Question

Calculate the `[OH^-]` and the `[H^+]` of a `0.0020` `"M"` solution of `"NaOH"`, a strong base?

I don't know if it's right, but I got `[OH^-] = 1.7 * 10^-12` `"M"` and `[H^+] = 0.0060` `"M"`.

Answer 1

`pH and pOH` are complementary. They must sum up to 14. The concentrations are related to the `"p"` values. In this case the respective molar concentrations are:

`[OH^-] = 0.0020( mol)/L`
`[H^+] = 5.01 xx 10^(-12)`mol/L

Explanation:

BOTH are defined as the "negative logarithm of the species (`OH^-` of `H^+`) concentration. We are given the `OH^-` concentration (molarity), so we will use it.

`pH + pOH = 14`

`pOH = -log[OH^-]`
`pOH = -log[0.0020]`
`pOH = 2.70` ; `pH = 14 - 2.70 = 11.3`

A strong base is fully dissociated, so the concentration of the ion is the same as the original compound - 0.0020M, or 0.0020 mol/L

The `[H^+]` is found from the calculated `pH` we obtained.
`pH = -log[H^+]`
`11.3 = -log[H^+]` ; `[H^+] = 5.01 xx 10^(-12)`mol/L