Base hoac oach2po4 hpo42nh4 nh3 buffer solutions

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Unformatted text preview: Weak Acid + Conj. Base HOAc + OAcH2PO4+ HPO42NH4+ + NH3 Buffer Solutions Consider HOAc/OAc- to see how buffers work. Consider ACID USES UP ADDED OHWe know that OAc- + H2O ⇄ HOAc + OH- ; Kb = 5.6 × 10-10 OAc base acid Therefore, if we add OH− to the solution, the reverse reaction of the WEAK ACID with added reverse OH- has OH Kreverse = 1/ Kb = 1.8 × 109 Kreverse is VERY LARGE, so HOAc - Buffer Solutions Consider HOAc/OAc- to see how buffers work. Consider CONJUGATE BASE USES UP ADDED H+ HOAc + H2O ⇄ OAc- + H3O+ Ka = 1.8 × 10-5 acid conj base Therefore, the reverse reaction of the Therefore, reverse WEAK BASE with added H+ has WEAK Kreverse = 1/ Ka = 5.6 × 104 Kreverse is VERY LARGE, so OAccompletely gobbles up the added H+!! completely Buffer Solutions Q: What is the pH of a buffer that has [HOAc] Q: = 0.700 M and [OAc-] = 0.600 M? 0.700 HOAc + H2O ⇄ OAc- + H3O+ Ka = 1.8 × 10-5 0.700 M HOAc has pH = 2.45 The pH of the buffer will have 1. pH < 2.45 2. pH > 2.45 3. pH = 2.45 Buffer Solutions Q: What is the pH of a buffer that has [HOAc] = 0.700 M and [OAc-] = 0.600 M? 0.700 HOAc + H2O ⇄ OAc- + H3O+ Ka = 1.8 × 10-5 [HOAc] [HOAc] initial change equilib 0.700 -x 0.700 - x [OAc-] 0.600 +x 0.600 + x [H3O+] 0 +x x Buffer Solutions Q: What is the pH of a buffer that has [HOAc] = 0.700 M Q: and [OAc-] = 0.600 M? and HOAc + H2O ⇄ OAc- + H3O+ Ka = 1.8 × 10-5 [HOAc] [HOAc] [OAc-] [H3O+] equilib 0.700 - x 0.600 + x x Assuming that x << 0.700 and 0.600, we have [H3O+ ](0.600) Ka = 1.8 x 10-5 = 0.700 [H3O+] = 2.16 × 10-5 and pH = 4.67 > 2.16 2.45 2.45 Buffer Solutions Notice that the expression for calculating the H+ (OH-) conc. of the buffer is [H3O + ] = [H3O + ] = Orig. conc. of HOAc Orig. conc. of OAc- [Acid] • Ka [Conj. base] [OH− ] = • Ka [Base] • Kb [Conj. acid] For a buffer of base B and acid BH+ Notice that the [H+] or [OH-] depend on (1) K depend and (2) the ratio of acid and base concentrations. concentrations. Henderson-Hasselbalch Equation [H3O + ] = [Acid] • Ka [Conj. base] Take the negative log of both sides of this Take negative equa...
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