# What is the molarity of a sample of phosphoric acid if 15.00 ml of 2 M KOH was required to titrate 12 ml of the H_3PO_4 sample?

Jun 8, 2017

$\left[{H}_{3} P {O}_{4}\right] = 1.25 \cdot m o l \cdot {L}^{-} 1$

Of course, there is a catch.............

#### Explanation:

This one is not so straightforward, and requires specialist knowledge.

It is a fact that phosphoric acid is a DIACID in aqueous solution, and reaches a stoichiometric endpoint at $H P {O}_{4}^{2 -}$:

${H}_{3} P {O}_{4} \left(a q\right) + 2 K O H \left(a q\right) \rightarrow {K}_{2} H P {O}_{4} \left(a q\right) + 2 {H}_{2} O$.

And thus moles of $K O H$ $=$ $15.00 \times {10}^{-} 3 \cdot L \times 2.0 \cdot m o l \cdot {L}^{-} 1$ $= 0.030 \cdot m o l$.

And thus the equivalent quantity of ${H}_{3} P {O}_{4}$ was $0.015 \cdot m o l$ which were present in a $12 \cdot m L$ volume of the acid........

$= \frac{\frac{0.030 \cdot m o l}{2}}{12 \times {10}^{-} 3 \cdot L} = 1.25 \cdot m o l \cdot {L}^{-} 1$.

Remember what I said with regard to the acid-base behaviour of phosphoric acid. It is NOT a triacid..........and you might win brownie points from your prof if you know this fact.