hitunglah PH larutan berikut ini : a. 0,005 mol KOH dalam 200mL larutan b. 0,01 mol CHCOOH dalam 100mL larutan (Kb-10^-5) c. 0,05 mol NH3 dalam 250 mL larutan (

a) n= 0.005 , v = 200 ml = 0.2L
M=n/v = 0.005/0.2=0.025M
[H+] = M.a =25×10^-3x1= 25x 10^-3
pH=3-log25
b) n=0.01mol v=100ml=0.1L
M=0.01/0.1=0.1
[H+]= akar ka.M
=akar 10^-5.10^-1 = 10^-3
pH = 3
c) n= 0.05mol v =250 mL =0.25L
M=0.05/0.25 =0.2
[OH-] = akar 0.2 x 10^-8 = 10^-4 akar 0.2
p[OH-] = - log (10^-4.0.2^1/2)
=4-1/2log 0.2
p[H] = 14-4+1/2log0.2 =10+1/2log0.2