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Density and Specific Gravity of Solutions Lecture Two Chemical Equilibrium Asst .Prof.Dr. Azhar A. Ghali Density and specific gravity of solutions. The numeric value for a density given as g/mL or kg/L is the same. Water with a density at 20oC of 0.998 g/mL has a density of 0.998 kg/L. The density of gases is usually given as g/L because at normal pressures the density is about a thousand times less than that of liquids. The density of dry air at 760 torr at 1 atm pressure is 1.185 g/L. The specific gravity is a ratio between the mass of a given volume and the mass of an equal volume of water at 4oC. Since the density of water at 4oC is practically 1.00 g/cc, the density and specific gravity of aqueous solutions are almost identical. The difference needs always to be considered for analytical work of high precision, however. DILUTIONS Whenever you need to go from a more concentrated solution [“stock”] to a less concentrated one, you add solvent [usually water] to “dilute” the solution. No matter what the units of concentration are, you can always use this one formula C1 V1 = C2 V2 [Concentration of the stock] x [Volume of the stock] = [Concentration of the final solution] x Volume of the final solution] p-Functions The p-function of a number X is written as pX and is defined as pX = –log(X) X= H + , Cl- , …….etc. PH= - log [H +] 9 Lecture Two Chemical Equilibrium Asst .Prof.Dr. Azhar A. Ghali POH = - log [OH-] + - -14 [H ] + [OH ] = 10 = Kw PH + POH = 14 Stoichiometric Calculations: Volumetric Analysis In this section we look at calculations involved in titration processes as well as general quantitative reactions. In a volumetric titration, an analyte of unknown concentration is titrated with a standard in presence of a suitable indicator. For a reaction to be used in titration the following characteristics should be satisfied: 1. The stoichiometry of the reaction should be exactly known. This means that we should know the number of moles of A reacting with 1 mole of B. 2. The reaction should be rapid and reaction between A and B should occur immediately and instantly after addition of each drop of titrant (the solution in the burette). 3. There should be no side reactions. A reacts with B only. 4. The reaction should be quantitative. A reacts completely with B. 5. There should exist a suitable indicator which has distinct color change. 6. There should be very good agreement between the equivalence point (theoretical) and the end point (experimental). This means that Both points should occur at the same volume of titrant or at most a very close volume. Three reasons exist for the disagreement between the equivalence and end points. The first is whether the suitable indicator was selected, the second is related to concentration of reactants, and the third is related to the value of the equilibrium constant. These factors will be discussed in details later in the course. Standard solutions A standard solution is a solution of known and exactly defined concentration. Usually standards are classified as either primary standards or secondary standards. There are not too many secondary standards available to analysts and standardization of other substances is necessary to prepare secondary standards. A primary standard should have the following properties: 10 Lecture Two Chemical Equilibrium Asst .Prof.Dr. Azhar A. Ghali 1. Should have a purity of at least 99.98% 2. Stable to drying, a necessary step to expel adsorbed water molecules before weighing 3. Should have high formula weight as the uncertainty in weight is decreased when weight is increased 4. Should be non hygroscopic 5. Should possess the same properties as that required for a titration NaOH and HCl are not primary standards and therefore should be standardized using a primary or secondary standard. NaOH absorbs CO2 from air, highly hygroscopic, and usually of low purity. HCl and other acid in solution are not standards as the percentage written on the reagent bottle is a claimed value and should not be taken as guaranteed. Molarity Volumetric Calculations Volumetric calculations involving molarity are rather simple. The way this information is presented in the text is not very helpful. Therefore, disregard and forget about all equations and relations listed in rectangles in the text, you will not need it. What you really need is to use the stoichiometry of the reaction to find how many mmol of A as compared to the number of mmoles of B. Example A 0.4671 g sample containing NaHCO3 (FW = 84.01 mg/mmol) was dissolved and titrated with 0.1067 M HCl requiring 40.72 mL. Find the percentage of bicarbonate in the sample. Solution We should