Complete Question
The complete question is shown on the first uploaded image
Answer:
The value is [tex]K_b = 1.89 *10^{-6}[/tex]
Explanation:
From the question we are told that
The voltage of the cell is [tex]V = 0.731 \ V[/tex]
Generally [tex]K_b[/tex] is mathematically represented as
[tex]K_b = \frac{K_w }{ K_a }[/tex]
Where [tex]K_w[/tex] is the equilibrium constant for this auto-ionization of water with a value [tex]K_w = 1.0 *10^{-14}[/tex]
Generally the [tex]E_{cell}[/tex] is mathematically represented as
[tex]E_{cell} = V - E_{SCE}[/tex]
=> [tex]E_{cell} = 0.731 - 0.241 [/tex]
=> [tex]E_{cell} = 0.49 V [/tex]
This [tex]E_{cell}[/tex] is mathematically represented as
[tex]E_{cell} = \frac{0.0592}{n} * log K_a [/tex]
Where n is the number of moles which in this question is n = 1
So
[tex]0.490 = \frac{0.0592}{1} * log K_a[/tex]
=> [tex]K_a = 5.30*10^{-9}[/tex]
So
[tex]K_b = \frac{ K_w}{ K_a}[/tex]
=> [tex]K_b = \frac{1.0 *10^{-14}}{ 5.30*10^{-9}}[/tex]
=> [tex]K_b = 1.89 *10^{-6}[/tex]
