What is the molar concentration of Ag^+ (aq) in a 1.00 M solution of Ag (NH_3)_2

Question

What is the molar concentration of Ag^+ (aq) in a 1.00 M solution of Ag (NH_3)_2^+ with no excess ammonia? (K_f = 1.70 times 10^7 for Ag (NH_3)_2^+) (a) 2.45 times 10^-3 M (b) 3.09 times 10^-3 M (c) 2.42 times 10^-4 M (d) 1.47 times 10^-8 M (e) 1.70 times 10^-8 M Calculate the equilibrium concentration of Zn^2+ (aq) in a solution that is initially 0.0125 M Zn (NO_3)_2 and 0.600 M NH_3. Zn^2+ (aq) + 4 NH_3 (aq) Zn (NH_3)_4^2+ (aq) k_f = 2.9 times 10^9 1) Assume that zinc nitrate fully dissolves in water. 2) If you set up a RICE table with -x being the change in concentration of Zn^2+ (aq), the only real root of the resulting equation is 1.24999999528956 times 10^-2. (a) 7.8 times 10^-12 M (b) 4.7 times 10^-11 M (c) 1.5 times 10^-8 M (d) 2.5 times 10^-9 M (e) 2.9 times 10^9 M For the reaction Fe (H_2 O)_6^3+ (aq) + H_2 O (l) Fe (H_2 O)_5 (OH)^2+ (aq) + H_3 O^+ (aq) + H_3 O^+ (aq), Fe (H_2 O)_6^3+ (aq) is a (n) ____ and water is a (n) ____ (a) base; acid. (b) acid; base. (c) acid; base. (d) catalyst; base. (e) Lewis base; Lewis base; Lewis acid.

Explanation / Answer

Solution:-The reaction is

Ag+ + 2NH3 Ag(NH3)2+

[Ag(NH3)2+]/ [Ag+][NH3]^2 = 1.70 x 10^7
Preparing the ICE table

[Ag+]……[NH3]……..[Ag(NH3)2+]
initial 0 0 1.00
change..+x +2x -x
equilibrium x 2x 1-x

(1-x)/ x(2x)^2 = 1.70 x 10^7
since the Kf is so large, we will neglect the x in (1-x)

1/4x^3 = 1.70 x 10^7

1 = 4x^3(1.70 x 10^7)

1 = x^3(6.8 x 10^7)

x^3= 1 /6.8 x 10^7

x^3 = 1.47 x 10^-8

x = 2.45 x 10^-3

[Ag+] = 2.45 x 10^-3

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