Question

Consider the titration of a 26.0-mL sample of 0.175 M CH3NH2 with 0.155 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.)

A. Determine the initial pH.

B. Determine the pH at 6.0 mL of added acid.

C. Determine the pH at one-half of the equivalence point.

D. Determine the pH after adding 4.0 mL of acid beyond the equivalence point.

Answer 1

A. For initial pH:

Dissociation of base:

CH₃NH₂ + H₂O <-> CH₃NH₃⁺ + OH^-

We make our ICE table:

	CH3NH2	<->	CH3NH3+	+	OH-
I	0.175 M		0		0
C	-x		+x		+x
E	0.175 - x		x		x

Now with the definition of constant:

Kb = [CH₃NH₃⁺] [OH^-] / [CH₃NH₂]

4.4 x 10^-4 = x² / (0.175 - x)

Isolating for x:

x = [OH^-] = 0.00855772 M

Calculating pOH and pH:

pOH = -log(0.00855772) = 2.0676

pH = 14 - 2.0676 = 11.9323

B. When adding 6 mL, we'll be adding:

0.155 mol / L * 0.006 L = 0.00093 moles of HBr

And we had:

0.175 mol / L * 0.026 L = 0.00455 moles of Base

Concentrations therefore will be:

[CH₃NH₂] = (0.00455 - 0.00093) / (0.032) = 0.113125 M

[CH₃NH₃Br] (produced by adding acid) = 0.00093 / 0.032 = 0.0290625 M

We know use henderson hasselbach:

pOH = - log (4.4 x 10^-4) + log [0.0290625 / 0.113125]

pOH = 2.7663

pH = 14 - 2.7663 = 11.2337

C. One-Half Equivalence Point:

At equivalence point, moles of acid = moles of base. At half equivalence point we'll therefore have:

Moles of base: 0.0034125 moles

Moles of acid: 0.0011375 moles

For getting that amount of moles of acid, we need:

0.0011375 mol / 0.155 mol / L = 0.007339 L

So we get new concentrations:

[Base] = 0.0034125 mol / 0.033339 L = 0.102357 M

[Acid] = 0.0011375 mol / 0.033339 L = 0.03412 M

pOH = - log (4.4 x 10^-4) + log [0.03412 / 0.102357]

pOH = 2.8794

pH = 14 - 2.8794 = 11.12

D. 4 mL after equivalence point:

As we did before, now we calculate volume to reach equivalence point:

Moles of base = 0.002275

Moles of acid = 0.002275

To get to those moles of acid:

0.002275 moles / 0.155 mol / L = 0.014677 L = 14.677 mL

So we'll need 18.677 mL, which will be:

0.018677 L * 0.155 mol / L = 0.002895 moles of acid

Which will result in:

Moles of Base = 0.001655

Moles of Acid = 0.002895

And hence, concentrations will be:

[Base] = 0.001655 / 0.044677 = 0.037 M

[Acid] = 0.002895 / 0.044677 = 0.0648 M

pOH = - log (4.4 x 10^-4) + log [0.0648 / 0.037]

pOH = 3.6

pH = 14 - 3.6 = 10.1663