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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:

$\ce{H_3PO_4 <=> H^{+} + H_2PO_4^{-}}\ \ \ K_\mathrm{a1} =7.5\times 10^{-3}\\$

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

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

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:

$\ce{H_2PO_4^{-} <=> H^{+} + HPO_4^{2-}}\ \ \ K_\mathrm{a2} =6.2\times 10^{-8}\\$

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:

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

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?

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Yes, x << $\mathrm{0.118 M}$ 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 $K_\mathrm{a2}$.

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