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 to K^(+)+OH^(-)`
`[KOH]=(0.561xx1000)/(56 xx 2000)=5.01 xx 10^(-2)M`
`therefore [OH^(-)]=5.01 xx 10^(-2)M`
`or POH=-log [OH^(-)]=-log 5.01 xx 10^(-2)`
`=1.3002 therefore pH =12.6998`
`[H^(+)]=1.996 xx 10^(-13), [K^(+)]=5.01 xx 10^(-2)`