Acids, Bases and Titrations Chapter 4 M. Shozi CHEM110 / 2014
ACID-BASE REACTIONS ACIDS Arrhenius: Compounds that contain an ionisable H and able to ionise in aqueous solution to form H + or H 3 O + Strong acids - ionise completely in solution - strong electrolytes HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl (aq) Weak acids - partially ionised in solution - weak electrolytes CH 3 CO 2 H(aq) + H 2 O(l) CH 3 CO 2 (aq) + H 3 O + (aq)
ACID-BASE REACTIONS BASES Arrhenius: Compounds that contain an OH group and are able to ionise in aqueous solution to form OH - Strong bases - ionise completely in solution - strong electrolytes NaOH(aq) Na + (aq) + OH (aq) Weak bases - partially ionised in solution - weak electrolytes NH 4 OH(aq) NH 4+ (aq) + OH (aq)
NEUTRALISATION REACTIONS When solutions of an acid and a base are combined, the products are a salt and water HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) (acid) (base) (salt) (water) Net ionic equation: H + (aq) + OH (aq) H 2 O(l)
GAS FORMING REACTIONS These are acid-base reactions which result in the formation of a gas Na 2 S(aq) + 2HCl(aq) H 2 S(g) + 2NaCl(aq) Net ionic equation: S 2- (aq) + 2H + (aq) H 2 S(g)
GAS FORMING REACTIONS Carbonates and hydrogen carbonates (or bicarbonates) will form CO 2 (g) when treated with an acid Na 2 CO 3 (aq) + 2H + (aq) H 2 O(l) + CO 2 (g) +2Na + (aq) Net ionic equation: 2H + (aq) + CO 3 2- (aq) H 2 O(l) + CO 2 (g)
MOLARITY Two solutions can contain the same compounds but be quite different because the proportions of those compounds are different. Molarity is one way to measure the concentration of a solution. Molarity (M) = moles of solute (mol) volume of solution in litres (L)
MIXING A SOLUTION
Exercise 1 Determine the molarity of a solution obtained by dissolving 2.000 g NaCl in sufficient water that the final volume is 10.00 ml. 9
Exercise 2 What are the calcium and nitrate concentrations in a 0.238 M Ca(NO 3 ) 2 (aq) solution? Chemical equation: Ca(NO 3 ) 2 (aq) Ca 2+ (aq) + 2NO 3 (aq) 10
SOLUTION DILUTION The preparation of a lower concentration solution from a higher concentration solution by the addition of solvent. Dilution does not change amount of solute in solution: n 1 = n 2 C 1 V 1 = C 2 V 2 11
SOLUTION DILUTION
Exercise 3 Calculate the molarity obtained by diluting 25.00 ml of 18.03 M H 2 SO to a final volume of 750.0 ml. 13
TITRATION THEORY The analytical technique in which one can determine the concentration of a solute in a solution.
SOME TERMINOLOGY Standard solution A solution of accurately known concentration. Analyte The substance being analysed in an analytical procedure. Indicator An intensely-coloured, organic dye that exhibits different colours in acidic & basic medium End Point the volume of titrant added at which the indicator changes colour Equivalence Point the volume of titrant at which stoichiometric amounts of reactant have been added.
Exercise 4 A 40.00 ml solution of sodium thiosulfate (Na 2 S 2 O 3 ) of concentration 0.1455 M is titrated with 26.36 ml of I 2 solution and the following reaction takes place: 2Na 2 S 2 O 3 (aq) + I 2 (aq) Na 2 S 4 O 6 (aq) + 2NaI(aq) Calculate the concentration of the I 2 solution in: a) mol L -1 b) g dm -3
Exercise 4 solution
Exercise 4 solution
Exercise 5 A piece of iron wire has a mass of 0.6201 g. It is dissolved in acid and then reduced from Fe 3+ to Fe 2+. The whole solution was titrated against 0.04050 M K 2 Cr 2 O 7 solution of which 42.50 ml was required for complete reaction. The balanced reaction equation is: Cr 2 O 7 2 (aq) + 14H + (aq) + 6Fe 2+ (aq) 2Cr 3+ (aq) + 6Fe 3+ (aq) + 7H 2 O(l) What is the percentage purity of the iron wire?
Exercise 5 solution
Exercise 5 solution
PRIMARY STANDARDS Highly purified compounds Reference material for titrimetric methods of analysis Preparation Accurate mass dissolved in accurate amount of H 2 O Resulting solution s concentration is accurately known Properties It is obtained in pure and stable form and dissolves completely It does NOT absorb H 2 O or CO 2 from the air It has an accurately known molar mass It reacts quickly and completely with the sample
Exercise 6 A particular solution of NaOH is supposed to be approximately 0.1 M. To determine the exact molarity of the NaOH(aq), a 0.5000 g sample of KHP (KHC 8 H 4 O 4, 204.2 g mol -1 ) is dissolved in water and titrated with 24.03 ml of the NaOH(aq). What is the accurate molarity of the NaOH(aq)? KHC 8 H 4 O 4 (aq) + NaOH(aq) KNaC 8 H 4 O 4 (aq) + H 2 O(l)
Exercise 6 solution
BACK TITRATION METHOD Indirect titration method Direct reaction is slow, lack of indicator, analyte is insoluble, analyte contains impurities Involves: Addition of excess reagent Analyzing the amount of this that remains unreacted Concentration of analyte is calculated backwards
Examples Calcium in antacid tablets or chalk Dissolved in excess acid for complete reaction Excess acid titrated with base Ammonia in cleaning agent Forms cloudy solution in water (NH 4 OH) Excess NH 3 titrated with acid
Exercise 7 A student reacts aspirin (C 9 H 8 O 4, 180.15 g mol -1 ) with 25.00 ml of 1.023 M NaOH. He then titrates the excess NaOH with 12.45 ml of 0.02489 M H 2 SO 4. Calculate the mass (in grams) of aspirin that originally reacted with NaOH. Reaction 1: C 9 H 8 O 4 + 2NaOH C 6 H 5 O 3- + CH 3 COO - + 2Na + + H 2 O Reaction 2: 2NaOH (excess) + H 2 SO 4 Na 2 SO 4 + H 2 O
Exercise 7 solution
Exercise 7 solution