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= 1 + 3 : 40
= 0,1
= - log 10
m 0,001 0,003
b 0,001 0,001 0,001 0,001
s - 0,002 0,001 0,001
M NaOH = 0,002 ÷ 0,04 = 0,05
OH^- = b × Mb = 1 × 5 × 10^-2 = 5 × 10^-2
pOH = 2 - log 5 ⇒ pH = 12 +l log 5