H2O =电离= H+ + OH-
设1mol H2O中,发生电离的水分子为x mol
x * 100% / 1 = 1.8×10^(-13)%
解出x = 1.8 * 10^-15
假如是酸溶液:
c(OH-) = 1.8 * 10^-15
pOH = -lg(c(OH-)) = 14
pH = 14 - 14 = 0
假如是碱溶液:
c(H+) = 1.8 * 10^-15
pH = -lg(c(OH-)) = 14
所以,溶液的pH值可能是0或14
H2O =电离= H+ + OH-
设1mol H2O中,发生电离的水分子为x mol
x * 100% / 1 = 1.8×10^(-13)%
解出x = 1.8 * 10^-15
假如是酸溶液:
c(OH-) = 1.8 * 10^-15
pOH = -lg(c(OH-)) = 14
pH = 14 - 14 = 0
假如是碱溶液:
c(H+) = 1.8 * 10^-15
pH = -lg(c(OH-)) = 14
所以,溶液的pH值可能是0或14