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A 0.231 M solution of acetate has a pOH of 4.90. What is the Kb of acetate?

User Idej
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Step-by-step explanation:

o solve this problem, we need to use the relation between pOH, pH, and the dissociation constant of the conjugate base of the weak acid, which is given by:

Kb = Kw / Ka

where Kb is the dissociation constant of the conjugate base, Ka is the dissociation constant of the weak acid, and Kw is the ion product constant of water, which is 1.0 x 10^-14 at 25°C.

First, we need to find the pH of the solution, since we know the pOH:

pH + pOH = 14

pH = 14 - 4.90 = 9.10

The weak acid in this case is the acetic acid (CH3COOH), which dissociates in water according to the equation:

CH3COOH + H2O ↔ CH3COO- + H3O+

The dissociation constant of acetic acid (Ka) is 1.8 x 10^-5 at 25°C. We can use this value and the relation between Ka and Kb to find Kb:

Kb = Kw / Ka = 1.0 x 10^-14 / 1.8 x 10^-5 = 5.56 x 10^-10

Therefore, the Kb of acetate is 5.56 x 10^-10.

User Chetan Bhasin
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