Friday
October 24, 2014

Homework Help: chemistry

Posted by slamj on Thursday, October 24, 2013 at 12:09am.

Determination of Kf of Cu(NH3)4 (aq) with given 10mL of 6M NH3 mix with 40mL of 1M Cu.

This is what I have so far:

nCu= .040L x 1= .040mol
nNH3= .010L x 6= .060mol

Cu^2+ + 4NH3 -> Cu(NH3)4
.040 .06 0
-.040 4(-.040) +.040
0 -.1 .040

[Cu(NH3)]= .040/(.010+.040)=.80M

for Kf=[Cu(NH3)]/[Cu][NH3]^4
= .80/(1*6^4)=6.17E-04
I'm not so quite sure with my answer. Help please??

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