0.561g KOH is dissolved in water to give 200 mL of solution at 298K. Calculate the concentrations of `K^(+),H^(+) and OH^(-)` ions. What is its pH?

Explanation

0.561g `KOH=(0.561)/(56)=0.01` mol of KOH
The molar concentration of KOH in 200 mL of solution
`=(0.01xx1000)/(200)=0.05(M)`
KOH, being a strong electrolyte, dissociate completely in solution: `KOH(aq)to K^(+)(aq)+OH^(-)(aq)`
So, in 0.05(M) solution KOH, `[K^(+)]=[OH^(-)]=0.05(M)`
Now, `[H^(+)][OH^(-)]=10^(-14)`
So, in 0.05 (M) solution of KOH,
`[H^(+)]=(10^(-14))/(0.05)(M)=2xx10^(-13)(M)`
For this solution, `pH=-log_(10)[H^(+)]`
`=log_(10)(2xx10^(-13))=12.69`.