1. Author: J R Reid Chapter 5 Volumetric Analysis

2. CONCEPT OF VOLUMETRIC ANALYSIS The reactants will react with the standard solution from burette of a known concentration called titrant. The analyte in which the concentration to be determined is placed in a conical flask. The appropriate indicator will add into the conical flaks to determine the end point, showed by the color changing of the indicator

3. Involves a neutralization reaction in which an acid is reacted with an equivalent amount of base Titration curve - pH of the solution versus volume of titrant added Titrant - usually strong standard acid or strong base Equivalence point - is the point of stoichiometric equivalence between the analyte and the reagent End point - a point when reaction is completed ACID BASE TITRATION

4. CALCULATIONS Write the balanced chemical equation for the reaction Extract all the relevant information from the question Check that data for consistency, for example, concentrations are usually given in M or mol L -1 but volumes are often given in mL. You will need to convert the mL to L for consistency. The easiest way to do this is to multiply the volume in mL x 10 -3 Calculate the moles of reactant (n) for which you have both the volume(V) and concentration(M) : n = M x V From the balanced chemical equation find the mole ratio known reactant : unknown reactant Use the mole ratio to calculate the moles of the unknown reactant From the volume(V) of unknown reactant and its previously calculated moles(n), calculate its concentration(M): M = n ÷ V

5. EXAMPLES Question: 30 mL of 0.10 M NaOH neutralised 25.0 mL of HCl. Determine the concentration of the acid Write the balanced chemical equation for the reaction NaOH(aq) + HCl(aq) -----> NaCl(aq) + H 2 O(l) Extract the relevant information from the question: NaOH V = 30 mL, M = 0.10M HCl V = 25.0 mL, M = ? Check the data for consistency NaOH V = 30 x 10 -3 L, M = 0.10 M HCl V = 25.0 x 10 -3 L, M = ? Calculate moles NaOH n(NaOH) = M x V = 0.10 x 30 x 10 -3 = 3 x 10 -3 moles From the balanced chemical equation find the mole ratio NaOH:HCl 1:1 Find moles HCl NaOH: HCl is 1:1 So n(NaOH) = n(HCl) = 3 x 10 -3 moles at the equivalence point Calculate concentration of HCl: M = n ÷ V n = 3 x 10 -3 mol, V = 25.0 x 10 -3 L M(HCl) = 3 x 10 -3 ÷ 25.0 x 10 -3 = 0.12 M or 0.12 mol L -1

6. SELF-EXERCISE 1.50mL of 0.2mol L -1 NaOH neutralised 20mL of H 2 SO 4. Determine the concentration of the acid (0.25 M) 2.25.0mL of 0.05M Ba(OH) 2 neutralised 40.0mL of nitric acid. Determine the concentration of the acid. (0.0625 M)

7. WATER DISSOCIATION Water is an amphoteric solvent that is having acidic and basic properties. H 2 O H + + OH - K w = [H + ][OH - ]= 1.0 x 10 -14 [H + ]= [OH - ]= 1.0 x 10 -7 M K w =molar equilibrium constant

8. pH CALCULATIONS pH= -log [H + ] pK w =pH + pOH At 25 o C, K w = 1.0 x 10 -14, then 14 = pH + pOH When [H + ]=[OH - ]=1.0 x 10 -7 M or pH=pOH=7, the solution is neutral. If [H + ]>[OH - ] or pH<7, acidic solution. If [H + ] 7, basic solution.

9. EXAMPLE: Calculate a) pK w for water ionization constant b) pH and pOH values for 1.5 x 10 -3 M hydrogen ion solution. Solution: a)pK w = -log K w = -log 1.0 x 10 -14 = 14.00 b) [H + ] = 1.5 x 10 -3 M pH= -log 1.5 x 10 -3 = 2.82 pOH= 14.00 - 2.82 =11.18

10. WEAK ACID AND WEAK BASE Weak acid and weak bases will ionize partially The acidity constant (K a ) is used to calculate the total H + that has ionized The basicity constant (K b ) is used to calculate the total amount of OH - ionized. e.g: acetic acid with a concentration of C CH 3 COOH H + + CH 3 COO - Initial (M) C 0 0 Equilibrium (M) C-x x x K a = [H + ][CH 3 COO - ] = 1.75 x 10 -5 [CH 3 COOH] (x)(x) = 1.75 x 10 -5 C-x If x at the denominator is neglected, the general formula to determine [H + ] for weak acid (HA) with a concentration of C HA : [H + ]= √K a C HA

11. For weak base, e.g NH 3 with basicity constant, K b will undergo the following reaction in water. NH 3 + H 2 O NH 4 + + OH- The same method can be applied to determine [OH - ] for a weak base (B) with a concentration of C B is: [OH - ] = √K b C B

