5.111 Lecture Summary #23

23.1 5.111 Lecture Summary #23 Acid/Base Equilibrium Continued Topic: Titrations From Friday’s handout Acid buffer action: The weak acid, HA, transfers protons to OH- ions supplied by strong base. The conjugate base, A-, of the weak acid accepts protons from the H3O+ ions supplied by a strong acid. A strong acid and the salt of its conjugate base don't make a good buffer. Why? Base Buffer Example: NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) When strong acid is added, NH3 accepts protons from incoming acid to make NH4+. When strong base is added, NH4+ donates a proton to form NH3 and H2O. pH remains the same. Base buffer action: The weak base, B, accepts protons supplied by strong acid. The conjugate acid, BH+, of the weak base transfers protons to the OH- ions supplied by a strong base. A buffer is a mixture of weak conjugate acids and bases that stabilize the pH of a solution by providing a source or sink for protons. Sample Buffer Problem: Suppose 1.00 mol of HCOOH and 0.500 mol of NaHCOO are added to water and diluted to 1.0 L. Calculate the pH. (Ka = 1.77 x 10-4) initial molarity change in molarity equilibrium molarity HCOOH + H2O 1.00 -x 1.00 -x H3O+ + HCOO0 0.500 +x +x +x 0.500 + x Ka = 1.77 x 10-4 = Using approximation that x is small compared to 1.00 and 0.500, x= Check assumption pH = Now - what if 0.100 mol of a strong acid (HCl) had been included in the 1.0 L solution. Because 0.100 mol of HCl reacts with equal number of moles of HCOO- to form equal moles of HCOOH: For HCOO-, 0.500 mol - 0.100 mol = 0.400 mol [HCOO-] = 0.400 mol/1.0 L =0.400 M For HCOOH, 1.00 mol + 0.100 mol = 1.10 mol [HCOOH] = 1.10 mol/1.0 L =1.10 M 23.2 H3O+ + HCOO- HCOOH + H2O initial molarity change in molarity equilibrium molarity ______ Ka = 1.77 x 10-4= Using approximation that x is small compared to 1.10 and 0.40, x= Check assumption (5% rule) pH = 3.31 So addition of 0.10 mol of strong acid only changed pH from 3.45 to 3.31 Designing a Buffer One must consider the relationship between the ratio of [HA] to [A-], the pKa, and the desired pH. HA (aq) + H2O H3O+ (aq) + A- (aq) Ka = [H3O+][A-] [HA] [H3O+] = Ka x [HA] [H3O+] = Ka x [HA] [A-] Rearrange: Rearrange: [A-] Take logarithms of both sides: Multiply by (-) : That is: log [H3O+] = log Ka + -log [H3O+] = -log Ka pH = pKa - log - log log [HA] [A-] [HA] [A-] [HA] [A-] eq The values of [HA] and [A-] in the equation are at equilibrium. However, a weak acid HA typically loses only a tiny fraction of its protons, so [HA] is negligibly different from the molarity of the acid 23.3 used to prepare the buffer. Likewise, only a tiny fraction of the weakly basic anions of A- accept protons, so [A-] is negligibly different from the molarity of A- used to prepare the buffer. So ~ pK - log pH = a [HA]0 [A-]0 Henderson-Hasselbalch Equation initial This assumption is valid when [H3O+] is small compared to [HA] and [A-] (i.e. less than 5%). Example: Design a buffer system with pH 4.60. A buffer solution is most effective in the range of pKa ±1 Acetic acid is suitable with a pKa of 4.75 pH = pKa - log [CH3COOH]0 [CH3COO-]0 log [CH3COOH]0 [CH3COO-]0 [CH3COOH]0 [CH3COO-]0 = pKa - pH = 4.75 - 4.60 = 0.15 = 100.15 = 1.4 The ratio is more important than the amounts used. However, the amounts used do affect the capacity of the buffer to resist changes in pH. Higher concentrations = more resistance to change. If you use too low concentrations, the Henderson-Hasselbalch equation won’t be valid. For pH 4.60, [H3O+] is 2.5 x 10-5. 