A titration of 25.0 mL of a solution of the weak base aniline, C6H5NH2, requires 25.67...

Question

Question

A titration of 25.0 mL of a solution of the weak base aniline, C₆H₅NH₂, requires 25.67 mL of 0.175 M HCl to reach the equivalence point.

C₆H₅NH₂(aq) + HCl(aq)--> C₆H₅NH₃⁺ (aq) + Cl- (aq)

The K_b for aniline, C₆H₅NH₂, is 4.0 x10^-10

a. What was the concentration of aniline in the original solution?

b. What is the concentration of C₆H₅NH₃⁺at the equivalence point (AND after equilibrium is established)?

c. What is the pH of the solution at the equivalence point?

science chemistry

Answer 1

Answer #1

Answer 2

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