# What volume, in milliliters, of 0.54 M Ca(OH)_2 is needed to completely neutralize 241.4 mL of a .26 M HI solution?

$C a {\left(O H\right)}_{2} \left(a q\right) + 2 H I \left(a q\right) \rightarrow C a {I}_{2} \left(a q\right) + 2 {H}_{2} O \left(l\right)$
$\text{Moles of HI}$ $=$ $241.4 \times {10}^{-} 3 \cdot L \times 0.26 \cdot m o l \cdot {L}^{-} 1$ $=$ $0.063 \cdot m o l \cdot H I$
$0.0313 \cdot m o l$ $C a {\left(O H\right)}_{2}$ are required for equivalence. Now of course I could use this molar quantity to calulate the volume of calcium hydroxide required. However, there is one problem.
Calcium hydroxide is very sparingly soluble. I don't think you could make a $0.5 \cdot m o l \cdot {L}^{-} 1$ solution of this beast (I might be wrong, but I am not going to go thru the calculation). This question should have been proposed with $N a O H$ solution