Question

oblicz ph roztworu NH4OH o stezeniu 0,05 mola/dm3 i alfa = 0,01

Answer 1

NH4OH ---> NH4(+) + OH(-)

α = 0,01

Cm=0,05mol/dm3

[OH-] = Cm*α

[OH-] = 0,05*0,01

[OH-] = 0,0005

pOH = -log[OH-]

pOH = 3,3

pH =14 - pOH

pH = 14-3,3

pH = 10,7

Answer 2

[OH-]=C*alfa= 0,05*0.01=0,0005 (5*10 do potęgi -4)

alfa=1%=0.01

pOH = 4

pH=10