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= 0,4 × (1000 ÷ 200)
= 2
[H^+] = a × M
= 2 × 2
= 4
b. [H^+] = α × M
= (1/100) × 0,01
= 10^-4
pH = 4
c. M = (gr ÷ Mr) × (1000 ÷ mL)
= (4 ÷ 40) × (1000 ÷ 200)
= 0,5
[OH^-] = b × M
= 1 × 5 × 10^-1
= 5 × 10-1
pOH = 1 - log 5 ⇒ pH = 13 + log 5
d. [OH^-] = √Kb × M
= √ 10-5 × 10-1
= 10^-3
pOH = 3 ⇒ pH = 11