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Using the value of the formation constant for the complex ion $\mathrm{Co}\left(\mathrm{NH}_{3}\right)_{6}^{2+},$ calculate the dissociation constant.

$7.7 \times 10^{-6}$

Chemistry 102

Chapter 15

Equilibria of Other Reaction Classes

Chemical Equilibrium

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Rice University

University of Kentucky

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Okay, let me start with writing a reaction. No 2 positive, plus 6 n h 3 gives you a complex compound, nh 362 positive and k f, which is a formation. Constant expression can be represented as concentration of products in this case divided by the concentration of reactants. So so k f value is provided as 1.310 raise to power 5. Now let me represent the dissociation for the formed products. The association reaction will be c, o n h, 362, positive c, o 2 positive, plus 6 n h. 3. Again, we have k d dissociation constant as concentration of products, that is to positive 2 n h 3, raised to power 6 divided by the concentration of reactants. We substitute the value of 1.310 raise to power negative 5 point. That is formation constant. So we know that formation is the reciprocal for dissociation, substituting the value of k f, which is already provided as 1.3 as reciprocal for k d, so substituting those values we have k. D, equals 1 divided by 1.310, raise to power 5 equals 7.69 multiply 10, raise to the power negative 6, which is equal to 7.710, raise to power negative 6. This is the dissociation constant.

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