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.
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.
Thanks! This was very helpful.
pH + pOH = pKw = 14 (and not pK2w)
(Kw has no K2).
a. To write the equilibrium dissociation equation, we need to understand that ethylamine (C2H5NH2) acts as a weak base in water. In water, ethylamine accepts a proton (H+) to form the conjugate acid, ethylammonium ion (C2H5NH3+). The equation can be written as follows:
C2H5NH2 (aq) + H2O (l) ⇌ C2H5NH3+ (aq) + OH- (aq)
b. The pH of a solution can be calculated using the concentration of hydroxide ions (OH-) in the solution. Since we know from the problem that the hydroxide ion concentration is 6.6x10^-3 M, we can use the equation:
pOH = -log[OH-]
pOH = -log(6.6x10^-3)
pOH ≈ 2.18
Since pH + pOH = 14, we can find the pH by subtracting the pOH value from 14:
pH = 14 - 2.18
pH ≈ 11.82
Therefore, the pH of the 0.075 M C2H5NH2(aq) solution is approximately 11.82.
c. The value of Kb (base dissociation constant) can be determined by using the concentration of products and reactants at equilibrium. We can use the equation:
Kb = [C2H5NH3+][OH-] / [C2H5NH2]
Since the problem states that the hydroxide ion concentration is 6.6x10^-3 M and the concentration of ethylamine is 0.075 M, we can substitute these values into the equation:
Kb = (6.6x10^-3)(6.6x10^-3) / 0.075
Kb ≈ 5.808x10^-4
Therefore, the value of Kb for the weak base ethylamine (C2H5NH2) is approximately 5.808x10^-4.