**The question and my incorrect response (for the first part, second part is cor

Question

**The question and my incorrect response (for the first part, second part is correct)

**The hint given to me:

General Chemistry 4th Edition this question has been customized by Hannah Morris at University of Pittsburgh Map A solution contains 0.0500 "(aq), 0.03 10 M S2-(aq), and 2.25 M NH3-Cobalt(III) ions in aqueous solutions complex with NHs to produce Co(NH3)83. (K 5.0 x 103) What will be the concentration of Co3*(aq) when Co(NH3)6 forms? Number 3+ 0 5.13x 10M Will Co2Ss precipitate? ( 4.0x 101) O O precipitate forms no precipitate forms

Explanation / Answer

initial

concentration

concentration at

completion

change in concentration

due to dissociation

final

concentration(at equilibrium)

now in the above table since according to the sticiometry of reaction one Co+3 molecule reacts with 6 NH3 molecules so the concentration of ammonia used to react with cobalt will be

6* .0500 = 0.3M

Hance the remaining concetration of ammonia = 2.25 - 0.3 =1.95M

since the value of formation constant is too high so reaction assume to be completed and dissociation will also takes place at equlibrium in small extent so at equlibrium the concentration of Co+3 will asume to be X .

Kf = [Co(NH3)6 ] / [Co+3] [NH3]6

5.0* 1031 = 0.0500-X / X ( 1.95- X) 6

in the above equation X will be very small beacause very less vakue of dissosiation constant so it can be negligible in the sum and substration term

So 5.0 *1031 = .0500 / X (1.95)6

X = 0.18 * 10-34 M

Co+3 6NH3 Co(NH3)63+

initial

concentration

0.0500M 2.25M 0

concentration at

completion

0 1.95 0.0500

change in concentration

due to dissociation

+x +6x -x

final

concentration(at equilibrium)

+x 1.95+6x 0.0500-x

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