The pOH of an aqueous solution of 0.480 M trimethylamine (a weak base with the formula (CH3)3N) is .
1 answer:
Answer:
Explanation:
Kb of (CH₃)₃N is 7.4 x 10⁻⁵
initial concentration of (CH₃)₃N a is .48 M
(CH₃)₃N + H₂O = (CH₃)₃NH⁺ + OH⁻
a - x x x
x² / (a - x ) = Kb
x is far less than a so a - x can be replaced by a .
x² / a = Kb
x² = a x Kb = .48 x 7.4 x 10⁻⁵ = 3.55 x 10⁻⁵ = 35.5 x 10⁻⁶
x = 5.96 x 10⁻³
pOH = - log ( 5.96 x 10⁻³ )
= 3 - log 5.96
= 3 - .775
= 2.225
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