Question

• A solution is made with 1.1×10-3 M Zn(NO3)2(aq) and 0.150 M NH3(aq) After the solution...

• A solution is made with 1.1×10-3 M Zn(NO3)2(aq) and 0.150 M NH3(aq)
After the solution reaches equilibrium, what concentration of “naked” Zn2+(aq) cation remains?
The complex Zn(NH3)42+ has Kf = 2.8×109.

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Answer #1

answer the reaction should be like this

Zn 2+   +    4 NH3      <------------>       ZN (NH3)42+   

as we are given with the ampunt of ZN(NO3)2 = 0.0011 so

0.0011   + 4(0.0011)    ------> 0.0011

extra NH3 = .150 -.0044 = 0.1456 M

Kf= [ ZN (NH3)42+ ] / [Zn 2+ ]   [ NH3]4

2.8 * 109 = 0.0011/ [Zn 2+ ]    [ 0.1456 M ]4

[Zn2+ ] = 0.0011 / 2.8 * 109   * 4.49 * 10-4

= 0.0011 / 12.57 *105 = 8.75 * 10-10        M

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