7 장 적정 Stirring bar One method in volumetric analysis is titration In titration: - substance to be analysed is known as the analyte - the solution added to the analyte is known as the titrant - titrant is usually delivered from a buret Volumetric analysis refers to any procedure in which the volume of reagent needed to react with the analyte is being measured.

- the point when the quantity of added titrant is the exact amount necessary for stoichiometric reaction with the analyte is called the equivalence point Example : 5HO-C-C-OH + 2MnO H + 10CO 2 + 2Mn H 2 O If the unknown contains 5.0 mmol of oxalic acid, the equivalence point is reached when 2.0 mmol of MnO 4 - has been added. Equivalence point is the ideal theoretical result we seek in a titration What we actually measure is the end point, which is marked by a sudden change in the physical property of the solution Titration

Methods for determining the end point are: -detecting a sudden change in the voltage or current between a pair of electrodes -observing an indicator colour change. An indicator is a compound which changes colour abruptly near the equivalence point. This change is caused by the disappearance of the analyte or appearance of excess titrant -monitoring the absorption of light

Estimation of titration error – blank titration : 5HO-C-C-OH + 2MnO H + 10CO 2 + 2Mn H 2 O -solution containing no oxalic acid was titrated with MnO 4 - to determined how much is needed to form an observable purple colour -this volume of MnO 4 - is then subtracted from the volume observed in the analytical titration What is the difference between equivalence point and end point? Titration error – by choosing a physical property whose change is easily observed one can minimise the titration error so that the end point is very close to the equivalence point

The concentration of titrant used can be derived if : -the titrant was prepared by dissolving a weighed amount of pure reagent in a known volume of solution -such a reagent is known as a primary standard because the reagent is pure enough (should be  99.9% pure)to be weighed and used directly. See Table 28-8; p836 What if a primary standard is not available ?

-titrate reagent (to be used as titrant in analysis) against a weighed primary standard in order to determine the concentration of titrant – process is known as standardization -after standardization, the reagent (to be used as titrant in analysis) is known as the standard solution 7-2 Titration Calculations Example : pp151 The calcuim content of urine can be determined by: 1. precipitating Ca 2+ as calcium oxalate in basic solution: Ca 2+ + C 2 O 4 2- Ca(C 2 O 4 ).H 2 O(s) 2. wash the precipitate and dissolve it in acid to give Ca 2+ and H 2 C 2 O 4 in solution 3. heat the dissolved oxalic acid to 60 o C and titrate with standardized KMnO 4 until the purple end point is observed -

Standardization : Suppose g of Na 2 C 2 O 4 is dissolved in a ml volumetric flask. Concentration of the oxalate solution is: (0.3562g Na 2 C 2 O 4 )( g/mol Na 2 C 2 O 4 ) ml = M Moles of C 2 O 4 2- in 10.0ml = ( )(0.010l) = x mol Since: 5 NaO-C-C-ONa + 2MnO H + 10CO 2 + 2Mn H 2 O 2 mol MnO 4 - requires 5 mol oxalate

moles of MnO 4 - = = mmol molarity of MnO 4 - = = x M Analysis of Unknown : Ca in a 5.0 ml urine sample was precipitated, redissolved and required 16.17ml of standard MnO 4 - solution. Find the concentration of Ca 2+ in the urine. (2mol MnO 4 - ) (5mol C 2 O 4 2- ) (mol C 2 O 4 2- ) mmol ml In 16.17ml of standard MnO 4 - solution, there are: ( l)(8.794 x M) = x mol MnO 4 -

1.422 x mol MnO 4 - will react with: mol C 2 O 4 2- = = mmol From : Ca 2+ + C 2 O 4 2- Ca(C 2 O 4 ).H 2 O(s) there is one oxalate ion for each Ca 2+ in Ca(C 2 O 4 ).H 2 O(s) there must be mmol Ca 2+ in 5.00ml urine [Ca 2+ ] = ( mmol )/5ml = M (5mol C 2 O 4 2- ) (2mol MnO 4 - ) (mol C 2 O 4 2- )

When titrant is added to the analyte until the reaction is complete direct titration Some reactions require an excess of the titrant for complete reaction with the analyte. In such cases, the first titration will be followed by a titration with a second standard reagent. The second standard reagent is used to titrate the excess of the first reagent back titration Back titration is also applicable if the end point of the back titration is clearer than the end point of the direct titration. Or Excess reagent is required for complete reaction.

