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Sagot :
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. (14-10.19)=3.81
antilog -3.81=1.55×10^-4/0.50=4.8×10^-8
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.
Learn more about Kb for this base here brainly.com/question/6465134
#4262
The [tex]K_b[/tex] of the base b is 4.805 × 10⁻⁸.
What is dissociation constant?
The dissociation constant is a type of equilibrium constant that measures the degree of dissociation of a salt.
[tex]K_a[/tex] is the dissociation constant for the acid which describes the strength of an acid.
[tex]K_b[/tex] is the dissociation constant of the base which describes the strength of a base.
pOH = 14 - pH = 14 - 10.19 = 3.81
pOH = -log[OH⁻]
3.81 = -log[OH⁻]
Antilog (-3.81) = [OH⁻]
[OH⁻] = 1.55 × 10⁻⁴
Here,
[tex]K_b= \frac{[BH^+][OH^-]}{[B]}[/tex]
= [tex]\frac{(1.55\times10^{-4})(1.55\times10^{-4})}{0.50}[/tex]
= 4.805 × 10⁻⁸
The Kb of the base b is 4.805 × 10⁻⁸
Learn more about dissociation constant:
https://brainly.com/question/9560811
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