2.5 x 10-5 x 100% < 5% [HA] or [A-] [HA] and [A-] need to be > 5.0 x 10-4 M Today’s material Acid Base Titrations An acid-base titration is the addition of a volume of base of known concentration to acid of unknown concentration (or addition of acid to base). This technique can be used to determine the concentration of an acid or base. Titrations involving a strong acid and a strong base In a titration of a strong acid with a strong base, or a strong base with a strong acid, the pH changes slowly initially, changes rapidly through pH 7 at the equivalence point and then changes slowly again. 23.4 Titration curves: 14 14 12 12 10 10 Strong acid Strong base pH 8 pH 8 S S 6 6 Strong base 4 4 Strong acid 2 2 0 0 10 20 0 30 0 Volume of base added (mL) 10 20 30 Volume of acid added (mL) Figures by MIT OpenCourseWare. Equivalence (stoichiometric, S) point = the theoretical volume at which the amount of base added equals the amount of acid that was originally present. End point = the experimentally measured volume at which the indicator changes color. Endpoint should equal equivalence point. Calculating points on a pH curve. Example: a strong base (0.250 M NaOH) is titrated with a strong acid (0.340 M HCl) 1. Calculating the pH before the equivalence point when 5.00 mL of 0.340 M HCl (aq) is added to 25.00 mL of 0.250 M NaOH (aq) A). Calculate moles of OH- present. (Base is strong so amount of NaOH added = amount of OHformed.) 0.02500 L x 0.250 mol/L = 6.25 x 10-3 moles B). Calculate moles of H3O+ supplied by titrant. (Acid is strong so HCl added = H3O+ formed.) 0.00500 L x 0.340 mol/L = 1.70 x 10-3 moles C). Find the moles of OH- remaining after the reaction with H3O+ ions. because stoichiometry is 1:1 6.25 x 10-3 moles - 1.70 x 10-3 moles = 4.55 x 10-3 mol of OH- left D). Calculate molarity of OH 4.55 x 10-3 mol/ 0.03000 L = 0.152 mol/L (Make sure to use total volume 5.00 mL + 25.00 mL) E). Calculate pH pOH = -log 0.152 = 0.818 pH = 14.00 -0.818 = 13.18 23.5 2. Calculating the volume of HCl needed to reach the equivalence point. Initially 6.25 x 10-3 mol of OH- were present. At the equivalence point, 6.25 x 10-3 mol of HCl will have been added (1:1 stoichiometry) 6.25 x 10-3 mol x 1L = 0.0184 L 0.340 mol What is the pH at the equivalence point? 3. Calculate the pH after 1.00 mL of HCl has been added after equivalence point has been reached. A). Find moles of H3O+ formed due to the 1.00 mL addition of HCl (H3O+ formed = amount of HCl added, since strong acid) 0.340 mol/L x (0.00100 L) = 3.40 x 10-4 mol of H3O+ B). Calculate molarity of H3O+ C). pH = -log (7.66 x 10-3) = 2.116 Titrations Curves for Weak acid/Strong base and for Weak base/Strong acid 14 V > Veq strong base in water 12 V = Veq (salt) conj. base of weak acid 10 pH equivalence equivalence point >7point >7 S 8 A- 4 0 H A 0 < V < Veq buffer buffering region half-eq point 2 0 10 20 30 Volume of base added (mL) Figure by MIT OpenCourseWare. A- V=Vhalf-eq special category of buffer buffering region half-eq point A- Strong base 7 6 A- H A Weak acid Start V=0 H A = A- A- H A A- H A weak acid in water H A H A H A H A 23.6 14 V=0 12 Strong acid buffering region 10 pH buffering region 8 7 6 equivalence point pH <7 4 equivalence point pH < 7 2 0 Weak base S 0 B B B B 0 < V < Veq buffer H B+ B B B V=Vhalf eq V = Veq 10 20 30 Volume of acid added (mL) weak base in water V > Veq special category of buffer H B+ H B+ = B (salt) conj. acid of weak base H B+ H B+ H B+ strong acid in water Figure by MIT OpenCourseWare. Example: Titration of weak acid with strong base 25.0 mL of 0.10 M HCOOH with 0.15 M NaOH (Ka = 1.77 x 10-4 for HCOOH) 1. Volume = 0 mL of NaOH added Before any NaOH is added, the problem is that of an ionization of a weak acid in water. HCOOH (aq) + H2O (l) initial molarity change in molarity equilibrium molarity H3O+ (aq) + HCO2- (aq) HCOOH (aq) 0.10 M -x 0.10-x B H3O+ (aq) + HCO2- (aq) 0 0 +x +x__ x x Ka = 1.77 x 10-4 = (x)2/(0.10-x) ~= (x)2/0.10 x = 0.00421 (check 0.00421 is 4.2% of 0.10) okay pH = -log [0.00421] = 2.38 (to how many sig figs?) 2. 