# What is the "normality" of a solution prepared by dissolving a 5*g mass of sodium hydroxide in a 1*L volume of water?

Aug 19, 2017

$\text{Normality} = 0.125 \cdot m o l \cdot {L}^{-} 1$.

#### Explanation:

Here $\text{molarity}$ $\equiv$ $\text{normality}$. Why? Because hydroxide anion does not speciate to any degree in aqueous solution.

And $\text{molarity}$ $\equiv$ $\text{moles of solute"/"volume of solution}$

$= \frac{\frac{5 \cdot g}{40.0 \cdot g \cdot m o {l}^{-} 1}}{1 \cdot L} = \frac{0.125 \cdot m o l}{1 \cdot L} = 0.125 \cdot m o l \cdot {L}^{-} 1$ with respect to $N a O H \left(a q\right)$.

What is $p H$ of this solution? We know that $p H + p O H = 14$ under standard conditions.