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sasho [114]
3 years ago
8

2.3*10^14 and 5.4*10^-3

Chemistry
1 answer:
Anvisha [2.4K]3 years ago
4 0

The first one would be 2.3e+14

and the second would be 0.0054

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

The complete question is shown on the first uploaded image  

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

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=>     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}

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