Question

Consider the titration of a 28.0 −mL sample of 0.175 M CH3NH2 with 0.155 M HBr. Determine each of the following. 1. the inital pH 2. the volume of added acid required to reach the equivalence point 3. the pH at 6.0 mL of added acid 4. the pH at one-half of the equivalence point 5. the pH at the equivalence point 6. the pH after adding 5.0 mL of acid beyond the equivalence point

Answer 1

First let's calculate the moles of the innitial reactant (Base)

Base (B) = CH3NH2; Acid (A) = HBr

moles B = 0.175 * 0.028 = 4.9x10^-3 moles

Now, let's determine the pH at given parts: (Kb = 5x10^-4 and Ka = 2x10^-11)

1. Innitial pH:

r: B + H₂O --------> BH⁺ + OH^-

i: 0.175 0 0

e: 0.175-x x x

5x10^-4 = x² / 0.175-x Kb is small so would be x so:

5x10^-4 * 0.175 = x²

x = 9.35x10^-3 M = [OH^-]

pOH = -log(9.35x10^-3) = 2.03

pH = 14-2.03 = 11.97

2. To reach the equivalence point:

M_aV_a = M_bV_b -----> V_a = M_bV_b / M_a

V_a = 0.175 * 28 / 0.155 = 31.61 mL

3. pH at 6 mL of acid used:

moles A = 0.155 * 0.006 = 9.3x10^-4 moles

reaction: B + A -------> S + H₂O

moles remanents of base with acid = 4.9x10^-3 - 9.3x10^-4 = 3.97x10^-3 moles

Now, using the HH equation:

pOH = pKb + log (S/B) ---> pKb = -log(5x10^-4) = 3.3

pOH = 3.3 + log(9.3x10^-4 / 3.97x10^-3) = 2.67

pH = 14-2.67 = 11.33

4. pH at half equivalence point.

In this point you have consumed half the volume of equivalence point, which means it's been consumed half the moles of acid, and at this point, pH = pKa

pKa = -log(2x10^-11) = 10.69

5. At the equivalence point, The methylamine and HCl react in equal molar quantities, so the equivalence point is reached when moles of HCl = moles of CH3NH2.

r:BH⁺ <--------> B + H⁺ Ka

moles of A = 0.155 * 0.03161 = 4.9x10^-3

r: BH⁺ <--------> B + H⁺

i: 4.9x10^-3   0 0

e: 4.9x10^-3 - x x x

2x10^-11 = x² / 4.9x10^-3-x

2x10^-11*4.9x10^-3 = x²

x = 3.13x10^-7 M = [H⁺]

pH = -log(3.13x10^-7)

pH = 6.5

To do the last part, use the following equation:

pH = pKa+ log (S/A)

Establish the eequilibrium as I did in part 4, and you will get the answer.

Hope this helps