Question

Help please acids and basis

Answer 1

Let's see

• [OH-] be x

`\\ \rm\Rrightarrow pOH=-log[OH-]`

`\\ \rm\Rrightarrow pOH=-logx`

`\\ \rm\Rrightarrow 4.86=-logx`

`\\ \rm\Rrightarrow logx=-4.86`

`\\ \rm\Rrightarrow x=10^(-4.86)`

`\\ \rm\Rrightarrow x=1.38M`