oblicz stezenie jonow ;[H+] i [OH-] w roztworze o ph =8
pH = -log[H+]
mol/dm³
pH+pOH=14 --> pOH = 14-pH = 14-8=6
pOH = -log[OH-]
pH = 8 czyli [H+] = 1*10^-8 mol/dm3
pH + pOH = 14 ------> 8 + pOH = 14 ------------> pOH = 6
Skoro pOH = 6 to [OH-] = 1*10^-6 mol/dm3
Odp. Steżenie jonów H+ wynosi 1*10^-8 mol/dm3 a stężenie jonów OH- 1*10^-6 mol/dm3
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pH = -log[H+]
pH+pOH=14 --> pOH = 14-pH = 14-8=6
pOH = -log[OH-]
pH = 8 czyli [H+] = 1*10^-8 mol/dm3
pH + pOH = 14 ------> 8 + pOH = 14 ------------> pOH = 6
Skoro pOH = 6 to [OH-] = 1*10^-6 mol/dm3
Odp. Steżenie jonów H+ wynosi 1*10^-8 mol/dm3 a stężenie jonów OH- 1*10^-6 mol/dm3