Why is ligand substitution only partial with copper(II) ions and ammonia? When studying ligand substitution (at UK year 13 level), the following example has been given:

\begin{align}
\ce{[Cu(H2O)_6]^2+ + 2NH3 &amp;&lt;=&gt; [Cu(OH)_2(H2O)_6] + 2NH4^+}\\
\ce{[Cu(OH)_2(H2O)_4] + 4NH3 &amp;&lt;=&gt; [Cu(NH3)_4(H2O)_2]^2+ +2OH- +2H2O}\\
\end{align}

There has been no explanation given of why the substitution is only partial, whereas the substitution of, say, $\ce{[Co(H2O)_6]^2+}$ is complete. 

To me, it seems that this partial substitution wouldn't be favourable, as it produces a slightly distorted octahedral shape, which is less stable than a regular octahedral shape, surely? Moreover my textbook states ammonia is a better ligand than water, so why would $\ce{NH3}$ not replace those final two aqua ligands?
Metal ion complexes have stepwise stability constants:
\begin{align}
\ce{[Cu(H2O)6]^2+ + NH3 &amp;&lt;=&gt;[$K_1$] [Cu(NH3)(H2O)5]^2+ + H2O}\\
\ce{[Cu(NH3)(H2O)5]^2+ + NH3 &amp;&lt;=&gt;[$K_2$] [Cu(NH3)2(H2O)4]^2+ + H2O}\\
\ce{[Cu(NH3)2(H2O)4]^2+ + NH3 &amp;&lt;=&gt;[$K_3$] [Cu(NH3)3(H2O)3]^2+ + H2O}\\
\ce{[Cu(NH3)3(H2O)3]^2+ + NH3 &amp;&lt;=&gt;[$K_4$] [Cu(NH3)4(H2O)2]^2+ + H2O}\\
\ce{[Cu(NH3)4(H2O)2]^2+ + NH3 &amp;&lt;=&gt;[$K_5$] [Cu(NH3)5(H2O)]^2+ + H2O}\\
\ce{[Cu(NH3)5(H2O)]^2+ + NH3 &amp;&lt;=&gt;[$K_6$] [Cu(NH3)6]^2+ + H2O}\\
\end{align}
with

\begin{align}
K_1 &amp;= \frac{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_2 &amp;= \frac{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]}}{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_3 &amp;= \frac{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]}}{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_4 &amp;= \frac{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]}}{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_5 &amp;= \frac{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]}}{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_6 &amp;= \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]\left[ \ce{NH3} \right]}}\\
\end{align}

The overall stability constant is:

$$K_\text{B} = \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}} = K_1 \cdot K_2 \cdot K_3 \cdot K_4 \cdot K_5 \cdot K_6$$

(In particular, note the exponent 6.)

Furthermore, usually $K_i/K_{i+1}&gt;1$; thus, $K_1 &gt; K_2 &gt; K_3 &gt; K_4 &gt; K_5 &gt; K_6$. Consequently, it gets more and more difficult to exchange further ligands.

Therefore, the predominating complex strongly depends on the concentration of $\ce{NH3}$.
And you probably work with dilute aqueous solutions.

Simply speaking, the missing $\ce{[Cu(NH3)6]^2+}$ in aqueous solution can be attributed to the high concentration of water which is competing with ammonia for the coordination sites.