If 0.561 g of KOH is dissolved in water to give 200 mL of solution at 298 K. Calculate the concentrations of potassium, hydrogen and hydroxyl ions. What is its pH ? (K=39,O=16,H=1)

Explanation

Molarity of KOH=`"Weight (in gm) x 1000"/"Molecular mass x Volume of solution mL"`
`M=(0.561 g xx 1000)/(56 "g mol"^(-1) xx "200 mL")`
=0.05 M = `5.0xx10^(-2)` M
KOH=39+16+1
=56 g `mol^(-1)`
[KOH]=`[K^+]=[OH^-]=5.0xx10^(-2)` M
`[H^+][OH^-]=1xx10^(-14) = K_w`
`therefore [H^+] = (1xx10^(-14))/[[OH^-]]=(1xx10^(-14))/(5.0xx10^(-2))=2.0xx10^(-13)` M
pH=-(log` [H^+]` =-log `(2.0xx10^(-13))`
`=-(log 2+ log 10^(-13))`
=-(0.3010-13.0)=13.0-0.3010
=12.6990 `approx` 12.7