pH of an acetate buffer after dilution 100 mL of a 0.1 mM buffer solution made from acetic acid and sodium acetate with pH 5.0 is diluted to 1 L. What is the pH of the diluted solution?

The pKa value for acetic acid is given as 4.75.

The answer provided in my textbook is pH = 5.53. The closest that I can get to that is pH = 5.507. I know that's an almost insignificant difference but I'm just confused how they got to their answer, so if anyone can give me some suggestions that would be awesome. I'll outline my approach below:

Using the Henderson-Hasselbalch equation to find the ratio of [A-] to [HA] in the 100ml: 
$$10^{5-4.75} = \frac{[A^-]}{[HA]}$$

$$[A^-] = \left(\frac{10^{0.25}}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 6.4{\times}10^{-5}M$$

$$[HA] = \left(\frac{1}{10^{0.25}+1}\right)\times\left(\frac{0.1mM}{1000}\right) = 3.6{\times}10^{-5}M$$

The addition of 900ml of water to 100ml of buffer is a 10x dilution:

$$[A^-] = 6.4{\times}10^{-5} \times 0.1 = 6.4{\times}10^{-6}M$$

$$[HA] = 3.6{\times}10^{-5} \times 0.1 = 3.6{\times}10^{-6}M$$

$$\ce{[H+]} = 10^{-5} \times 0.1 = 10^{-6}M$$

ICE table to determine equilibrium concentrations after dilution:

$$
\begin{array}{c|c@{hello}c@{}c@{}c@{}c@{}c@{}c}
   &amp;   [CH_3 COOH] &amp; + &amp;[H_2 O] &amp; {}\rightleftharpoons{} &amp; [H_3 O^+] &amp; + &amp;[CH_3 COO^-] \\
\hline
I &amp; 3.6 \times 10^{-6} &amp;&amp; - &amp;&amp; 10^{-6} &amp;&amp; 6.4 \times 10^{-6} \\
C &amp; -x &amp;&amp; - &amp;&amp; +x &amp;&amp; +x \\
E &amp; 3.6 \times 10^{-6} -x &amp;&amp; - &amp;&amp; 10^{-6} +x &amp;&amp; 6.4 \times 10^{-6} +x \\
\hline
\end{array}
$$

$$K_a = \frac{\ce{[H3O+][CH3COO-]}}{\ce{[CH3COOH]}}$$ 
Then substituting in Ka and the equilibrium values from the ICE table: $$(10^{-4.75}) = \frac{(10^{-6}+x)(6.4 \times 10^{-6} + x)}{(3.6 \times 10^{-6} -x)}$$

$$x^2 + 2.518 \times 10^{-5}x -5.762 \times 10^{-11} = 0$$

$$x = 2.111 \times 10^{-6}$$ $$ [H^+] = 10^{-6} + 2.111 \times 10^{-6} = 3.111 \times 10^{-6}$$

$$ pH = -log_{10}(3.111 \times 10^{-6}) = 5.507 $$
Depending on your first calculations before dilution:
\begin{align}
[\ce{CH3COO-}]&amp;=[\ce{CH3COONa}]=\pu{6.4\times{10^{-5} M}}=[\ce{Na+}]\\ 
[\ce{Na+}]_\text{(After dilution)}&amp;=\pu{6.4\times{10^{-6} M}}\\ 
[\ce{CH3COO-}]+[\ce{CH3COOH}]&amp;=\pu{1\times{10^{-4} M}}~\text{ (Before dilution)}\\
[\ce{CH3COO-}]+[\ce{CH3COOH}]&amp;=\pu{1\times{10^{-5} M}}~\text {(After dilution)}
\end{align}
So,you have the the following $5$ independent equations in order to specify the unknown concentrations of $5$ species as present in an aqueous solution consisting of $\pu{1\times{10^{-5} M}}$  buffer solution made from acetic acid and sodium acetate :

\begin{align}
6.4 \times 10^{−6} &amp;= [\ce{Na+}] \tag{1}\\
K_\mathrm{a} &amp;= \frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} 
     = 1 \times 10^{−4.75} \tag{2}\\
K_\mathrm{w} 
    &amp;= [\ce{H+}][\ce{OH-}] 
     = 1 \times 10^{−14} \tag{3}\\
1\times{10^{-5}}
    &amp;=[\ce{CH3COOH}] + [\ce{CH3COO-}]\tag{4}\\
[\ce{H+}]+[\ce{Na+}] 
    &amp;= [\ce{OH-}] + [\ce{CH3COO-}] \tag{5}
\end{align}