If `0.561 g` of `(KOH)` is dissolved in water to give. `200 mL` of solution at `298 K`. Calculate the concentration of potassium, hydrogen and hydroxyl ions. What is its `pH`?

Explanation

`[KOH]=(0.561)/(56)xx(1000)/(200)M=0.050M`
As `KOH rarr K^(+)+OH^(-), :. [K^(+)]=[OH^(-)]=0.05M`
`[H^(+)]=K_(w)//[OH^(-)] = 10^(-14)//0.05 = 10^(-14)//(5xx10^(-2))=2.0xx10^(-13)M`.
`pH = - log [H^(+)]=- log (2.0xx10^(-13))=13-0.3010=12.699`