Question

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 1

`KOH rarr K^(+)+OH^(-)`
`[KOH]=(0.561xx1000)/(56xx200)`
`= 5.01xx10^(-2)M`
`:. [OH^(-)]=5.01xx10^(-2)M`
or `pOH= -log[OH^(-)]`
`= -log 5.01xx10^(-2)`
`= 1.3002`
`:. pH=12.6998`
`[H^(+)]=1.996xx10^(-13)`,
`[K^(+)]=5.01xx10^(-2)`