Does adding water to a reaction mixture shift equilibrium?
Yes, the equilibrium does indeed readjust after a change in concentration. This can be seen by careful examination of the formula of the equilibrium constant $K$. Consider an arbitrary reaction that produces two products from a single reactant and is in equilibrium:

$$\begin{align}\ce{A &amp;&lt;=&gt; B + C}\tag{1}\\[0.5em]
K &amp;= \frac{[\ce{B}][\ce{C}]}{[\ce{A}]}\tag{2}\end{align}$$

Let’s assume initial concentrations of $\pu{2M}$ for each species; that gives us $K = 2$ in dimensionless units (you can also use M as a unit here). Remember that $K$ is a constant so we can examine whether equilibrium has been reached by examining $K$.

Now let’s double the volume of solvent. The new concentration of each reactant is $\pu{1M}$. Therefore:

$$K' = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]} = \frac{1\times 1}{1} = 1\tag{2'}$$

Since $1\ne2$, we are no longer at equilibrium. Thus the concentrations must readjust. How? Obviously, the value of the fraction is too low. To increase its value, we can increase the numerator or decrease the denominator — luckily for us, both of these mean that we must perform the reaction in a forward direction to get think right again. Since $[\ce{B}]=[\ce{C}]$, we can actually calculate the ratio of concentrations and from $2[\ce{A}]+[\ce{B}]+[\ce{C}]=\text{const.}=4$, we can calculate exact values.

$$\begin{align}&amp;&amp;K &amp;= \frac{[\ce{B}]^2}{[\ce{A}]}\\[0.5em]
\Longrightarrow&amp;&amp;[\ce{A}] &amp;= \frac{[\ce{B}]^2}2\\[1.3em]
&amp;&amp;2\times\frac{[\ce{B}]^2}{2} + [\ce{B}] + [\ce{B}]&amp;=4\\
&amp;&amp;[\ce{B}]^2 + 2[\ce{B}]-4 &amp;= 0\\
\Longrightarrow&amp;&amp;[\ce{B}]_{1/2} &amp;= \frac{-2\pm\sqrt{2^2+4\times4}}{2}\\[0.5em]
&amp;&amp;[\ce{B}]_{1/2} &amp;= -1\pm\sqrt{5}\\[0.5em]
&amp;&amp;[\ce{B}]_{1/2} &amp;\approx 1.24\end{align}$$

Thus, the product side has increased.



A more real-world example is maybe the dimerisation of hydrogenchromate $\ce{HCrO4-}$ according to $(3)$:

$$\ce{2 HCrO4- &lt;=&gt; Cr2O7^2- + H2O}\tag{3}$$

This is also pH-dependent, since hydrogenchromate is a weak acid and can deprotonate to give chromate $\ce{CrO4^2-}$. However, it is fairly easy to try this experiment yourself: take an orange coloured dichromate solution and add water. A colour change to yellow should be observed indicating that the reaction has equilibrated to the reactants’ side.