The formation of silver ammonium ion can be illustrated by the following chemical reaction
Ag+ (aq) + 2NH3 (aq) ---------> [Ag(NH3)2]+ (aq)
the formation constant can be given as follows :
Kf = ([Ag(NH3)2+]) / ([Ag]+ [NH3]2 )
= 1.6 X 107
the chemical reaction for the dissociation of AgCl at equilibrium is shown below:
AgCl(s) -------> Ag+(aq) + Cl-(aq)
substitute the given value of Ksp in above equation and the solubility product of AgCl at equilibrium can be determined as follows :
Ksp = [Ag+] [Cl- ]
= 1.6 X 10-10
the overall chemical reaction is shown below:
AgCl(s) + 2NH3 (aq) ---------> [Ag(NH3)2]+(aq) + Cl-(aq)
therefore the overall formation constant for the above reaction can be given as follows
Keq = ([Ag(NH3)2+])[Cl-] / [NH3]2
multiply and divide the above equation by [Ag+]
Keq = ([Ag(NH3)2+]) [Cl-] [Ag+] / ( [NH3]2 X [Ag+] )
=([Ag(NH3)2+]) /([Ag+] [NH3]2) X [Ag+][Cl-]
= Kf X Ksp
substitute the given value of Kf and Ksp in the above equation
Keq = (1.6) X 107 X 1.6 X 10-10
= 2.56 X 10-3
therefore the overall equilibrium is 2.56 X 10-3
An ICE table for the overall chemical reaction is tabulated as shown below
AgCl(s) + 2NH3 (aq) ---------> [Ag(NH3)2]+(aq) + Cl-(aq)
I (M) 0.10 0 0
C(M) -2x x x
E(M) 0.10 - 2x x x
here the concentration of products so formed is x
the equilibrium constant can also be given in terms of x which is shown below:
Keq = ([Ag(NH3)2+])[Cl-] / [NH3]2
2.56 X 10-3 = x X x / (0.10 - 2x)2
2.56 X 10-3 = ( x / (0.10 - 2x))2
x / (0.10 - 2x) = square root of (2.56 X 10-3)
x / (0.10 - 2x) = 0.0506
(0.10 - 2x) / x =`19.76
0.10/x -2 = 19.76
0.10 / x = 21.76
x = 4.59 X 10-3
therefore the solubility product of AgCl is 4.59 X 10-3 M
hope it is clear
give thumbs up if you like
thank you
can you teach me how to solve this problem? Calculate the molar solubility of silver chloride...
H. Complex Ions 1. Silver chloride has very low solubility in water (Ks of 1.6 x 10-19), however, it is highly soluble in an ammonia solution due to the formation of the complex ion [Ag(NH3)21 (Kr" 1.7 X 10) Given this information, calculate the molar solubility of AgCl in 2.25 M ammonia. 2. Calculate the molar solubility of Cul in 0.92 M KCN. (Ksp of Cul is 1.1 x 10-12, Kp of [Cu(CN)2] is 1.0 x 1016).
please explain how you got all your answers
20. Calculate the MOLAR SOLUBILITY of silver bromide (Kp - 5.40 x 1013) in 1.0 M NH. Complex ion [Ag(NHs)2] can be formed (K, 1.70 x 10). Ag(NH,) (aq)+ Br' (aq) Overall Reaction: AgBr(s)+2 NH,(aq) 21. The pH of a 0.0412 M solution of a monoprotic acid is 1.39. Is this a STRONG ACID? a. Yes b. No c. Not enough information d. Unknown
Calculate the solubility of silver chloride in 10.0 M ammonia given the following information: K_sp (AgCI) = 1.6 times 10^-10 Ag^+ (aq) + NH_3(aq) = AgNH_3^+ (aq) K_f1 = 2.2 times 10^3 AgNH_3^+ (aq) NH_3(aq) = Ag(NH_3)_2^+ K_f2 = 8.2 times 10^3
Testbank, Question 17.074
In an aqueous solution silver ions can form
Ag(NH3)2+(aq) in the
presence of ammonia. For
Ag(NH3)2+(aq): (see
photo)
Testbank, Question 17.074 In an aqueous solution silver ions can form Ag(NH3)2+ (aq) in the presence of ammonia. For Ag(NH3)2+(aq) Ag(NH3)2+(aq) is a complex ion; NH3 is a donor atom; Ag+ is a ligand Ag(NH3)2+(aq) is an acceptor ion; NH3 is a ligand; Ag+ is a complex ion Ag(NH3)2+(aq) is a complex ion; NH3 is a ligand; Ag+ is...
please help!!
We were unable to transcribe this imageConsider the insoluble compound silver hydroxide, AgOH.The silver ion also forms a complex with ammonia. Write a balanced net ionic equation to show why the solubility of AgOH(s) increases in the presence of ammonia and calculate the equilibrium constant for this reaction. For Ag(NH3)2* K = 1.1*10'. Use the pull-down boxes to specify states such as (aq) or (s). Knet = Consider the insoluble compound iron(II) hydroxide. Fe(OH)2. The iron(II) ion also...
Part B please.
Part A Silver chloride is only slightly soluble in pure water at 25 C AgCI(s) Ag (ag)+CI (aq) K-1.8 x 10 1 Calculate the concentration of Ag and Cl in a solution that is saturated with AgCl e, the system is at equilibrium and there is still solid AgCl visible). Express your answers using two significant figures, separate your answers by a comma. [Ag+],[Cl-)- 1.3×10-5 1.3×10-5 mol Li Previous An Correct Part B The addition of ammonia...
20 points water is 1.3 x 10-16mol/L. The solubility of silver sulfide can be greatly increased by formation of a complex ion such as Ag(NH3)2. Calculate the molar solubility of AgaS in 10M NH3. Show your work for full credit. Sulfides salts are generally very insoluble. The solubility of silver sulfide, AgaS, in pure Ag2S Ksp= 8.0 x 1048 Ag(NHs)2 Kr 1.7x 10
2. With the precipitate (AgCl) still present in the test tube, add aqueous ammonia. The precipitate should disappear. This means that silver chloride is soluble in ammonia solution. It is soluble because silver ion and ammonia form a complex called diamminesilver (I) chloride, [Ag(NH3)2]Cl which in solution exists as [Ag(NH)2]* and Cl ions. You could have guessed that this would have happened if you looked over the Table A values of Kr at the end of this lab (and you...
4) Calculate the molar solubility of AgCl (Ksp = 1.8 x 10-) at 25°C in: (a) Pure water (b) 3.0 M NH3 [Hint: AgCl(s) + 2NH3(aq) Ag(NH3)2(aq) + Cl(aq) K = 3.1 x 10
Silver chloride, AgCl, is a sparingly soluble solid. Answer the following questions about a saturated solution prepared by placing solid silver chloride in a 1.94 x 10-5 M NaCl(aq) solution. At some temperature, the silver ion concentration, [Ag+], was found to be 6.24 x 10-6 M. (a) What is the concentration of chloride ions, [CI – ], in the resulting solution? XM (b) What is the molar solubility of silver chloride, AgCl, in 1.94 x 10-5 M NaCl? 4.9 6.24e-6...