# Titration of H3PO4 with NaOH

I came across an ionic equilibrium problem stating:

Find the pH when 150 ml 1 M $\ce{NaOH}$ has been added to 100 ml 1 M $\ce{H3PO4}$.

I'm stuck with this question. What I know: $$\ce{H3PO4 + NaOH -> NaH2PO4 + H2O}$$ $$\ce{NaH2PO4 + NaOH -> Na2HPO4 + H2O}$$

Finally 50 millimoles $\ce{Na2HPO4}$ and 50 millimoles $\ce{NaH2PO4}$ remains. Should I apply acid buffer equation?

You have got your 1 molar phosphoric acid that will get deprotonized by the 1 molar sodium hydroxide solution that is added. As both solutions have got the same concentration, you will reach the first inflection point after 100 mL and the second inflection point after 200 mL.

Now what is between those two volumes, at your desired volume of 150 mL?
A simple buffer system that can be described using the Henderson–Hasselbalch equation: $$\mathrm{pH}=\mathrm pK_\mathrm a-\log_{10}\frac{\ce{c(H2A- )}}{\ce{c(HA^2- )}}$$

As both concentrations $\ce{c(H2A- )}$ and $\ce{c(HA^2- )}$ are equivalent, the logarithmic expression gets zero and your final pH equals the second $\mathrm pK_\mathrm a$ value of phosphoric acid, which is 7.21 (according to the German Wikipedia).

$\hskip3.5cm$ This is a simple question involving a mixture of strong acid (not that strong) and a strong base. Since $k_a$ values are not given the book might have wanted you to solve it like this. They want you to assume complete dissociation of both compounds. That is, 5 moles of $\ce{NaOH}$ gives 5 moles of $\ce{OH^-}$ ions after complete dissociation. Find the equivalents of $\ce{H^+}$ from phosphoric acid (a) and the equivalents of $\ce{OH^-}$ from $\ce{NaOH}$ (b). Also calculate the total volume. Number of equivalents=$NV$. But be careful since concentration is given in molarity.

If $b>a$, then the resulting solution is basic. Calculate the equivalents of $b$ that remain. From this you will get pOH. $$[\ce{OH^-}]=\frac{b-a}{V_1+V_2}$$ If $a>b$, then resulting solution is acidic. Calculate the amount of $a$ that remains. From this, you will get pH. $$[\ce{H^+}]=\frac{{a-b}}{V_1+V_2}$$

Where N is normality and equivalents of acid is greater than equivalents of base.