Answer:
Step-by-step explanation:
[ H₃O⁺¹] = 4.8 x 10⁻¹¹ mol / l
H₃O⁺¹ = H₂O + H⁺
[ H₃O⁺¹] = [H⁺]
pH = -log[H⁺]
= -log[ 4.8 x 10⁻¹¹]
= -log 4.8 + 11
= - 0 .68 + 11
= 10.32
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