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The voltage for the following cell is +0.731 V. Find Kb for the organic base RNH2. Use E_SCE = 0.241V.
Pt(s)│H2 (1.00 atm)│RNH2 (0.100M), RNH3 + (0.0500M)║SCE

Respuesta :

nuhulawal20

Answer:  Kb = 1.89 × 10⁻⁶

Explanation:

R-NH₃+ (aq) <----------> R-NH₂ (aq) + H^+ (aq)

Ecell = - ( 0.0592 / n ) log Kₐ

Where, Ecell = 0.731 - 0.241 = 0.490 V

Therefore, 0.490 = - (0.0592 / 1 ) log Kₐ

Therefore, Kₐ = 5.30 × 10⁻⁹

Thus, Kb = Kw / Ka = ( 1.0 × 10⁻¹⁴ / 5.30 × 10⁹ ) = 1.89 × 10⁻⁶