Question

Use the rules for logarithms and exponents to solve for [OH-] in terms of pOH.

Answer 1

`\text{ pOH = - log }\lbrack\text{OH}^-\rbrack`
`-\text{pOH = log }\lbrack\text{OH}^-\rbrack`
`\text{ 10}^{-\text{pOH}}\text{ = 10}^{\text{ log}\lbrack\text{OH}^-\rbrack}`

logarithm and power 10 are inverse operations

`\lbrack\text{OH}^-\rbrack\text{ = 10}^{-\text{pOH}}`

The answer is [OH-] = 10^-(pOH)

and, if pOH = 2.77

`\lbrack\text{ OH}^-\rbrack\text{ = 10}^{-2.77\text{ }}\text{ = 0.0017 M}`