Question #236975
A 400ml sample of 0.4375 M of ethyl amine (Kb=4.5m×10^-4) reacted with2.50L of HCN ( Ka=6.2×10^-10) at 298 k and 1.2 atm. Assuming the volume of the solutions remain constant. Calculate concentrations of all species present and their PH at equilibrium.
1
Expert's answer
2021-09-15T02:41:09-0400

Molar mass of ethyl amine = 45.08 g/mol

= 0.4375 × 45.08 = 19.7225

400/19.7225 = 20.28 mol

20.28/2.5 = 8.112molL-1



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