Chem33-1, C-hour Problem Set 4

1.A buffer solution is prepared by dissolving 1 mole of formic acid (HCOOH, pKa = 3.75) and 0.5 mole of sodium formate (HCOONa) in enough water to make 1 L. a. b. c. Calculate the pH of this solution. 100 mL of a 1 M solution of HCl is added to the buffer solution. What is the pH? 100 mL of a 1 M solution of NaOH is added to the buffer solution. What is the pH? ANSWER: Recall that thus, − +[ HCOO ][ H 3 O ] Ka = [ HCOOH ] [ H 3 O+ ] = Ka [ HCOOH ] −[ HCOO ] + ⎛ [ HCOOH ] ⎞ [HCOOH ] [ HCOOH ] pH = − lg[H 3O ] = − lg⎜ ⎜ Ka − ⎟ ⎟ = − lg Ka − lg − = pKa − lg −⎝ [ HCOO ] ⎠ [ HCOO ] [ HCOO ] The pH is [ HCOOH ] 1 pH = pKa − lg = 3.75 − lg = 3.45 −[ HCOO ] 0.5 b. After adding 100 ml of a 1M solution of HCl, the molar number of HCOOH in the buffer is 1 L - 1L 1 mol + 1M ⋅ 100ml ⋅ = 1.1 mol , and the molar number of HCOO is 0.5mol − 1 M ⋅ 100ml ⋅ = 0.4 mol , 1000ml 1000ml therefore [HCOOH ] 1.1 pH = pKa − log = 3.75 − lg = 3 .31 − [HCOO ] 0.4 c. After adding 100 ml of a 1M solution of NaOH, the molar number of HCOOH in the buffer is 1 L - 1 mol − 1 M ⋅100ml ⋅ = 0.9mol , and the molar number of HCOO is 1000 ml 1 L 0.5mol + 1M ⋅100ml ⋅ = 0.6mol , therefore 1000ml [HCOOH ] 0.9 pH = pKa − log = 3 .75 − lg = 3. 57 − [HCOO ] 0.6 This study source was downloaded by 100000780339733 from CourseHero.com on 06-03-2022 03:45:10 GMT -05:00 https://www.coursehero.com/file/169047/probset4withanswers/ 2. A sample of solid ammonium chloride was placed in an evacuated container and then heated so that it decomposed to ammonia gas and hydrogen chloride gas. After heating, the total pressure in the container was found to be 4.4 atm. Calculate Kp at this temperature for the decomposition reaction NH4Cl(s) ⇔ NH3(g) + HCl(g) ANSWER: NH4Cl(s) ⇔ NH3(g) + HCl(g) Kp = PNH3 x PHCl For this system to reach equilibrium, some of the NH4Cl(s) decomposes to form equal moles of NH(g) and HCl(g) at equilibrium. Since mol HCl = mol NH3, then the partial pressures of each gas must be equal to each other. At equilibrium: Ptotal = PNH3 + PHCl and PNH3 = PHCl. Ptotal = 4.4 atm = 2 PNH3 , 2.2 atm = PNH3 = PHCl; Kp = (2.2 atm)(2.2 atm) = 4.8 atm2 3. Changing the pressure in a rigid reaction vessel by changing the volume may shift the position of a gas-phase equilibrium, but changing the pressure by adding an inert gas will not. Why? ANSWER: When we change the pressure by adding an unreactive gas, we do not change the partial pressures of any of the substances in equilibrium with each other. In this case the equilibrium will not shift. If we change the pressure by changing the volume, we will change the partial pressures of all the substances in equilibrium by the same factor. If there are unequal number of gaseous particles on the two sides of the equation, then the reaction is no longer at equilibrium and must shift to either products or reactants to return to equilibrium.