Acids and Bases; The Basics_II

COMPLETE CHAPTER 10 : COMPLETE CHAPTER 10 ACIDS & BASES

REACTIONS : REACTIONS Acid + Base -------? Water + Salt Neutralization Kc >> 1......one arrow s acid, s base s acid, w base w acid, s base Kc < 1....double arrow w acid, w base

REACTIONS : REACTIONS Acid + Base -------? Water + Salt Neutralization Kc >> 1......one arrow s acid, s base HCl + NaOH s acid, w base HCL + H2O w acid, s base CH3CO2H + NaOH Kc < 1....double arrow w acid, w base NH4Cl + H2O

BALANCING : BALANCING Apply Bronsted-Lowry E.g., HCl + Ba(OH)2 ------? H2O + BaCl2 1 mole HCl has 1 mole H+ 1 mole Ba(OH)2 has 2 moles HO- Thus, 2:1 mole-mole factor for HCl:Ba(OH)2 2:2 for HCl:H2O Thus: 2HCl + Ba(OH)2 ------? 2H2O + BaCl2

BALANCING : BALANCING HCl + Al(OH)3 -------? H2O + AlCl3 1 mole HCl has 1 mole H+ 1 mole Al(OH)3 has 3 moles HO- Thus, 3:1 HCl:Al(OH)3 3:3 HCl:H2O 3:1 HCl:AlCl3 3HCl + Al(OH)3 -------? 3H2O + AlCl3

CARBONATES, METALS : CARBONATES, METALS CARBONATES H2CO3 “carbonic acid” Unstable: H2CO3(aq) --------? H2O(l) + CO2(g) “decarboxylation” A ‘decomposition reaction’

CARBONATES : CARBONATES HCl + K2CO3 -------? H2O + CO2 + KCl 1 mole HCl has 1 mole H+ 1 mole K2CO3 has 1 mole CO32- = two base equivalents Thus, 2:1 HCl:K2CO3 Thus, 2:2 HCl:KCl 2HCl + K2CO3 -------? H2O + CO2 + 2KCl

CARBONATES : CARBONATES HCl + K2CO3 -------? H2O + CO2 + KCl 1 mole HCl has 1 mole H+ 1 mole K2CO3 has 1 mole CO32- = two base equivalents Thus, 2:1 HCl:K2CO3 Thus, 2:2 HCl:KCl 2HCl + K2CO3 -------? H2O + CO2 + 2KCl only 1 H2O because of carbonic acid

METALS : METALS HCl(aq) + M (I, II)(s) -----? H2(g) + MCl(aq) 2HCl(aq) + 2Na(s) -----? H2(g) + 2NaCl(aq) 2HCl(aq) + Ca(s) ----? H2(g) + CaCl2(aq) Really redox

Slide 10 : ACID-BASE TITRATION

TITRATION : TITRATION ‘The careful addition of a precisely measured volume of acid or base of a known molarity, to a known volume of base or acid of unknown molarity, for the purpose of determining the molarity of the unknown.’ When the #moles base = #moles acid....neutralization......endpoint Visualize endpoint with indicators

TITRATION : TITRATION Two types: Aqueous base + solid acid (NaOH standardization) Aqueous base + aqueous acid (unknown acid) Aqueous base + aqueous acid M1V1 = M2V2......# moles base = # moles acid MbVb = MaVa Ma = MbVb/ Va

AQUEOUS BASE + SOLID ACID : AQUEOUS BASE + SOLID ACID NaOH(aq) + KHP(s) ------? H2O(l) + NaHP(aq) MNaOH = (#g acid/MM acid)/VNaOH(L) MNaOH = (#g KHP/MM KHP)/VNaOH(L)

AQUEOUS BASE + SOLID ACID : AQUEOUS BASE + SOLID ACID NaOH(aq) + KHP(s) ------? H2O(l) + NaHP(aq) MNaOH = (#g acid/MM acid)/VNaOH(L) MNaOH = (#g KHP/MM KHP)/VNaOH(L) # moles acid

AQUEOUS BASE + AQUEOUS ACID : AQUEOUS BASE + AQUEOUS ACID NaOH(aq) + HCl(aq) ------? H2O(l) + NaCl(aq) M1V1 = M2V2 MbVb = MaVa MHCl = MNaOHVNaOH/ VHCl

TITRATION EXAMPLES : TITRATION EXAMPLES

TITRATION EXAMPLES : TITRATION EXAMPLES # moles NaOH < # moles HCL # moles NaOH >~ # moles HCL ENDPOINT

STANDARDIZATION : STANDARDIZATION 0.2134 g KHP 204.22 g/mole Vinit (NaOH) = 12.43 mL Vfin = 26.54 mL ? M NaOH

STANDARDIZATION : STANDARDIZATION 0.2134 g KHP 204.22 g/mole Vinit (NaOH) = 12.43 mL Vfin = 26.54 mL (0.2134g/204.22 g/mol)/0.01411L = 0.07406M NaOH

UNKNOWN ACID : UNKNOWN ACID Vinit (NaOH) = 26.54 mL Vfin = 38.35 mL VHCl = 10.00 mL (0.07406M NaOH)(11.81 mL NaOH)/10.00 mL HCl = 0.08746M HCl

Slide 21 : BUFFERS

BUFFERS : BUFFERS ‘A buffer is a solution comprised of a weak acid and a salt of its conjugate base. When the quantities and concentrations are properly adjusted, a buffer solution will resist dramatic changes in pH due to addition of strong acid or base.’

EXAMPLES : EXAMPLES You have a solution

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