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)

If 0.561 g of KOH is dissolved in water to give 200 mL of solution at

1.	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)
Answer» Solution :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