# What is pH for a solution prepared from a 5.7*g mass of sodium hydroxide that is dissolved in a 100*mL volume of water?

Sep 30, 2016

$p H$ is exceptionally high; $p O H$ is negative.

#### Explanation:

$\left[N a O H\right]$ $=$ $\frac{5.7 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1}$ $=$ $0.143 \cdot m o l$. And thus $\left[H {O}^{-}\right]$ $=$ $\frac{0.143 \cdot m o l}{0.100 \cdot L}$ $=$ $1.43 \cdot m o l \cdot {L}^{-} 1$.

Since sodium hydroxide gives quantitative hydroxide ion in aqueous solution, $\left[H {O}^{-}\right] = 1.43 \cdot m o l \cdot {L}^{-} 1$.

Now $p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$ $=$ $- {\log}_{10} \left(1.43\right)$ $=$ $- 0.154$; this is an exceptionally high concentration of hydroxide ion.

Now $p H + p O H = 14$ (by definition; see here for the derivation, thus $p H$ $=$ $14 - p O H$ $=$ $14.15$