Question

Consider the titration of a 28.0-mL sample of 0.180 M CH3NH2 with 0.145 M HBr. (The value of Kb for CH3NH2 is 4.4×10−4.)

Determine the pH after adding 6.0 mL of acid beyond the equivalence point.

Express your answer using two decimal places.

Answer 1

Beyond the equivalence point, the pH is controlled by the excess amount of HBr.

Thus, pH due to 6.0 mL of HBr can be calculated as follows:

Number of moles of H⁺ = Number of moles of HBr

Number of moles of H⁺ = Volume (L) x Molarity (mol/L)

Number of moles of H⁺ = (6.0 mL x 1L/1000mL) x (0.145 mol/L)

Number of moles of H⁺ = (0.006 L) x (0.145 mol/L)

Number of moles of H⁺ = 8.7 x 10^-4 mol

Volume of HBr solution needed till equivalence point can be calculated as follows:

At equivalence point, M1V1 = M2V2

Where, M1 and V1 are molarity and volume of HBr

M2 and V2 are molarity and volume of CH3NH2.

Thus, V1 = (28.0mL) x (0.180M) / (0.145M)

V1 = 34.8mL

Thus, total volume after addition of 6.0mL of HBr beyond equivalence point can be calculated as follows:

Total volume = 28.0mL + 34.8mL + 6.0 mL

Total volume = 68.8mL

Thus, concentration of H+ beyond equivalent point after addition of 6.0mL HBr can be calculated as follows:

[H+] = No. of moles / volume

[H+] = (8.7 x 10^-4 mol) / (68.8mL x 1L/1000mL)

[H+] = (8.7 x 10^-4 mol) / 0.0688L

[H+] = 12.64 x 10^-3 M

Thus, pH = -log [H+]

pH = -log (12.64 x 10^-3 M)

pH = 1.90

Therefore, pH of the solution after adding 6.0mL of HBr beyond equivalence point is 1.90