Calculate the pOH of an aqueous solution
of 8 mL of 0.00048 M HClO4(aq) after dilution
to 51 mL.
CHEM HELP ASAP PLEASE!?!?
2020-06-01 8:58 pm
回答 (1)
2020-06-01 9:43 pm
Original: C₁ = 0.00048 M, V₁ = 8 mL
New: C₂ = ? M, V₂ = 51 mL
For Dilution: C₁V₁ = C₂V₂
Then, C₂ = C₁ × (V₁/V₂)
[HClO₄] after dilution, C₂ = 0.00048 × (8/51) M = 7.53 × 10⁻⁵ M
HClO₄ is a strong acid which completely ionized in water to give H₃O⁺ ions.
Hence, [H₃O⁺] = 7.53 × 10⁻⁵ M
pH = -log[H₃O⁺] = -log(7.53 × 10⁻⁵) = 4.12
pOH = pKw - pH = 14.00 - 4.12 = 9.88
New: C₂ = ? M, V₂ = 51 mL
For Dilution: C₁V₁ = C₂V₂
Then, C₂ = C₁ × (V₁/V₂)
[HClO₄] after dilution, C₂ = 0.00048 × (8/51) M = 7.53 × 10⁻⁵ M
HClO₄ is a strong acid which completely ionized in water to give H₃O⁺ ions.
Hence, [H₃O⁺] = 7.53 × 10⁻⁵ M
pH = -log[H₃O⁺] = -log(7.53 × 10⁻⁵) = 4.12
pOH = pKw - pH = 14.00 - 4.12 = 9.88
收錄日期: 2021-04-18 18:32:00
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20200601125813AA0m0Yn