# What is the pH of a 1.0 x 10^-5 M solution of NaOH?

Dec 2, 2016

In water $p H + p O H = 14$

$p H = 14 - p O H = 14 - 5 = 9$

#### Explanation:

$p O H = - {\log}_{10} \left[H {O}^{-}\right]$ by definition.

Thus $p O H = - {\log}_{10} \left(1 \times {10}^{-} 5\right) = - \left(- 5\right) = 5$

$p H + p O H = 14$

$p H = 14 - p O H = 14 - 5 = 9$

When we write ${\log}_{a} b = c$, we ask to what power we raise the base, $a$, to get $b$; here ${a}^{c} = b$. The normal bases are $10$, $\text{(common logarithms)}$, and $e$, $\text{(natural logarithms)}$.

Thus when we write ${\log}_{10} \left({10}^{-} 5\right)$, we are asking to what power we raise $10$ to get ${10}^{-} 5$. Now clearly the answer is $- 5$, i.e. ${\log}_{10} \left({10}^{-} 5\right) = - 5$, alternatively ${\log}_{10} \left({10}^{5}\right) = + 5$.

What are log_10(100), log_10(1000), log_10(1000000)??

You shouldn't need a calculator, but use one if you don't see it straight off.