# What is the molarity of 1.5 liters of an aqueous solution that contains 52 grams of lithium fluoride, LiF?

$\text{Molarity}$ $=$ $\text{Moles of solute"/"Volume of solution}$
And thus, here, $\text{Molarity}$ $=$ $\frac{\frac{52 \cdot \cancel{g}}{25.9 \cdot \cancel{g} \cdot m o {l}^{-} 1}}{1.5 \cdot L} = 1.3 \cdot m o l \cdot {L}^{-} 1$, with respect to lithium fluoride
Now it is a fact that this solution would have a $p H > 7$. Given that $H F \left(a q\right)$ is a fairly weak acid, why should this be so?