What is the pH of an aqueous solution prepared from a 600*mg mass of NaOH dissolved in 75*mL of water?

Oct 27, 2016

$p H \cong 13$

Explanation:

We need to find $\left[H {O}^{-}\right]$ $=$ $\text{Moles of hydroxide"/"Volume of water}$ $=$ $\frac{0.600 \cdot g}{40.00 \cdot g \cdot m o {l}^{-} 1} \times \frac{1}{75.0 \times {10}^{-} 3 L}$ $=$ $0.20 \cdot m o l \cdot {L}^{-} 1$.

Now $p O H$ $=$ $- {\log}_{10} \left[H {O}^{-}\right]$ $=$ $- {\log}_{10} \left(0.20\right)$ $=$ $- \left(- 0.70\right)$ $=$ $0.70$.

But we know (or should know) that in water at $298 \cdot K$, $p H + p O H = 14$.

Thus pH=14-0.70=?????????