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]=0.561/56xx1000/200 M=0.050 M`
As `KOH rarr K^(o+)+overset(Θ)(O)H, :.[K^(o+)]=[overset(Θ)(O)H]=0.05 M`
`[H^(o+)]=K_(w)//[overset(Θ)(O)H]=10^(-14)//0.05=10^(-14)//(5xx10^(-2))`
`=2.0xx10^(-13) M, pH=12.7`