更新1:
A 148-mL strontium hydroxide solution with a pH of 8.31 is diluted by adding 902 mL of water. Determine the pOH of the diluted solution
Chem 20 help please!?
回答 (1)
2019-03-06 1:34 pm
Sr(OH)₂ is a strong base which completely dissociates in aqueous solution to give OH⁻ ions.
In the original solution:
pOH = pKw - pH = 14.00 - 8.31 = 5.69 M
[OH⁻] = 10⁻⁵˙⁶⁹ M
Dilution:
C₁V₁ = C₂V₂
(10⁻⁵˙⁶⁹ M) × (148 mL) = C₂ × {(148 + 902) mL}
[OH⁻] in the diluted solution = 10⁻⁵˙⁶⁹ × (148/1050) M
pOH in the diluted solution
= -log{10⁻⁵˙⁶⁹ × (148/1050)}
= 5.69 - log(148/1050)
= 6.54
In the original solution:
pOH = pKw - pH = 14.00 - 8.31 = 5.69 M
[OH⁻] = 10⁻⁵˙⁶⁹ M
Dilution:
C₁V₁ = C₂V₂
(10⁻⁵˙⁶⁹ M) × (148 mL) = C₂ × {(148 + 902) mL}
[OH⁻] in the diluted solution = 10⁻⁵˙⁶⁹ × (148/1050) M
pOH in the diluted solution
= -log{10⁻⁵˙⁶⁹ × (148/1050)}
= 5.69 - log(148/1050)
= 6.54
收錄日期: 2021-05-01 22:39:48
原文連結 [永久失效]:
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