✔ 最佳答案
計算的方法:
(1) pH 與 [H^+] 的關係:
pH = -log[H^+] .... 即[H^+] = 10^-pH
(2) pOH 與 [OH^-] 的關係:
pOH = -log[OH^-] .... 即[OH^-] = 10^-OH
(3) [H^+] 與 [OH^-] 的關係:
[H^+] x [OH^-] = 1 x10^-14
(4) pH 與 pOH 的關係:
pH + pOH = 14
1. pH=10的鹼液,求〔H+〕〔OH-〕=?
pH = 10
[H^+] = 10^-pH = 10^-10 =1 x 10^-10 M
pOH = 14 - pH = 14 - 10 = 4
[OH^-] = 10^-pOH = 10^-4 = 1 x 10^-4 M
2. 在25度C時人體血液之pH為7.41,試著計算其pOH〔H+〕〔OH-〕各為多少?
pOH = 14 - pH = 14 - 7.41 = 6.59
[H^+] = 10^-pH = 10^-7.41 = 3.89 x 10^-8 M
[OH^-] = 10^-pOH = 10^-6.59 = 2.57 x 10^-7 M