należy obliczyć stężenie molowe jonów OH-
m OH- = 0,008 g
M OH- = 17g/mol
n OH- = 0,008/17 = 4,706 *10-4 mola
V roztworu = 1 dm3 czyli
c mol OH- = 4,706*10-4 mol/dm3
pOH = - log (OH-)= - log (4,706*10-4)= 3,33
pH + pOH = 14
pH = 14 - 3,33 = 10,67
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należy obliczyć stężenie molowe jonów OH-
m OH- = 0,008 g
M OH- = 17g/mol
n OH- = 0,008/17 = 4,706 *10-4 mola
V roztworu = 1 dm3 czyli
c mol OH- = 4,706*10-4 mol/dm3
pOH = - log (OH-)= - log (4,706*10-4)= 3,33
pH + pOH = 14
pH = 14 - 3,33 = 10,67