University:
College
Course:
College Chemistry
Academic year:

2022
Views:

216

Pages:

15
Author:

CometConductor

Problem: 40.00 mL of 0.350 M CH3NH2 is titrated with 0.280 M HCl until the end point is reached. Calculate the pH of the solution at the end point. (Kb for CH3NH2 = 5.0 x 10^-4) Solution: Step 1: Determine the moles of CH3NH2 and HCl. Moles of CH3NH2 = 0.0400 L * 0.350 mol/L = 0.0140 mol Moles of HCl = 0.0500 L * 0.280 mol/L = 0.0140 mol Step 2: Set up an ICE table for the reaction between CH3NH2 and HCl. CH3NH2 + HCl -> CH3NH3+ + Cl- Initial: 0.0140 0.0140 0 0 Change: -0.0140 -0.0140 +0.0140 +0.0140 Equil.: 0 0 0.0140 0.0140 Step 3: Calculate the concentration of CH3NH3+ [CH3NH3+] = 0.0140 mol / 0.0700 L = 0.200 M Step 4: Set up an ICE table for the hydrolysis of CH3NH3+ CH3NH3+ + H2O -> CH3NH2 + H3O+ Initial: 0.200 0 0 Change: -x +x +x Equil.: 0.200-x x x Step 5: Calculate the Ka for CH3NH3+ Ka = Kw / Kb = (1.0 x 10^-14) / (5.0 x 10^-4) = 2.0 x 10^-11 Step 6: Set up the Ka expression and solve for x Ka = [CH3NH2][H3O+] / [CH3NH3+] 2.0 x 10^-11 = (x)(x) / (0.200 - x) Assuming x is small compared to 0.200, we can approximate: 2.0 x 10^-11 ≈ x^2 / 0.200 x ≈ 2.0 x 10^-6 Step 7: Calculate the pH pH = -log[H3O+] = -log(2.0 x 10^-6) ≈ 5.74 Answer: 5.74 2) How many moles of HNO3 must be added to a 1.00 liter of 0.370 M NaNO2 to give a buffer of pH 4.20? (Ignore any volume change due to the addition of HNO3) (Ka for HNO2 is 4.5 x 10^-4) Setup: HNO2 + H2O ⇌ H3O+ + NO2- Initial conc. C 0.370 0 Change in conc. +x +x -x Equil. conc. C+x x 0.370-x Ka = [H3O+][NO2-] / [HNO2] 4.5 x 10^-4 = (x)(0.370) / (C+x) C ≈ 0.052 moles Answer: 0.052 moles 3) a) Is NaHCO3 (aq) acidic, basic, or neutral? You must show your work to justify your answer. Setup: HCO3- + H2O ⇌ H2CO3 + OH- Kb = Kw / Ka = (1.0 x 10^-14) / (4.3 x 10^-7) = 2.3 x 10^-8 Answer: Kb for HCO3- is larger than Ka for H2CO3; NaHCO3 (aq) is basic. b) Is NaHCO3 (aq) a buffer? (You must show your work to prove that your answer is not a guess.) Setup: HCO3- + H+ → H2CO3 (added) HCO3- + OH- → CO3^2- + H2O (added) A solution of HCO3- reacts with any H+ or OH- added. A solution of HCO3- resists the change in pH when a small amount of H+ or OH- is added. Answer: Yes 3) What is the pH at the equivalence point when 27.0 mL of 0.200 M CH3NH2 are titrated with 0.350 M HCl? Kb for CH3NH2 is 4.4 x 10^-4 Setup: CH3NH2 + HCl -> CH3NH3+ + Cl- 0.200 mol/L * 0.0270 L = 5.40 x 10^-3 moles (react) 5.40 x 10^-3 moles (formed) 5.40 x 10^-3 moles (formed) (1) Find vol of HCl added: 0.200 mol CH3NH2 * 0.0270 L CH3NH2 * (1 mol HCl / 1 mol CH3NH2) * (1 L HCl / 0.350 mol HCl) = 0.0154 L HCl = 15.4 mL HCl (2) Find total volume after end point is reached: 27.0 mL CH3NH2 + 15.4 mL HCl = 42.4 mL = 0.0424 L (3) Hydrolysis problem: CH3NH3+ + H2O ⇌ CH3NH2 + H3O+ Initial conc. 5.40 x 10^-3 moles / 0.0424 L = 0.127 M Change in conc. -x +x +x Equil. conc. 0.127 - x x x KaCH3NH3+ = Kw / Kb = 1.0 x 10^-14 / 4.4 x 10^-4 = 2.3 x 10^-11 Ka = [CH3NH2][H3O+] / [CH3NH3+] 2.3 x 10^-11 = (x)(x) / (0.127 - x) x ≈ [H+] = 1.7 x 10^-6 pH = 5.76 Answer: 5.76