12. EXAMPLE Calculate the pH of 1.0 x 10 -2 M acetic acid solution. K a for acetic acid is 1.75 x 10 -5. Solution CH 3 COOH H + + CH 3 COO - Initial(M) 1.00 x 10 -2 0 0 Equilibrium (M) 1.00 x 10 -2 -x x x Ka = [H + ] [CH 3 COO - ]= 1.75 x 10 -5 [CH 3 COOH] (x)(x) = 1.75 x 10 -5 1.00 x 10 -2 -x X at the denominator can be neglected since C > 100 K a (x)(x) = 1.75 X 10 -5 1.00 x 10 -2 x = 4.18 x 10 -4 M = [H + ] pH= -log 4.18 x 10 -4 = 3.38

13. SELF-EXERCISE Calculate the pH and pOH of a 1.00 x 10 -3 M solution of acetic acid. (answer: pH=3.88;pOH=10.12)

14. TITRATION OF STRONG ACID WITH STRONG BASE The titrant and the analyte ionize completely since strong acid and strong base dissociate completely e.g: titration between HCl and NaOH HCl + NaOH H 2 O + NaCl

15. NaOH (aq) + HCl (aq) H 2 O (l) + NaCl (aq) TITRATION CURVE STRONG ACID WITH STRONG BASE

16. Before addition of NaOH - pH = 1.00 When the NaOH added - pH increase slowly at first Near the equivalence point (the point which equimolar amounts of acid and base have reacted) - the curve rises almost vertically Beyond the equivalence point - pH increases slowly

17. CALCULATION OF pH AT EVERY STAGE OF TITRATION 1)After addition of 10.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl Total volume = 35.0 mL i)Moles of NaOH in 10.0 mL ii) Moles of HCl in 25.0 mL M = mol L = 0.1 x (10 mL/1000 mL) = 1.00 x 10 -3 mol M = mol L = 0.1 x (25 mL/1000 mL) = 2.50 x 10 -3 mol

18. Amount of HCl left after partial neutralization = (2.50 x 10 -3 ) - (1.00 x 10 -3 ) = 1.50 x 10 -3 mol Concentration of H + ions in 35.0 mL [H + ] = 0.0429 M, pH = -log 0.0429 = 1.37 2) After addition of 25.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 M HCl [H + ] = [OH - ] = 1.00 x 10-7 pH = 7.00 M = mol L = 1.50 x 10 -3 mol (35/1000) = 0.0429 M

19. 3) After addition of 35.0 mL of 0.100 M NaOH to 25.0 mL of 0.100 mL of HCl Total volume = 60.0 mL Moles of NaOH added Moles of HCl in 25.0 mL solution = 2.50 x 10 -3 mol After complete neutralization of HCl, no of moles of NaOH left = (3.50 x 10 -3 ) - (2.50x10 -3 ) = 1.00 x 10 -3 mol M = mol L = 0.1 x (35 mL/1000 mL) = 3.50 x 10 -3 mol

20. Concentration of NaOH in 60.0 mL solution [OH - ] = 0.0167 M pOH = -log 0.0167 = 1.78 pH = 14.00 – 1.78 = 12.22 M = mol L = 1.00 x 10 -3 mol (60/1000) = 0.0167 M

21. WEAK ACID versus STRONG BASE WITH BUFFER REGION The titration between acetic acid and NaOH. CH 3 COOH + NaOH H 2 O + CH 3 COONa The acetic acid which is only a few percent ionized is neutralized to water and equivalent amount of CH 3 COONa salt

22. For the first part of the graph, we have an excess of sodium hydroxide. The curve will be exactly the same as when we add hydrochloric acid to sodium hydroxide. Once the acid is in excess, there will be a difference. Past the equivalence point - a buffer solution containing sodium acetate and acetic acid.

23. As soon as the titration is started, some of the acetic acid is converted to sodium acetate As the titration proceeds, the pH slowly increased as the ratio [Acetate]/[Acetic acid] changes. At the midpoint of the titration, [Acetate]/[Acetic acid] and the pH is equal to pKa. At the equivalence point, we have a solution of sodium acetate Since this is a Bronsted base, the pH at the equivalence point will be alkaline. The pH will depend on the concentration of sodium acetate The greater the concentration, the higher the pH.

24. WEAK BASE versus STRONG ACID WITH BUFFER REGION The titration between hydrochloric acid and ammonia NH 3 + HCl NH 4 + + Cl -

25. At the very beginning of the curve, the pH starts by falling quite quickly as the acid is added, but the curve very soon gets less steep (some of the NH 3 is converted to NH 4 + ). This is because a buffer solution is being set up - composed of the excess ammonia and the ammonium chloride being formed At the midpoint of the titration, [NH 4 + ] equals [NH 3 ] The equivalence point (solution of NH 4 Cl form) is somewhat acidic (less than pH 5), because pure ammonium chloride isn't neutral. However, the equivalence point still falls on the steepest bit of the curve. Beyond the equivalence point, the pH is determined by the concentration of H + added in excess.