Question

100ml蒸馏水需要加多少ml浓度为1.5mol/L的氨水使得溶液PH值为10

Answer 1

Kb = 1.77*10^(-5)

pH=10,[H+] = 10^(-10) mol/L,[OH-] = 10^(-14)/[H+] = 10^(-4) mol/L

[OH-] = (cKb)^(1/2) 开平方

[OH-]^2 = cKb

c = 10^(-8)/(1.77*10^(-5)) = 5.65*10^(-4) mol/L

100ml蒸馏水需要加浓度为1.5mol/L的氨水的体积为x ml：

(100 + x)* 5.65* 10^(-4) = x*1.5

x = 0.038 ml