write the equation in order to identify the stoichiometry NaHCO3 + HCl = NaCl + H2CO3 Now it is clear that the number of mmol of bicarbonate is equal to the number of mmol HCl mmol NaHCO3 = mmol HCl mmol = M x VmL mmol NaHCO3 = (0.1067 mmol/ml ) x 40.72 mL = 4.345 mmol Now get mg bicarbonate by multiplying mmol times FW Mg NaHCO3 = 4.345 mmol x (84.01 mg/mmol) = 365.01 11 Lecture Two Chemical Equilibrium Asst .Prof.Dr. Azhar A. Ghali -3 % NaHCO3 = (365.01 x 10 g/0.4671 g) x 100 = 78.14% We can use dimensional analysis to calculate the mg NaHCO3 directly then get the percentage as above. ? mg NaHCO3 = (0.1067 mmol HCl/ml) x 40.72 mL x (mmol NaHCO3/mmol HCl) x (84.01 mg NaHCO3/ mmol NaHCO3) = 365.0 mg Example A 0.4671 g sample containing Na2CO3 (FW = 106mg/mmol) was dissolved and titrated with 0.1067 M HCl requiring 40.72 mL. Find the percentage of carbonate in the sample. Solution The equation should be the first thing to formulate Na2CO3 +2 HCl = 2NaCl + H2CO3 mmol Na2CO3 = ½ mmol HCl Now get the number of mmol Na2CO3 = ½ x ( MHCl x VmL (HCl) ) Now get the number of mmol Na2CO3 = ½ x 0.1067 x 40.72 = 2.172 mmol Now get mg Na2CO3 = mmol x FW = 2.172 x 106 = 230 mg -3 % Na2CO3 = (230 x 10 g/0.4671 g ) x 100 = 49.3 % Example How many mL of 0.25 M NaOH will react with 5.0 mL of 0.10 M H2SO4. H2SO4 + 2 NaOH = Na2SO4 + 2 H2O Solution mmol NaOH = 2 mmol H2SO4 mmol NaOH = 2 {M (H2SO4) x VmL (H2SO4)} mmol NaOH = 2 x 0.10 x 5.0 = 1.0 mmol mmol NaOH = MNaOH x VmL (NaOH) 1.0 = 0.25 x VmL VmL = 4.0 mL 12 Lecture Two Chemical Equilibrium Asst .Prof.Dr. Azhar A. Ghali We can also calculate the volume in one step using dimensional analysis ? mL NaOH = (mL NaOH/0.25 mmol NaOH) x (2 mmolNaOH/mmol H2SO4) x (0.10 mmol H2SO4 / mL H2SO4) x 5.0 mL = 4.0 mL In a previous lecture you were introduced to calculations involved in titrimetric reactions. We agreed that you do not have to memorize relations listed in the text and the only relation you need to remember is the mmol. A mmol is mg/FW or molarity times volume (mL). We have also agreed that writing the equation of the reaction is essential and is the first step in solving any problem where a reaction takes place. Let us look at more problems and develop a logic solution using the simple mmol concept. General Concepts of Chemical Equilibrium analytical chemistry a branch of chemistry that deals with the identification of compounds and mixtures (qualitative analysis) or the determination of the proportions of the constituents (quantitative analysis): techniques commonly used are titration, precipitation, spectroscopy, chromatography, etc. In this chapter you will be introduced to basic equilibrium concepts and related calculations. The type of calculations you will learn to perform in this chapter will be essential for solving equilibrium problems in later chapters. The ease of solving equilibrium problems will depend on the skills you will master in this chapter. The Equilibrium Constant For chemical reactions that do not proceed to completion, an equilibrium constant can be written as the quotient of multiplication of molar concentrations of products divided by that of reactants, each raised to a power equal to its number of moles. aA + bB = cC + dD Where, the small letters represent the number of moles of substances A, B, C, and D. The equilibrium constant is written as: c d a b Keq = ([C] [D] )/([A] [B] ) In a chemical reaction, as reactants start to react products start to form. Therefore, reactants continuously decrease and products continuously increase till a point is reached where eventually no change in concentrations can be detected. This is the point of equilibrium which is a point where the rate of the forward reaction (product formation) equals the rate 13 Lecture Two Chemical Equilibrium Asst .Prof.Dr. Azhar A. Ghali O f backward reaction (reactants formation). In fact equilibrium implies continuous transformation between infinitesimally small amounts of products and reactants. Calculation of Equilibrium Constants We know from thermodynamics that a reaction occurs spontaneously if it has a negative G (Gibbs free energy) where: G = H - TS Where, H is the change in enthalpy of the reaction and S is the change in entropy. At standard conditions of temperature and pressure (standard state) we have the standard free energy Go where: Go = Ho - TSo The standard free energy is related to equilibrium constant by the relation: Go = - RT ln K K = e-Go/RT R is the gas constant (8.314 deg K-1mol-1) It should be clear that Go gives us good information about the spontaneity of the reaction but it offers no clue on the rate at which the reaction may occur.
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