Answer to Question #88656 in General Chemistry for Elena

Question

In 200cm3 of water are added 10 cm3 of KOH and the solvate formed shows pH = 12.2. How much was the initial concentration of KOH?

Answer 1

$c(KOH)=c(OH-)=10^{pH-14}=10^{-1.8}=0.01585 mol/L$

$n(KOH)=c*V= 0.01585 mol/L*0.21L =0.0033mol$

$c(KOH)= n/V = 0.0033mol / 0.01L=0.33mol/L$