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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