A 25.0-mL sample of 0.10 M C₂H₃NH₂ (ethylamine) is titrated with 0.15 M HCl. What is the pH of the solution after 9.00 mL of acid have been added to the amine? K_b = 6.5 *10¯⁴ A) 11.08 B) 10.88 C) 10.74 D) 10.55 E) 10.49

Question

Quizplus · Accepted Answer

First, write the balanced chemical equation for the reaction between ethylamine and HCl:
C2H5NH2 + HCl → C2H5NH3+Cl-

The initial concentration of ethylamine is 0.10 M, and the volume used is 25.0 mL, so the initial moles of ethylamine are:
(0.10 M) x (0.0250 L) = 0.00250 mol

When 9.00 mL of 0.15 M HCl is added, the number of moles of HCl added is:
(0.15 M) x (0.00900 L) = 0.00135 mol

Using stoichiometry, we can see that one mole of HCl reacts with one mole of ethylamine, so the remaining moles of ethylamine are:
0.00250 mol - 0.00135 mol = 0.00115 mol

The final volume is 25.0 mL + 9.00 mL = 34.0 mL, so the final concentration of ethylamine is:
0.00115 mol / 0.0340 L = 0.0338 M

Now we can use the equation for the ionization constant of ethylamine:
Kb = ([C2H5NH3+][OH-]) / [C2H5NH2]

At equilibrium, the concentration of [C2H5NH3+] is equal to the concentration of [Cl-] (from the reaction between ethylamine and HCl), which is:
0.00135 mol / 0.0340 L = 0.0397 M

Next, we can use the fact that Kb x Ka = Kw (the ion product constant of water) to solve for Kb:
Kb x 6.5 x 10^-11 = 1.0 x 10^-14
Kb = 1.54 x 10^-4

Now we can set up an ICE table to solve for the concentration of [OH-] at equilibrium:

[C2H5NH2] + [H2O] ⇌ [C2H5NH3+] + [OH-]
I: 0.0338 M   0 M         0 M        0 M
C: -x          +x           +x         +x
E: 0.0338-x  x            0.0397+x  0.0397+x

Using Kb, we can set up the equation:
Kb = ([C2H5NH3+][OH-]) / [C2H5NH2]
1.54 x 10^-4 = (0.0397+x)(x) / (0.0338-x)
Solving for x using the quadratic equation gives x = 4.44 x 10^-6 M.

Now we can use the fact that pH + pOH = 14 to find the pH of the solution:
pOH = -log[OH-] = -log(4.44 x 10^-6) = 5.35
pH = 14 - pOH = 14 - 5.35 = 8.65

However, we need to remember that we started with a solution containing ethylamine, which is a weak base. Therefore, the pH of the solution will be higher than what we calculated, since the ethylamine will also partially dissociate to release OH-. We can estimate the pH using the Henderson-Hasselbalch equation:
pH = pKa + log([A-] / [HA])

The pKa of ethylamine can be calculated using pKa + pKb = 14:
pKa = 14 - pKb = 14 - 3.81 = 10.19

The concentration of [A-] (the ethylamine conjugate base) is equal to the concentration of [Cl-], which is 0.00135 mol / 0.0340 L = 0.0397 M.

The concentration of [HA] (the undissociated ethylamine) is equal to the concentration of ethylamine remaining after the reaction with HCl, which is 0.00115 mol / 0.0340 L = 0.0338 M.

Plugging in these values, we get:
pH = 10.19 + log(0.0397 / 0.0338) = 10.88

Therefore, the pH of the solution after 9.00 mL of HCl have been added is 10.88, which is answer choice B.

A 250-ML Sample of 0