Question

for a solution of the weak base, 0.075 M ethylamine C2H5NH2(aq), the hydroxide ion concentration (OH-) is 6.6x10^-3 M. Assume that temperature is constant and volumes are additive.

a. Write the equilibrium dissociation equation.

b. What's the pH of 0.075 M C2H5NH2(aq)?

c. What's the value of Kb?


Please show your work on how you calculate it so I can see how you figured it out. Thanks.

Answer 1

C2H5NH2 + HOH ==> C2H5NH3^+ + OH^-

You know OH^-; therefore,
pOH = -log(OH^-) and
pH + POH = pK2w = 14. Solve for pH.

Write the Kb expression
Substitute OH^- = 0.0066
(C2H5NH3^+) = 0.0066
(C2H5NH2) = 0.075-0.0066
Solve for Kb.

Answer 2

Thanks! This was very helpful.

Answer 3

pH + pOH = pKw = 14 (and not pK2w)
(Kw has no K2).