Lalu tentukan molaritas M= n/V M= 0,01 mol/0,5L M= 0,02
--> = 0,04 =
pOH = - log [OH] = - log =2 - log 4
pH = 14 - pOH pH = 12+ log 4
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nayohansandy54
Mencari konsentrasi (M) mol(n)=gram/Mr =Mr:1 x Ca + 2 x O + 2 x H =1 x 40 + 2 x 16 + 2 x 1 =74 mol=0,74/74 =0,01 M=n/V =0,01/0,5 =0,02 = 2 x 10^ -2M [OH-]= [basa] x jumlah OH = [2 x 10^ -2] x 2 =4 x 10^ -2 pOH= -log OH- = -log 4 x 10^ -2 =2 -log 4 pH=14-( 2 -log 4) =12 + log 4
n= gr/mr
n= 0,74/74
n= 0,01 mol
Lalu tentukan molaritas
M= n/V
M= 0,01 mol/0,5L
M= 0,02
-->
= 0,04 =
pOH = - log [OH]
= - log
=2 - log 4
pH = 14 - pOH
pH = 12+ log 4
mol(n)=gram/Mr
=Mr:1 x Ca + 2 x O + 2 x H
=1 x 40 + 2 x 16 + 2 x 1
=74
mol=0,74/74
=0,01
M=n/V
=0,01/0,5
=0,02 = 2 x 10^ -2M
[OH-]= [basa] x jumlah OH
= [2 x 10^ -2] x 2
=4 x 10^ -2
pOH= -log OH-
= -log 4 x 10^ -2
=2 -log 4
pH=14-( 2 -log 4)
=12 + log 4