Question

An analytical chemist is titrating 66.8 mL of a 0.7300 M solution of ammonia (NH3) with a 0.5100 M solution of HNO3. The p K, of ammonia is 4.74 Calculate the pH of the base solution after the chemist has added 109.2 mL of the HNO2 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO2 solution added. Round your answer to 2 decimal places. pH= ?

Answer 1

Number of moles of NH3 = M*V = 0.7300 M* 66.8 mL = 48.764 mmol Number of moles of HNO3 = M*V = 0.5100 M* 109.2 mL = 55.692 mmol This is after the equivalence point. At this point, the pH of the solution is determined by the amount of unreacted HBr in the solution. Net moles of HNO3 = 55.692 – 48.764 = 6.928 mmol Total volume of the solution = 66.8 + 109.2 = 176 mL Conc. of HNO3 = [H+] = moles/ volume = 6.928 mmol / 176 mL = 0.039364 M pH =- log [H+] = -log[0.039364] = 1.40