 0.2 M 0.4 M
Equilibrium constant, Kc = [Ag]+ [HCN]2 / [Na[Ag(CN)2]] [H+]2
0.2 M silver complex reacts with 2 eq of acid (0.4 M) and forms 0.2 M silver ions and 0.4 M NaCl. Accordingly
Kc = 0.2 x (0.4)2 / 0.2 x (0.4)2
= 0.032 / 0.032
Kc = 1.0
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.