Question

$25mL$ of $0.107$ $H_3{PO}_4$ was titrated with $0.115M$ solution of $NaOH$ to the endpoint identified by indicator bromocresol green. This required $23.1mL$ of the $0.115M$ $NaOH$. The titration was repeated using phenolphthalein as an indicator. This time $25mL$ of $0.107M$ $H_3{PO}_4$ required $46.2mL$ of the $0.115M$ $NaOH$. What is the coefficient of $n$ in this equation for each reaction?

$H_3{PO}_4+n{OH}^{-}⟶n{H}_{2}O+{\left[H_{3-n}{PO}_4\right]}^{n-}$

Answer 1

In the first titration, only first H atom of phosphoric acid is removed.
In the second titration, 2 H atoms of phosphoric acid are removed.
The number of millimoles of phosphoric acid

$=25$ mL $\times 0.107mol/L=2.7$ millimol

The number of millimoles of NaOH required for first titration

$=23.1$ mL $\times 0.115mol/L=2.7$ millimol

The number of millimoles of NaOH required for second titration

$=46.2$ mL $\times 0.115mol/L=5.4$ millimol

Hence, the coefficient n for the first titration is 1 and for the second titration is 2.

$H_3{PO}_4+{OH}^{-}⟶H_{2}O+{\left[H_2{PO}_4\right]}^{-}$

$H_3{PO}_4+2{OH}^{-}⟶2H_{2}O+{\left[H_1{PO}_4\right]}^{2-}$