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How would a solution made of $\pu{20 mL}$ of $\ce{NaOH}$, $\pu{10 mL}$ of $\ce{HCl}$, $\pu{10 mL}$ $\ce{NaH2PO4}$ and $\pu{5 mL}$ of $\ce{Na3PO4}$, all of them in a concentration of $\pu{10 mmol/L}$, behave? I already tried every way i can think of, but none give the solution of the $\mathrm{pH}$ of this system.

Ways I tried:

  1. After reacting the $\ce{HCl}$ with $\ce{NaOH}$, I use the $\ce{NaOH}$ that i still got to react with $\ce{NaH2PO4}$, and to calculate the $[\ce{H+}]$ I used $[\ce{Na2HPO4}]$ + $\ce{[Na3PO4]}$, but i didnt got any answers.

  2. I used this equation $\ce{NaH2PO4 + 2NaOH -> Na3PO4 + 2H2O}$ to find the concentration and then applied to the amphoteric equation, using $K_{a_2}$ and $K_{a_3}$.

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$ \begin{align} (n_{\ce{NaOH}})_i= \pu{0.2 mmol}\\ (n_{\ce{HCl}})_i= \pu{0.1 mmol}\\ (n_{\ce{NaH2PO4}})_i= \pu{0.1 mmol}\\ (n_{\ce{Na3PO4}})_i= \pu{0.05 mmol} \end{align} $

At first the $\ce{NaOH}$ will react with $\ce{HCl}$ as per the following reaction:

$$ \begin{align} \begin{array}{ccccc} \ce{NaOH} & + & \ce{HCl} & \ce{->} & \ce{NaCl} & + & \ce{H2O} \\ \pu{0.2 mmol} & & \pu{0.1 mmol} & & \pu{0 mmol} & & \\ \pu{0.1 mmol} & & \pu{0 mmol} & & \pu{0.1 mmol} & & \\ \end{array} \end{align} $$ Now, the remaining $\ce{NaOH}$ will react with $\ce{NaH2PO4}$ as per following:

$$ \begin{align} \begin{array}{ccccc} \ce{NaOH} & + & \ce{NaH2PO4} & \ce{->} & \ce{Na2HPO4} & + & \ce{H2O} \\ \pu{0.1 mmol} & & \pu{0.1 mmol} & & \pu{0 mmol} & & \\ \pu{0 mmol} & & \pu{0 mmol} & & \pu{0.1 mmol} & & \\ \end{array} \end{align} $$

Now the resulting solution is a buffer of $\ce{HPO4^2-}$ and $\ce{PO4^3-}$. So the Henderson equation can be used.

$ [\ce{PO4^3-}]=\frac{0.05}{45} \pu{M}\\ [\ce{HPO4^2-}]=\frac{0.1}{45} \pu{M}\\ $

$$ \begin{align} \begin{array}{ccc} \mathrm{pH} & = & \mathrm{pK_{a_3}} + \log(\frac{[\ce{PO4^3-}]}{[\ce{HPO4^2-}]}) \\ \mathrm{pH} & = & 12.35 + \log(\frac{0.05}{0.1})\\ \mathrm{pH} & = & 12.35 + \log(\frac{1}{2})\\ \mathrm{pH} & = & 12.04 \\ \end{array} \end{align} $$

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