Confusion about 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 $K_\mathrm{a2}$.