# Question b256c

Apr 24, 2017

You have the stoichiometric equation:

#### Explanation:

${H}_{3} P {O}_{4} + 3 K O H \rightarrow {K}_{3} P {O}_{4} + 3 {H}_{2} O$.

$\text{Moles of KOH}$ $=$ $\frac{89.5 \cdot g}{56.11 \cdot g \cdot m o {l}^{-} 1} = 1.60 \cdot m o l$

And thus, given the equation, $0.531 \cdot m o l$ ${H}_{3} P {O}_{4}$ will react, i.e. $0.531 \cdot m o l \times 98.08 \cdot g \cdot m o {l}^{-} 1 = 52.1 \cdot g$.

Of course, there is a catch. In aqueous solution, phosphoric acid acts as a diacid, and NOT AS A TRIACID. Only 2 equiv of the potassium hydroxide will react to give K^(+)""^(-)HPO_4(aq)#