Question

Cameco in Port Hope, Ontario uses hydrofluoric acid to make a uranium hexafluoride product, which is used as a fuel for nuclear reactors. A waste drum containing 90.0 L of 5.6 mol/L hydrofluoric acid needs to be neutralized so that it isn't hazardous. Calculate the mass of potassium hydroxide pellets that would be required to completely neutralize the acid.

Answer 1

Given:

$V(HF)=90L$

$c(HF)=5.6\frac{mol}{L}$

First, we can calculate amount of HF in waste drum:

$n(HF)=c(HF) \cdot V(HF)=5.6\frac{mol}{L} \cdot 90L = 504 mol$

According to the reaction of neutralization:

$HF + KOH=KF + H_2O$

the amount of KOH in moles should be the same as HF.

$n(KOH)=n(HF)=504mol$

The mass of pure KOH is

$m(KOH)=n(KOH) \cdot M(KOH)=504mol \cdot 56\frac{g}{mol}=28224g$

Taking into account that reagent grade pellets of potassium hydroxide contain ~85% KOH, we can calculate the mass of potassium hydroxide pellets:

$m(pellets \ KOH)=\frac{m(KOH)}{0.85}=33.2 kg$