Question

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

Pt(s)H2(1.00 atm) |RNH2 (0.100 M), RNH (0.0500 M) || SCE

Answer 1

Complete Question

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Answer:

The value is
`K_b  =   1.89 *10^(-6)`

Step-by-step 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  =  (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) =  (0.0592)/(n) *  log K_a`

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

So

`0.490 =  (0.0592)/(1)  *  log K_a`

=>
`K_a  =  5.30*10^(-9)`

So

`K_b  =  ( K_w)/( K_a)`

=>
`K_b  =  (1.0 *10^(-14))/( 5.30*10^(-9))`

=>
`K_b  =   1.89 *10^(-6)`