A 24.9 mL sample of 0.227 M ethylamine, C2H5NH2, is titrated with 0.321 M perchloric acid.

Question

A 24.9 mL sample of 0.227 M ethylamine, C₂H₅NH₂, is titrated with 0.321 M perchloric acid.

(1) At the titration midpoint, the pH is . 

(2) At the equivalence point, the pH is . 

Answer 1

Describe the initial solution with the info we have:

C2H5NH2 acts as a base in water.

C2H5NH2 + H2O ---> C2H5NH3+ + OH-

From the information you provided, I can also find the total number of moles of C2H5NH2 we start with:

.227 mol C2H5NH2/1L solution x .0249 L solution = .0056523 mol C2H5NH2

Next we need to identify what type of reaction we have. Ethylamine is a weak base (because the amine group) and perchloric acid is a strong acid (because the affinity of the perchlorate ion for an additional electron. So we have a weak base-strong acid reaction.

To answer pH questions you need to know the equilibrium constant of the reaction of ethylamine reacting with water (chemical equation above):

So I looked up the equilibrium constant for this reaction and got Kb = 4.3 X 10^-4.

Let's run an ICE chart to see what the constituents are initially in the solution (remember water does not contribute to the equilibrium constant):

C2H5NH2 C2H5NH3+ OH-

Initial .227 0 0

Change -x +x +x

Equilibrium (.227-x) +x +x

Kb = [C2H5NH3+] * [OH-] / [C2H5NH2]

4.3 *10^-4 = (x)(x)/(.227 - x)

Solving for x using the quadratic equation, I get x = .009667. This means that the molarity of OH- in solution is .009667 mol OH-/L initially, .009667 mol C2H5NH3+/L, and I have .2173 mol C2H5NH2 /L

Your question asks to find the pH at the midpoint and the equivalence point. One has to know what the definitions of these are to proceed.

The midpoint is the point at which the number of moles of strong acid = 1/2 the number of moles of base initially in the solution.

The equivalence point is the point at which the number of moles of strong acid = the number of moles of base initially in the solution.

Since perchloric is a monoprotic acid (meaning it can only donate one H+ ion), the number of moles of strong acid that need to be added are to reach the midpoint is .00282615 moles of HClO4. The number of moles of strong acid that need to be added are to reach the equivalence point is .0056523 moles of HClO4.

Now we need to determine the the volume of .321 M HClO4 needed to make .00282615 moles of HClO4 at the midpoint and .0056523 moles of HClO4 at the equivalence point:

MV = # of moles

V = # of moles/M

At the midpoint:

V = .00282615 moles of HClO4/ .321 M HClO4 = 0.0088 L

At the equivalence point:

V = .0056523 moles of HClO4/ .321 M HClO4 = 0.01761 L

If I add .0088L of .321 M HClO4, I have increase the volume of the solution. Therefore I need to recalculate the molarities of my components:

New Molarity C2H5NH2 = .2173 mol C2H5NH2 /L * .0249 L / (.0249 L + .0088 L) = .1606 M C2H5NH2

New Molarity C2H5NH3+ = .009667 mol C2H5NH3+ /L * .0249 L / (.0249 L + .0088 L)= 0.00714 M C2H5NH3+

New Molarity OH- = .009667 mol OH- /L * .0249 L / (.0249 L + .0088 L) = 0.00714 M OH-

Addition of H+ Molarity = .00282615 mol H+ L/(.0249 L + .0088 L) = 0.000737 M H+

To answer pH questions you need equilibrium constant of the reaction of ethylamine reacting with water (chemical equation above) again, which is Kb = 4.3 x 10^-4:

Now we need to recalculate the molarities of our components. Since to reach the midpoint of the titration I added .0088 L of perchloric acid and I have eliminated half of the moles of ethylamine initially started with my new molarity of ethylamine is:

M = .00282615 moles C2H5NH2 / (.0249 + .0088) = 0.08386 M C2H5NH2

running through the ICE Chart: I have:

C2H5NH2 C2H5NH3+ OH-

Initial .1606 0.00714 0.00714

Change -x +x -x

Equilibrium .0839 .0839 .00714 - x

Again we use the same equilibrium constant equation

Kb = [C2H5NH3+] * [OH-] / [C2H5NH2]

4.3 *10^-4 = (.0839)(.00714 - x)/(.0839)

Notice that equation reduces to:

4.3 *10^-4 = .00714 - x

x = .00671

This says that change in OH- concentration is .00671 M, so to find the pOH, we must find the final molarity of OH-:

M OH- = .00714 - .00671 = 0.00043

pOH = -log[OH-] = 3.37

pH = 14-pOH = 14-3.37 = 10.63

So the pH at the midpoint is 10.63

Because we have a weak base-strong acid titration, at the equivalence point, we will have a ph < 7. We have to convert our Kb value to a Ka value.

