Confusion about the ICE table Calculate the pH of a $\pu{2M}$ $\ce{H_3PO_4}$.Calculate the equilibrium concentrations of $\ce{H_3PO_4 , H_2PO_4^{-} , HPO_4^{2-}}$.


First equation:

$\ce{H_3PO_4  &lt;=&gt; H^{+} + H_2PO_4^{-}}\ \ \ K_\mathrm{a1} =7.5\times 10^{-3}\\$

$\begin{array}{|c|c|c|c|} \hline
\ &amp; [\ce{H_3PO_4}] &amp; [\ce{H^{+}}] &amp; [\ce{H_2PO_4^{-}}]\\ \hline
\text{I} &amp; 2 &amp; 0 &amp; 0 \\
\text{C} &amp; -x &amp; +x &amp; +x \\
\text{E} &amp; 2-x &amp; +x &amp; x\\ \hline \end{array}\\$

$$K_\mathrm{a1}=\frac{[\ce{H_2PO_4^{-}}]*[\ce{H^{+}}]}{[\ce{H_3PO_4}]}=\frac{x^2}{2-x}=7.5\times 10^{-3}$$

Solving for $x$ we get $x=\pu{0.118M}$.

$[\ce{H_3PO_4}]=2-x=2-0.118=\pu{1.882M}$

$[\ce{H_2PO_4^{-}}]=x=\pu{0.118M}$

Second equation:

$\ce{H_2PO_4^{-}  &lt;=&gt; H^{+} + HPO_4^{2-}}\ \ \ K_\mathrm{a2} =6.2\times 10^{-8}\\$

The solution manual says:$$K_\mathrm{a2}=\frac{[\ce{HPO_4^{2-}}]*[\ce{H^{+}}]}{[\ce{H_2PO_4^{-}}]}=\frac{0.118\times[\ce{HPO_4^{2-}}]}{0.118}$$

$[\ce{HPO_4^{2-}}]=K_\mathrm{a2}$

However I'm not sure why didn't we use an ICE table for the second equation as follows:

$\begin{array}{|c|c|c|c|} \hline
\ &amp; [\ce{H_2PO_4^{-}}] &amp; [\ce{H^{+}}] &amp; [\ce{HPO_4^{2-}}]\\ \hline
\text{I}&amp; 0.118 &amp; 0.118 &amp; 0 \\
\text{C}&amp; -x &amp; +x &amp; +x \\
\text{E}&amp; 0.118-x &amp; 0.118+x &amp; x\\ \hline \end{array}\\$

Is it because $x$ is negligible with respect to 0.118? Or should the concentrations precalculated above be in the equilibrium row of the ICE table?
Yes, x &lt;&lt; $\mathrm{0.118 M}$ that it is not gonna hurt even if it is ignored. But for a rigid perfectionist, 
$$K_\mathrm{a2}=\frac{[\ce{HPO_4^{2-}}][\ce{H^{+}}]}{[\ce{H_2PO_4^{-}}]}=\frac{x\times (0.118+x)}{0.118-x}$$
Hence we get a quadratic equation,
$$x^2 + 0.118000062 - 7.316\times 10^{-9}=0$$
Solving for $x$, we get $x = 6.199993484584665\times 10^{-8}$ which is approximately equal to the value of $K_\mathrm{a2}$.