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pH=-log[H+] pOH=-log[OH-]
10^(-1,64)=[H+] 10^(-12,36)=[OH-]
[H+]= 1,64 mol/dm^3 [OH-]= 4,365*10(-13)
reszta analogicznie
b) pH= 2,58 ----> pOH=11,42
[H+]= 2,6* 10^(-3) mol/dm^3 [OH-]= 3,8*10^(-12) mol/dm^3
c) pH= 13,59 ----> pOH= 0,41
[H+]= 2,57* 10^(-14) mol/dm^3 [OH-]= 0,39 mol/dm^3
d) pH=11,5 ----> pOH= 2,5
[H+]= 3,16*10^(-12) mol/dm^3 [OH-]= 3,1* 10^(-3) mol/dm^3