0 < V < Veq In this range, the acid has been partly ionized by the strong base (buffering region). Calculate the pH of the solution resulting from the addition of 5.0 mL of 0.15 M NaOH. Because OH- is a stronger base than HCO2-, it reacts almost completely with HCOOH. HCOOH (aq) + OH- (aq) → H2O (l) + HCO2- (aq) K>>1 H B+ 23.7 Initial Moles For HCOOH, (25.0 x 10-3 L)(0.10M) = 2.5 x 10-3 moles For OH-, (5.0 x 10-3 L)(0.15M) = 0.75 x 10-3 moles Moles after Reaction 2.5 x 10-3 moles - 0.75 x 10-3 moles = 1.75 x 10-3 moles of HCOOH left 0.75 x 10-3 moles OH- produces 0.75 x 10-3 moles of HCO2Molarity 1.75 x 10-3 moles of HCOOH/ (0.0250 + 0.0050 L) = 0.0583 M HCOOH 0.75 x 10-3 moles of HCO2- / (0.0250 + 0.0050 L) = 0.0250 M HCO2Option 1 H3O+ + HCO2- HCOOH initial molarity 0.0583 change in molarity -x +x +x equilibrium molarity 0.0583 -x +x 0.0250 + x Ka = 1.77 x 10-4 = (0.0250 + x )(x) 0 0.0250 assume x is small ~= 0.0250x (0.0583-x) 0.0583 x = 4.13 x 10-4 Check assumption: 4.13 x 10-4 is 1.65% of 0.025 and is 0.7% of 0.0583 okay pH = -log [4.13 x 10-4] = 3.38 Option 2 pH ~= pKa - log ([HA]/[A-]) pH ~= 3.75 - log ([0.0583]/[0.0250]) = 3.75 - 0.368 = 3.38 check assumption: for a pH of 3.38, [H3O+] = 4.2 x 10-4 and that is <5% of 0.0583 and is <5% of 0.0250. Okay If the 5% assumption is not valid, than option 1 must be used and Ka = 1.77 x 10-4 = (0.0250 + x)(x)/(0.0583 - x) can not be simplified. Must solve by quadratic equation. Note: when the volume of NaOH added is between 0 and the equivalence volume Veq, the problems are similar to buffer problems. This region of the titration curve is called the "buffering region." 23.8 Half-equivalence point When the volume of NaOH added is equal to half the equivalence volume, [HA] = [A-]. pH ~= pKa - log ([HA]/[A-]) pH ~=pKa -log (1) pH ~=pKa 3. V = Veq At the equivalence point, the amount of NaOH added is equal to the amount of HCOOH. The pH is not 7 as it is for a strong acid and a strong base. The pH is >7 when a weak acid is titrated with a strong base. The pH depends on the properties of the salt formed during the neutralization process. HCOOH and NaOH form NaHCO2 and H2O. Na+ has no effect on pH and HCO2- is a base. Thus at the equivalence point, the pH is >7. Calculate the pH at the equivalence point Calculate total volume at equivalence point moles of HCOOH = 2.5 x 10-3 moles = moles of HCO2- formed = moles of OH- added 2.5 x 10-3 moles of OH- x 1L =1.67 x 10-2 L of NaOH added 0.15 mol Total volume = 0.0250 L + 0.0167 L = 0.0417 L Molarity of HCO22.5 x 10-3 moles of HCO2-/ (0.0417 L) = 0.0600 M HCO2This is an ionization of weak base in water problem. HCO2- (aq) + H2O (l) initial molarity change in molarity equilibrium molarity HCOOH (aq) + OH- (aq) HCO2- (aq) 0.0600 -x 0.0600 -x HCOOH (aq) + OH- (aq) 0 0 +x +x +x +x You can take it from here. Simplify if x is small compared to 0.0600 M. Calculate x, which is equal to [OH- ] = 1.83 x 10-6 M. Then calculate pOH = 5.74. From pOH, calculate pH. pH = 8.26 (which is >7) 23.9 4. V > Veq Beyond the equivalence point, NaOH is added to the solution of the conj. base HCO2-. Since HCO2does not give rise to much OH- in solution (1.83 x 10-6 M), the pOH and pH are determined by the amount of excess NaOH added. This problem is similar to a strong acid/strong base problem. At 5.00 mL past the equivalence point: 0.00500 L x 0.15 M = 7.5 x 10-4 moles excess OH 7.5 x 10-4 moles OH- /(0.00500 L + 0.0250 L+ 0.0167 L) = 0.016 M OH pOH = -log [0.16] = 1.79 pH = 12.21