the next three problems deal with titration of 125 mL of 1.75 M methylamine, CH3
ID: 617133 • Letter: T
Question
the next three problems deal with titration of 125 mL of 1.75 M methylamine, CH3NH2 (Kb - 4.4 x10-4) with 0.25 M HCl Water will be a major species throughout the titration. The chemical species, in addition to water, that can be found in this reaction mixture during the titration are: I. H+, II. OH-, III. Cl-, IV. CH3NH2, V. CH3NH3+ . What is the pH of the solution at equivalence point? How many mL of HCl will need to be added to the methlamine, CH3NH2 solution to reach a pH of 10.64? What will the major species in solution be when 1198.75 mL of HCl has been added to the base solution?Explanation / Answer
At end point, Volume of HCl added V = 125 * 1.75 / 0.25 = 875 ml total volume = 1L Moles of salt formed = 125 * 1.75 / 1000 = 0.21875 moles Concentration of salt = 0.21875 Molar Hydrolysis will take place, [Hydrolysis of CH3NH3+] CH3NH3+ + H2O -----------> H3O+ + CH3NH2 c - nc nc nc nc * nc / c(1-n) = Kw/Kb ; n = 1.02*10^-5 [H3O+] = nc = 2.23 * 10^-6 M ; pH = 5.65 ---------------------------------------------------------------------------------------------- pOH = pKb + log ([salt]/[base]) for buffer 14 - 10.64 = - log Kb + log (salt/base) log (salt/base) = 0 =====>>> [salt] = [base] acid added is half of base present. Volume of HCl added = 875 / 2 = 437.5 ml ----------------------------------------------------------------------------------------------- moles of HCl = 1198.75 * .25 = .2996875 mole moles of base = 0.21875 mole Moles of HCl left = 0.0809375 Molarity = moles / volume = 0.0809375/(1198.75+125) = 6.114 * 10^-5 M [H+] = 6.114 * 10^-5 M pH = 4.21
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