Question

0.001mol/L的硫酸的PH

Answer 1

第二步为弱酸解离,

HSO4- H+ + SO42-

[HSO4-]=0.001-X

[SO42-]=X

[H+]=0.001+X

K=1.2x10^-2=[SO42-][H+]/[HSO4-]=X(0.001+X)/(0.001-X)

X=8.7x10^-4 mol/L.

[H+]=0.00187 mol/L

pH=2.73