Diluting acetic acid to obtain a solution of pH 5
This is a dilution problem. So in this instance, you want to add deionized water to your acetic acid. However, the question is, how much water do you add?

It is necessary first to find the molarity of the acetic acid. On PubChem, we see that the molecular weight is $60.05196 \ \frac{\text{g}}{\text{mol}}$ and density is $1.0446 \ \frac{\text{g}}{\text{cm}^3}$ @ $25\ ^\circ \text{C}$.

If we have 1 L of acetic acid, then

$$ 1.0446 \ \frac{\text{g}}{\text{cm}^3} \cdot \frac{1000\ \text{cm}^3}{1\ \text{L}} \cdot \frac{1\ \text{mol}}{60.05196\ \text{g}} = 17.394\ \frac{ \text{mol}}{\text{L}}$$ 

We know that,

$$ \text{pH} = -\log{\ce{[H3O+]}} $$

which implies that the concentration of hydronium ions to make the solution with pH equal to 5 is, $\ce{[H3O+]} = 1 \cdot 10^{-5} \ \text{M}$. We make our RICE table,

$$\ce{HC2H3O2 + H2O -&gt; C2H3O2- + H3O+}$$

\begin{array} {|c|c|c|c|c|}
\hline
\text{Initial conc.} &amp; x \ \text{mol} &amp; - &amp; \text{0 mol} &amp; \text{0 mol}\\
\hline
\text{Change conc.} &amp; -1 \cdot 10^{-5}\ \text{M} &amp; - &amp; 1 \cdot 10^{-5}\ \text{M} &amp; 1 \cdot 10^{-5}\ \text{M}\\
\hline
\text{End conc.} &amp; x - 1 \cdot 10^{-5}\ \text{M} &amp; - &amp;  1 \cdot 10^{-5}\ \text{M} &amp; 1 \cdot 10^{-5}\ \text{M}\\
\hline
\end{array}

$$K_\text{a} = \frac{[\ce{H3O+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = \frac{(0.00001)(0.00001)}{(x - 0.00001)} = 1.75 \cdot 10^{-5} $$

$$\frac{(1 \cdot 10^{-10})}{(x - 0.00001)} = 0.0000175$$
$$1 \cdot 10^{-10} = 0.0000175x - 1.75 \cdot 10^{-10} $$ 
$$2.75 \cdot 10^{-10} = 0.0000175x$$ 
$$x = 1.6 \cdot 10^{-5}\ \text{M} = [\ce{CH3COOH}] $$

Thus, you go from a $17.394\ \text{M}$ solution of acetic acid, to a $0.000016\ \text{M}$, which is a factor of $1.09\cdot10^{6}$. You can then dilute the solution by the appropriate amount by placing the amount of acid you want in a volumetric flask, and then successively dilute it to the required pH.

Two assumptions,


Deprotonation is small enough that the equilibrium concentration of the acid is approximately equal to the same as its initial concentration. 
Autoprotolysis of water does not significantly contribute to pH.