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`?

Answer

`[K^(+)] = [OH^(-)] = 0.05M, [H^(+)] = 2.0 xx 10^(-13) M`

Explanation

`[KOH_(aq)] = (0.561)/((1)/(5)) g//L`
`= 2.805 g//L`
`= 2.805 xx (1)/(56.11) M`
`= .05 M`
`KOH_((aq)) rarr K_((aq))^(+) + OH_((aq))^(-)`
`[OH^(-)] = .05 M = [K^(+)]`
`[H^(+)] [H^(-)] = K_(w)`
`[H^(+)] = (K_(w))/([OH^(-)])`
`= (10^(-14))/(0.05) = 2 xx 10^(-13) M`
`:. pH = 12.70`