Question

Question #db69c

Answer 1

`[OH^-]=0.32"mol/l"`

Explanation:

`pK_w=pOH+pH`

`:.14=pOH+13.5`

`:.pOH=14-13.5=0.5`

`:.-log[OH^-]=0.5`

`:.log[OH^-]=-0.5`

`10^(-0.5)=0.32`

`:.[OH^-]=0.32"mol/l"`