Question

The voltage for the following cell is +0.731 V. Find Kb for the organic base RNH2. Use EscE 0.241V.

Answer 1

Complete Question

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Answer:

The  value is   $K_b = 1.89 *10^{-6}$

Explanation:

From the question we are told that

   The  voltage of the cell is  $V = 0.731 \ V$

Generally $K_b$ is mathematically represented as  

           $K_b = \frac{K_w }{ K_a }$

Where  $K_w$  is the equilibrium constant for this auto-ionization of water with a value  $K_w = 1.0 *10^{-14}$

Generally the $E_{cell}$ is mathematically represented as

       $E_{cell} = V - E_{SCE}$

=>     $E_{cell} = 0.731 - 0.241$

=>       $E_{cell} = 0.49 V$

This  $E_{cell}$ is mathematically represented as

             $E_{cell} = \frac{0.0592}{n} * log K_a$

Where n is the number of moles which in this question is  n = 1

         So  

         $0.490 = \frac{0.0592}{1} * log K_a$

=>      $K_a = 5.30*10^{-9}$

So  

     $K_b = \frac{ K_w}{ K_a}$

=>   $K_b = \frac{1.0 *10^{-14}}{ 5.30*10^{-9}}$

=>    $K_b = 1.89 *10^{-6}$