Q-Acids and Bases Prem Sattsangi Copyright 2007

Acids Bases Phosphoric (tri-basic) H 3 PO 4 Sulfuric (di-basic) H 2 SO 4 Nitric (mono-basic) HNO 3 Acetic HC 2 H 3 O 2 Chloric HClO 3 Hydrochloric HCl Ammonia/ammonium hydroxide NH 3 /NH 4 OH Sodium hydroxide NaOH Calcium hydroxide Ca(OH) 2 Aluminum hydroxide Al (OH) 3 Sodium carbonate Na 2 CO 3

Definitions Arrhenius (1880s) Acid: Produce H + ions in solutions. Base: Produce OH - ions in solutions. Lowry Bronsted (1932) Acid: Proton donor Base: Proton acceptor Lewis (1915) Acid: Electron pair acceptor Base: Electron pair donor

Commercial products Acidic Cola (Phosphoric acid) Jams and jellies (malic and citric acid) Toilet bowl cleansers and lime deposit removers. Basic Detergents (sodium polyphosphate, Anionic surfactants) Drain cleaners, oven cleaners (NaOH) Window, floor cleaners (ammonia) Antacids (aluminum hydroxide, magnesium hydroxide, calcium carbonate)

Neutralization reactions Net ionic equation: H + (aq) + OH - (aq)  H 2 O(l) NaOH + HCl  NaCl + H 2 O CaCO 3 + 2HCl  CaCl 2 + H 2 O + CO 2 2NaOH + H 2 SO 4  Na 2 SO 4 + 2H 2 O CaCO 3 + H 2 SO 4  CaSO 4 + H 2 O + CO 2 3NaOH +H 3 PO 4  Na 3 PO 4 + 3H 2 O 3CaCO 3 +2H 3 PO 4  Ca 3 (PO 4 ) 2 +3H 2 O + 3 CO 2

Dilution of sulfuric acid Chemtrek, p.8-11 Heat is evolved when a concentrated solution is diluted. Explain. Cations and anions are farther separated in a dilute solution. Which will have a higher heat of dilution? (a) H 2 SO 4 ~98% Explain. Amount of heat generated is directly proportional to the amount of acid present. (a) H 2 SO 4 ~98% or (b) HCl ~38%

Dilution of sulfuric acid Chemtrek, p.8-11 Proper procedure for dilution is: (a) Pour Concentrated acid into water. (b) Pour water into concentrated acid. (a) Acid into water. Write equation for dilution of Conc. H 2 SO 4 H 2 SO 4 (aq) + H 2 O(l)  H 3 O + (aq) + HSO 4 - (aq) Step-1 HSO 4 - (aq) + H 2 O(l)  H 3 O + (aq) + SO 4 2- (aq) Step-2

Handling of bases NH 3 (g) has a pungent JOVIAN odor which quickly attacks eyes and lungs. Concentrated Ammonia (~30%) solution is opened, after cooling in ice, inside a hood. NaOH is kept closed tight. Explain 1. NaOH(s) is hygroscopic. Give a reason for packaging NaOH in plastic bottles as opposed to glass. NaOH reacts with glass to form sodium silicate Na 4 SiO 4 (aq). NaOH also reacts with atmospheric CO 2 (g). Complete the equation: NaOH(s) + CO 2 (g)  NaHCO 3

Questions p Chemtrek Q.8 For HCl(aq) and NaOH(aq) titration, write: The over-all equation: (Answer in slide 5) The net ionic equation: Q.10 Calculate M U.K. Data: U.K. HCl 2 drs = M NaOH 3 drs M HCl x 2 Drs HCl = M NaOH x 3Drs NaOH M of U.K. HCl = 0.100M x 3Drs/2Drs = 0.150M Q.11 Why is it unnecessary to use calibrated micropipet in this section. Same micropipet. Same size drops. “Drs” cancel out in calculation.

Questions p.8-18 Chemtrek Define, Standard solution. A solution of known concentration. Q. 12 What is meant by “Standardize” a NaOH solution? Use a solution of known concentration to titrate and find the Exact M of the NaOH solution. Q.15 What is the advantage of back titration in egg shell titration? Reaction CaCO 3 + 2HCl  CaCl 2 + CO 2 + H 2 O is very slow at 25 o C. It is markedly faster to dissolve the CaCO 3 in excess of HCl at higher temperature and then titrate the un-reacted HCl(aq), as in back titration.

Prequiz - 1 Q3. NaOH 2.5 mL, 1.00 M = ? NaOH (g) Strategy: n = M x V  mol  g = 1.00 mol x (2.5 x L) x 40.0g 1L 1 mol NaOH = g NaOH

Prequiz - 2 Facts: 1.Egg shell, (a) Provides protective covering, (b) is a source of Calcium, and (c) is a respiratory membrane for the embryo. 2.Average weight of a chicken egg: 5 g (Ca = 2 g) 3.Shell formation takes 15 hrs. 4.Ca deposits at 133mg/hr (2000 mg/15 hrs) 5.Total Ca content of a chicken blood is 25 mg 6.Every 11 minutes, total Ca in blood is used up. 7.Deficit of “Ca” is made up by breakdown of marrow bone growths. The released phosphate is excreted in urine. 8.What disrupts these biochemical processes? Chlorinated hydrocarbons, e.g. PBB, PCB, DDT, Dieldrin

%Ca in egg shell-1 Data: MP calibration: 1 dr = g or mL, 1g = 1mL (Density of water) egg shell = g, HCl 2.00M, 20 drs., needed NaOH 1.05 M, 11 drs, Moles of HCl added: 20 drs  mL  L x 2.00M = 1.44 x mol HCl.

%Ca in egg shell-1 Moles of NaOH required to titrate unreacted HCl: 11 drs  mL  x 1.05M = 4.16 x mol NaOH 4.16 x mol NaOH = 4.16 x mol HCl Moles of HCl that reacted: = Moles HCl added – moles of unreacted HCl = 1.44 x x = 1.02 x mol

%Ca in egg shell-2 Previously calculated: Moles of HCl that reacted with CaCO 3 = 1.02x10 -3 mol [CaCO 3 + 2HCl  CaCl 2 + H 2 O + CO 2 ] CaCO 3 (g): =1.02x10 -3 molHCl x 1 mol CaCO 3 x 100 g CaCO 3 2mol HCl 1 mol CaCO 3 = 5.12x10 -2 g CaCO 3 %Ca in Egg shell: = 5.12x10 -2 g CaCO 3 x g egg shell =72.6%