Question

An analytical chemist is titrating 89.0mL of a 0.4400M solution of ethylamine C2H5NH2 with a 0.7900M solution of HNO3. The pKb of ethylamine is 3.19. Calculate the pH of the base solution after the chemist has added 55.1mL of the HNO3 solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of HNO3 solution added. Round your answer to 2 decimal places.

Answer 1

Mi = 0.4400M ; V, = 89.0ml M2 = 0.79001 ; V2 = 55.1ml Now Molarity of solution M = Miv, + M2 V2 - Vit V₂ = 0.4400 x 89.0 + 0.7900 x 55.1 89.0 + 55.7 = 39.16 + 43.53 144.1 M = 0.57M pH = -log (Holdin) = -log(0.57) pH = 0.25