What is #["Ni"^(2+)]# at equilibrium upon formation of #"Ni"("NH"_3)_6^(2+)# from #"1 mmol"# of #"Ni"^(2+)# in #"230.0 mL"# aqueous solution containing #"0.300 M NH"_3#?

Wouldn't it be almost zero? This is pretty much the exact opposite of a strong acid, in the sense that it associates pretty much completely.

I got #["Ni"^(2+)] ~~ 1.87 xx 10^(-8) "M"#.

Well, what you should always do is write out the reaction. The concentration of #"Ni"^(2+)# is initially

#"1 mmol"/"230.0 mL" = "0.004348 M"#,

and the reaction is:

#"Ni"^(2+)(aq) " "+" " 6"NH"_3(aq) -> ["Ni"("NH"_3)_6]^(2+)(aq)#

#"I"" "0.004348" "" "" "0.300" "" "" "" "" "0#
#"C"" "-x" "" "" "" "-6x" "" "" "" "" "+x#
#"E"" "0.004348-x" "0.300-6x" "" "" "x#

Given the formation constant for the above complexation reaction, hopefully you can realize why I didn't write this as an equilibrium. Nevertheless, the formation "equilibrium" constant expression is:

#K_f = 5.5 xx 10^8 = ([["Ni"("NH"_3)_6]^(2+)])/(["Ni"^(2+)]["NH"_3]^6)#

#= x/((0.004348 - x)(0.300 - 6x)^6)#

It's highly likely that #["Ni"^(2+)]_(eq) ~~ "0 M"#. Why? Because #K_f# is huge... We could still solve this, under the approximation that #0.004348 - x ~~ 0#, but we would need to NOT evaluate that term.

To maintain the maximum accuracy, consider taking the #ln# of both sides (instead of plugging in #x ~~ "0.004348 M"# right away).

#ln(5.5 xx 10^8) = lnx - ln(0.004348 - x) - 6ln(0.300 - 6x)#

Now, plug in #x ~~ 0.004348# (EXCEPT in the #ln#'s that go to #pmoo#!) to get:

#20.125 = -5.438 - ln(0.004348 - x) - (-7.7697)#

Therefore:

#color(blue)(["Ni"^(2+)]) = 0.004348 - x ~~ e^(-17.7933)#

#~~ color(blue)(1.87 xx 10^(-8) "M")#