H3BO3+H2O=H†+[B(OH)4]–
pKa=9.24; Ka=5,8*10^-10 M
Because Kacid<Kawater
[H+]=√(Ka*CHA+Kw)
[H+]=√(5.8*10^-10*1,5M+10^-14)
[H+]=2,94*10^-5
pH=-lg[H+]
pH=4,53
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