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

Molar concentration of the solution `=("Amount of KOH "//"litre of solution")/("Molar mass of KOH")`
`=(0.561)/(200) xx (1000)/(56) = 0.05 M`
Since KOH is a strong electrolyte, it gets completely dissociated in aqueous solution :
`KOH (s) overset((aq))(to) K^(+) underset(0.05 M)((aq)) + OH^(-)underset(0.05 M)((aq))`
Concentration of `K^(+) , [K^(+)] =0.05 M`
Concentration of `OH^(-) , [OH^(-)] =0.05 M`
Concentration of `H^(+) , [H_(3)O^(+)] = (K_(w))/[[OH^(-)]] = ((10^(-14) M^(2)))/((5xx 10^(-2)M)) =2 xx 10^(-13)M`
pH of the solution may be calculated as follows :
`pH =- log [H_(3)O^(+)] =- log (2xx 10^(-13))`
`=13 log 10 - log 2 = 3 - (0.3010) = 12.69`