Question

An analytical chemist is titrating 75.7 ml. of a 0.4000 M Solution of propylamine (c,H,NH,) with a 0.6900 M solution of HNO,. The pK, of propylamine is 3.46. Calculate the pH of the base solution after the chemist has added s2.5 mL. of the HNo, solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO, solution added. Round your answer to 2 decimal places. PH

Answer 1

Given:

M(HNO3) = 0.69 M

V(HNO3) = 52.5 mL

M(C3H7NH2) = 0.4 M

V(C3H7NH2) = 75.7 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.69 M * 52.5 mL = 36.225 mmol

mol(C3H7NH2) = M(C3H7NH2) * V(C3H7NH2)

mol(C3H7NH2) = 0.4 M * 75.7 mL = 30.28 mmol

We have:

mol(HNO3) = 36.225 mmol

mol(C3H7NH2) = 30.28 mmol

30.28 mmol of both will react

excess HNO3 remaining = 5.945 mmol

Volume of Solution = 52.5 + 75.7 = 128.2 mL

[H+] = 5.945 mmol/128.2 mL = 0.0464 M

use:

pH = -log [H+]

= -log (4.637*10^-2)

= 1.3337

Answer: 1.33