A beaker contains 25.0 mL of 0.80 M propylamine (C_3H_7NH_2; K_B = 5.1 times 10^

Question

A beaker contains 25.0 mL of 0.80 M propylamine (C_3H_7NH_2; K_B = 5.1 times 10^-4). Is propylamine classified as a weak or strong, acid or base? Write the equation showing the equilibrium established by propylamine in water. Solid C_3H_7NH_3Br is added to the 0.80 M C_3H_7NH_2 solution. Write the equation showing the ionization of the C_3H_7NH_3Br. Does this equation proceed 100% to completion? What is the identity of the common ion? How does addition of this common ion to the 0.80 M affect (increases, decreases, remains the same) each of the following? Explain. Edition of solid KBr to the 0.80 M C_3H_7NH_2 affect the pH and [H_3O^+]? Why or why not?

Explanation / Answer

a) Propylamine+water gives hydroxide and congugate acid (shown below)

C3H7NH2+H2O GIVES OH+C3H7NH3...Propylamine is a weak base

b)C3H7NH3Br+C3H7NH2 gives C3H7NH3+C3H7NH2Br

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