Kw = Ka*Kb

Ka = Kw/Kb

Ka = 1*10^-14/4.3 *10^-4

Ka = 2.36 * 10-11

We have a new reaction as well. Because there is no more [OH-] in solution and we have converted all of our C2H5NH2 to C2H5NH3+, here is the new reaction:

C2H5NH3+ + H2O --> C2H5NH2 + H+

We again have to calculate a new molarity for C2H5NH3+, since we have added HClO4 to the solution

New Molarity C2H5NH3+ = .0056523 moles C2H5NH2 / (.0249 L + .01761 L) = .133 M C2H5NH3+

Our new ICE Chart then is:

C2H5NH3+ C2H5NH2 H+

Initial 0.133 0 0

Change -x +x +x

Equilibrium 0.133 - x x x

So,

Ka = (x) * (x) / (.133 - x)

2.36 *10^-11 =x² / (.133 - x)

Solving using the quadratic equation we get:

x = 1.772 * 10^-6

Which means that we have a concentration of 1.772 x 10^-6 [H+]

Solving for pH we get:

pH = -log[H+]

pH = log (1.772 x 10^-6)

pH = 5.75 at equivalence point

Answer 2

First, you should always write the balanced equation for the reaction of interest:

C₂H₅NH₂ + HClO₄ ==> C₂H₅NH₃ + ClO4^-

At the midpoint of titration, 1/2 of the C₂H₅NH₂ has been converted to C₂H₅NH₃ and so you now have a BUFFER with a weak base C₂H₅NH₂ , and the conjugate acid, C₂H₅NH₃ each present in equal amounts.

0.0249 L x 0.227 mol/L = 0.00565 moles C₂H₅NH₂

0.00565 moles C₂H₅NH₂ = (x L)(0.321 mol/L) = 0.0176 L C₂H₅NH₃⁺ = 17.6 ml

Total volume = 24.9 ml + 17.6 ml = 43.5 ml

At midpoint, you'll have 0.00283 moles of each

The pH at this point will be equal to the pKa of the conjugate acid, because according to Henderson Hasselbalch equation: pH = pKa + log [conjugate base]/[acid] = pKa + log 1 = pKa = 10.65. Showing this the long way, you can use an alternate form of the HH equation:

pOH = pKb + log [conj. acid]/[base] and looking up Kb of ethylamine, I find 4.5x10^-4 and pKb = 3.35

At mid point you have pOH = 3.35 + log (0.002835/0.00283) = 3.35 + log 1 = 3.35 + 0 = 3.35

pH = 14 - pOH = 14 - 3.35

pH = 10.65 @ half way point

At the equivalence point of titration, all of the C₂H₅NH₂ (0.00565 moles)has been converted to 0.00565 moles of C₂H₅NH₃⁺ and then you have..

C₂H₅NH₃⁺ + H2O ==> C₂H₅NH₂ + H₃O⁺ (this shows the hydrolysis of the conjugate acid, C₂H₅NH₃⁺)

Since the volume is now 43.5 ml (0.0435 L) you have final concentration of C₂H₅NH₃⁺ = 0.00565 mol/0.0435 L

[C₂H₅NH₃⁺] = 0.130 M

Ka = [C₂H₅NH₂][H₃O⁺]/[C₂H₅NH₃⁺] and since Ka = Kw/Kb, we have Ka = 1x10^-14/4.5x10^-4 = 2.2x10^-11= Ka

2.2x10-11 = (x)(x)/0.130

x² = 2.86x10^-12

x = 1.69x10-6 = H₃O⁺

pH = -log 1.69x10^-6

pH = 5.77 @ equivalence point