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]=`0.561/56 xx1000/200` =0.050 M
KOH=`K^(+)+OH^-`
`[K^+]` =0.05 M , `[OH^-]` =0.05 M
`[H^+]=K_w/([OH^-])=(1xx10^(-14))/0.05 =2.0xx10^(-13)`
pH=-log `(2.0xx10^(-13))` =12.699