# Titration of multi-basic acid [closed]

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 $$K_{a_3}$$.

\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}