Kjeldahl Nitrogen Analysis -developed in one of the most accurate and widely used method for determining nitrogen in substances Procedure : -substance is decomposed and dissolved (digested) in boiling H 2 SO 4 to which K 2 SO 4 has been added. Selenium-coated boiling chips are used to catalyze the reaction organic C H N NH CO 2 + H 2 O -after digestion, the solution containing NH 4 + is made basic NH OH - NH 3 + H 2 O -the liberated NH 3 is distilled into a known amount of HCl NH 3 + H + NH 4 + -Excess unreacted HCl is titrated with standard NaOH to determine how much HCl has been consumed H + + OH - H 2 O boiling H 2 SO 4

7-4. Precipitation Titration Curve A titration curve is a graph showing how the concentration of one of the reactants varies as titrant is added Titration curve tells us how concentrations of analyte and titrant vary during a titration. From it we can: - understand the chemistry that occurs during a titration - learn how experimental control can be exerted to influence the quality of an analytical titration - since concentration varies over many orders of magnitude, the graph plots the p function { pX = -log[X]} against the volume of X Example : Consider the titration of ml of M I - with M Ag +, given: AgI(s) I - + Ag + K sp = [Ag + ][I - ] = 8.3 x

Each aliquot of Ag + reacts completely with I -. At the equivalence point, there will be a sudden increase in the Ag + concentration because all the I - has been consumed and now unreacted Ag + exists in the solution What volume of Ag + is needed to reach the equivalence point ? Let V e denote the volume of Ag + is needed to reach the equivalence point Since 1 mol of Ag + reacts with 1 mol of I -, thus ( l)( mol I - /l) = V e ( mol I - /l) V e = l = ml Since the equilibrium constant for the titration reaction is large, this means the equilibrium lies far to the right. What does this mean experimentally? This implies: I - + Ag + AgI(s) K = 1/ K sp =1.2 x 10 16

Before the equivalence point : Consider the point when 10.00ml Ag + has been added to the I - solution: - solution has more moles of I - than Ag + - moles of I - =original moles of I - - moles of Ag + added =( l)(0.100 mol/l) – ( l)( mol/l) = mol I - Since the volume of solution is now ml, [I - ] == M [Ag + ] = = 1.4 x M pAg + = -log[Ag + ] = mol I l K sp [I - ]

At the equivalence point : All the AgI precipitates and some may redissolve to give equal concentrations of Ag + and I -. Hence let [Ag + ] = [I - ] = x K sp = [Ag + ][I - ] = x 2 = 8.3 x x = 9.1 x pAg + = -log [Ag + ] = -log(9.1 x ) = 8.04 After the equivalence point : When V > V e, the concentration of Ag + is determined by the excess Ag + added from the buret Suppose that V = 52.00ml mol Ag + =( l)( mol Ag + /l) = mol Ag +

Since the total volume = 77.00ml, [Ag + ] = ( mol)( l) = 1.30 x M pAg + = -log (1.30 x ) = equivalence point is the steepest point of the curve - concentration of reactant affects the titration - for stoichiometries other than 1:1, the equivalence point is not at the centre of the steepest section of the curve - in practice, conditions are chosen such that the titration curves are steep

K sp also affects the steepness of the titration curve At the equivalence point, the titration curve is steepest for the least soluble precipitate The larger the equilibrium constant for any titration reaction, the more pronounced will be the change in concentration near the equivalence point

Ag End Point Detection

Ag +