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the ph of a 0.50 m solution of base b is found to be 10.19. what is the kb of the base? the equation described by the kb value is b(aq) h2o(l)↽−−⇀bh (aq) oh−(aq)

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The pH of a 0.50 M solution of base B is found to be 10.19. what is the Kb of the base? the equation described by the Kb value is B(aq) + H2O(l) ⇄ BH(aq) + OH⁻(aq).

The pH of a 0.50 M solution of base B is found to be 10.19. The Kb of the base is 4.8 × 10⁻⁸.

Let's consider the following basic reaction.

B(aq) + H2O(l) ⇄ BH(aq) + OH⁻(aq)

First, we will calculate the pOH of the solution using the following expression.

[tex]pH + pOH = 14\\\\pOH = 14 - pH = 14 - 10.19 = 3.81[/tex]

Then, we will calculate the concentration of OH⁻ using the definition of pOH.

[tex]pOH = -log [OH^{-} ]\\\\\[[OH^{-} ] = antilog -pOH = antilog -3.81 = 1.55 \times 10^{-4} M[/tex]

Given [OH⁻] = 1.55 × 10⁻⁴ M and concentration of the base Cb = 0.50 M, for a weak base we can calculate Kb using the following expression.

[tex]Kb = \frac{[OH^{-} ]^{2} }{Cb} = \frac{(1.55 \times 10^{-4})^{2} }{0.50} = 4.8 \times 10^{-8}[/tex]

The pH of a 0.50 M solution of base B is found to be 10.19. The Kb of the base is 4.8 × 10⁻⁸.

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