CH16 × T Chemistry question lche X > d www.webassign.net/vveb/Student/Assignment

Question

CH16 × T Chemistry question lche X > d www.webassign.net/vveb/Student/Assignment. Responses/submit?dep-1681 8156 0 as follows: Question: 1 2 3 2 The Hg ion forms complex ions with Hgl. + r-, Hgl2 K2 = 1.0 x 105 Hgl2Hgla K 1.0 x 10 Hgls Hgl K 1.0 x 108 A solution is prepared by dissolving 0.067 mol Hg(NO3)2 and 5.00 mol Nal in enough water to make 1.0 L of solution Calculate the equilibrium concentration of [Hg Submit 11.34e-34 Your answer is incorrect. Please try again. Show Hints Submit AnswerSave Submit Assignment Save Assignment Progress 9:49 AM Type here to search 116/2017

Explanation / Answer

Solution-

We have From the question

Hg2+ + I– <------> HgI+, K1 = 1.0 x 10^8
HgI+ + I– <---------> HgI2 , K2 = 1.0 x 10^5
HgI2 + I– <-----------> HgI3–, K3 = 1.0 x 10^9
HgI3– + I– <---------> HgI42–, K4 = 1.0 x 10^8

So by adding all

Hg+2 + 4I- -------------------> HgI4-2   

for this equilibrium constant Kc = K1 x K2 x K3 x K4 = 10^30

Kc = 10^ 30

Since Kc is very very high so assume that all the Hg2+ is converted into the product , leaving a trace of Hg2+ to maintain equilibrium.

Hg+2 + 4 I- <-----------------------> HgI4-2

x 4.752 0.067 (here I- = 5 - 4 x 0.067 = 4.732)

10^30 = 0.067 / x (4.732)^4

x = 1.34 x 10 ^ - 34 .

Hence the equilibrium concentration of Hg2+ = 1.34 x 10 ^ - 34 M

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