Ag+(aq) + 2 NH3(aq) Ag(NH3)2+ I 5.0�10-4 1.0 0 C -x -2x +x E 5.0�10-4-x 1.0-2x x

$K_{c12}=1.7\times 10^7 =\frac{[\mathrm{Ag}(\mathrm{NH}_3)_2^+]}{[\mathrm{Ag}^+][\mathrm{NH}_3]} = \frac{x}{(5.0\times 10^{-4}-x)\times(1.0-2x)^2}$

We assume that x is small compared to 1.0.  Thus, we simplify to the following:

$K_{c12}=1.7\times 10^7 = \frac{x}{(5.0\times 10^{-4}-x)}$

$8.5\times10^3 = (1 + 1.7\times10^7)x$

At this point we look a the left-hand side and realize that 1 << 1.7�107, so we replace (1 + 1.7�107) with just 1.7�107. This whole thing reduces to

x = 5.0�10-4.

Since the amount of change, x, is exactly the same as the amount of initial Ag+,  We would conclude that there is exactly a 100% reaction (no silver ions left over). To 2 sig. fig., that is the best we can do.  This MUST be incorrect since there is a measured equilibrium constant that is not infinity (some silver ions left over).  Hence, this method is not the best way to solve this particular equilibrium problem.

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