Question

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

Answer 1

In water `pH+pOH=14`

`pH=14-pOH=14-5=9`

Explanation:

`pOH=-log_10[HO^-]` by definition.

Thus `pOH=-log_10(1xx10^-5)=-(-5)=5`

`pH+pOH=14`

`pH=14-pOH=14-5=9`

When we write `log_ab=c`, we ask to what power we raise the base, `a`, to get `b`; here `a^c=b`. The normal bases are `10`, `"(common logarithms)"`, and `e`, `"(natural logarithms)"`.

Thus when we write `log_10(10^-5)`, we are asking to what power we raise `10` to get `10^-5`. Now clearly the answer is `-5`, i.e. `log_10(10^-5)=-5`, alternatively `log_10(10^5)=+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.