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pH= -log [H+]= -log [10^-2]= 2
b. M=n/v--> n=gr/mr=3,65/35=0,10mol
M= n/v=0,10/100=0,001
[H+]= a.Ma= 1.10^-3=10^-3
pH= -log [H+]= -log [10^-3]= 3
c. [H+]= a.Ma= 2.3×10^-2=6×10^-2
pH= -log [H+]= -log [6×10^-2]= 2-log 6
d. [OH-]= b.Mb= 1.10^-1=10^-1
pOH= -log [OH-]= -log [10^-1]= 1
pH= 14-1= 13
e. [OH-] = b.Mb= 2.5×10^-3= 10^-2
pOH= -log [OH-]= -log [10^-2]= 2
pH= 14-2= 12