Question

1. A 24.9 mL sample of 0.332 M methylamine, CH₃NH₂, is titrated with 0.264 M perchloric acid.

At the titration midpoint, the pH is .

Use the Tables link in the References for any equilibrium constants that are required.

2. A 20.3 mL sample of 0.278 M methylamine, CH₃NH₂, is titrated with 0.224 M hydroiodic acid.

The pH before the addition of any hydroiodic acid is .

Use the Tables link in the References for any equilibrium constants that are required.

3.  A 24.0 mL sample of 0.242 M triethylamine, (C₂H₅)₃N, is titrated with 0.344 M hydrochloric acid.

After adding 23.6 mL of hydrochloric acid, the pH is .

Use the Tables link in the References for any equilibrium constants that are required.

Answer 1

1) At the midpoint of titration, the pOH = pKb = 3.36

then we calculate the pH:

pH = 14 - pOH = 14 - 3.36 = 10.64

2) The reaction that occurs is:

CH3NH2 + H2O = CH3NH3 + + OH-

Calculate Kb:

Kb = 10 ^ -3.36 = 4.37x10 ^ -4

From the expression of Kb we have:

Kb = X ^ 2 / 0.278 - X = 4.37 x10 ^ -4

We cleared:

X ^ 2 + 4.37x10 ^ -4 X - 1.21x10 ^ -4 = 0

Applying equation of the second degree we have:

X = [OH-] = 0.0108 M

We calculate pOH and pH:

pOH = - Log (0.0108) = 1.97

pH = 14 - 1.97 = 12.02

3) We calculate the equivalence point with the neutralization relation:

Va = Cb * Vb / Ca = 0.242 M * 24 mL / 0.344 M = 16.88 mL

We note that the titration has already exceeded the equivalence point, then we calculate the added moles of HCl:

n HCl = [HCl] * V = 0.344 M * 0.0236 L = 8.12x10 ^ -3 moles

We calculate the neutralized base moles:

n Base = 0.242 M * 0.024 L = 5.81x10 ^ -3 mol

We calculate the remaining HCl moles:

n Final HCl = 8.12x10 ^ -3 - 5.81x10 ^ - 3 = 2.31x10 ^ -3 moles

We calculate the concentration of HCl in solution:

[HCl] = [H3O +] = 2.31x10 ^ -3 mol / 0.0476 L = 0.049 M

We calculate the pH:

pH = - Log (0.049 M) = 1.31

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