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Topic 22 Topic 22 Topic 22: Chemical Equilibrium Basic Concepts Additional Concepts Table of Contents Topic 22 Topic 22.

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Presentation on theme: "Topic 22 Topic 22 Topic 22: Chemical Equilibrium Basic Concepts Additional Concepts Table of Contents Topic 22 Topic 22."— Presentation transcript:

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2 Topic 22 Topic 22

3 Topic 22: Chemical Equilibrium Basic Concepts Additional Concepts Table of Contents Topic 22 Topic 22

4 Consider the reaction for the formation of ammonia from nitrogen and hydrogen. What is equilibrium? Chemical Equilibrium: Basic Concepts Topic 22 Topic 22

5 What is equilibrium? Chemical Equilibrium: Basic Concepts Note that the equation for the production of ammonia has a negative standard free energy, G°. Recall that a negative sign for G° indicates that the reaction is spontaneous under standard conditions. Topic 22 Topic 22 Standard conditions are defined as 298 K and one atmosphere pressure.

6 What is equilibrium? Chemical Equilibrium: Basic Concepts But spontaneous reactions are not always fast. When carried out under standard conditions, this ammonia-forming reaction is much too slow. Topic 22 Topic 22 To produce ammonia at a rate that is practical, the reaction must be carried out at a much higher temperature than 298 K and a higher pressure than one atmosphere.

7 What is equilibrium? Chemical Equilibrium: Basic Concepts What happens when one mole of nitrogen and three moles of hydrogen, the amounts shown in the equation, are placed in a closed reaction vessel at 723 K? Topic 22 Topic 22 Because the reaction is spontaneous, nitrogen and hydrogen begin to react.

8 What is equilibrium? Chemical Equilibrium: Basic Concepts The concentrations of the reactants (H 2 and N 2 ) decrease at first while the concentration of the product (NH 3 ) increases. Topic 22 Topic 22 Then, before the reactants are used up, all concentrations become constant.

9 What is equilibrium? Chemical Equilibrium: Basic Concepts The reactants, H 2 and N 2, are consumed in the reaction, so their concentrations gradually decrease. Topic 22 Topic 22 After a period of time, however, the concentrations of H 2, N 2, and NH 3 no longer change.

10 What is equilibrium? Chemical Equilibrium: Basic Concepts All concentrations become constant, as shown by the horizontal lines on the right side of the diagram. Topic 22 Topic 22

11 Reversible reactions Chemical Equilibrium: Basic Concepts When a reaction results in almost complete conversion of reactants to products, chemists say that the reaction goes to completion. Topic 22 Topic 22 But most reactions, including the ammonia- forming reaction, do not go to completion. They appear to stop. The reason is that these reactions are reversible.

12 Reversible reactions Chemical Equilibrium: Basic Concepts A reversible reaction is one that can occur in both the forward and the reverse directions. Topic 22 Topic 22

13 Reversible reactions Chemical Equilibrium: Basic Concepts Chemists combine these two equations into a single equation that uses a double arrow to show that both reactions occur. Topic 22 Topic 22

14 Reversible reactions Chemical Equilibrium: Basic Concepts When you read the equation, the reactants in the forward reaction are on the left. Topic 22 Topic 22 In the reverse reaction, the reactants are on the right.

15 Reversible reactions Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 In the forward reaction, hydrogen and nitrogen combine to form the product ammonia. In the reverse reaction, ammonia decomposes into the products hydrogen and nitrogen.

16 Reversible reactions Chemical Equilibrium: Basic Concepts Just as the reaction begins at a definite, initial rate; no ammonia is present so only the forward reaction can occur. Topic 22 Topic 22

17 Reversible reactions Chemical Equilibrium: Basic Concepts As hydrogen and nitrogen combine to form ammonia, their concentrations decrease. Topic 22 Topic 22 The rate of a reaction depends upon the concentration of the reactants. The decrease in the concentration of the reactants causes the rate of the forward reaction to decrease.

18 Reversible reactions Chemical Equilibrium: Basic Concepts As soon as ammonia is present, the reverse reaction can occur, slowly at first, but at an increasing rate as the concentration of ammonia increases. Topic 22 Topic 22

19 Reversible reactions Chemical Equilibrium: Basic Concepts As the reaction proceeds, the rate of the forward reaction continues to decrease and the rate of the reverse reaction continues to increase until the two rates are equal. Topic 22 Topic 22

20 Reversible reactions Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 At that point, ammonia is being produced as fast as it is being decomposed, so the concentrations of nitrogen, hydrogen, and ammonia remain constant.

21 Reversible reactions Chemical Equilibrium: Basic Concepts The system has reached a state of balance or equilibrium. Topic 22 Topic 22 The word equilibrium means that opposing processes are in balance.

22 Reversible reactions Chemical Equilibrium: Basic Concepts Chemical equilibrium is a state in which the forward and reverse reactions balance each other because they take place at equal rates. Topic 22 Topic 22 Rate forward reaction = Rate reverse reaction

23 Reversible reactions Chemical Equilibrium: Basic Concepts You can recognize that the ammonia-forming reaction reaches a state of chemical equilibrium because its chemical equation is written with a double arrow like this. Topic 22 Topic 22 At equilibrium, the concentrations of reactants and products are constant. However, that does not mean that the amounts or concentrations of reactants and products are equal.

24 Equilibrium Expressions and Constants Chemical Equilibrium: Basic Concepts You have learned that some chemical systems have little tendency to react and others go readily to completion. Topic 22 Topic 22 Click box to view movie clip.

25 Equilibrium Expressions and Constants Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 In between these two extremes are the majority of reactions that reach a state of equilibrium with varying amounts of reactants unconsumed. If the reactants are not consumed, then not all the product predicted by the balanced chemical equation will be produced.

26 Equilibrium Expressions and Constants Chemical Equilibrium: Basic Concepts According to the equation for the ammonia- producing reaction, two moles of ammonia should be produced when one mole of nitrogen and three moles of hydrogen react. Topic 22 Topic 22 Because the reaction reaches a state of equilibrium, however, fewer than two moles of ammonia will actually be obtained. Chemists need to be able to predict the yield of a reaction.

27 Equilibrium Expressions and Constants Chemical Equilibrium: Basic Concepts In 1864, the Norwegian chemists Cato Maximilian Guldberg and Peter Waage proposed the law of chemical equilibrium, which states that at a given temperature, a chemical system may reach a state in which a particular ratio of reactant and product concentrations has a constant value. Topic 22 Topic 22

28 Equilibrium Expressions and Constants Chemical Equilibrium: Basic Concepts For example, the general equation for a reaction at equilibrium can be written as follows. Topic 22 Topic 22 A and B are the reactants; C and D the products.

29 Equilibrium Expressions and Constants Chemical Equilibrium: Basic Concepts The coefficients in the balanced equation are a, b, c, and d. Topic 22 Topic 22 If the law of chemical equilibrium is applied to this reaction, the following ratio is obtained.

30 Equilibrium Expressions and Constants Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 This ratio is called the equilibrium constant expression. The square brackets indicate the molar concentrations of the reactants and products at equilibrium in mol/L.

31 Equilibrium Expressions and Constants Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 The equilibrium constant, K eq, is the numerical value of the ratio of product concentrations to reactant concentrations, with each concentration raised to the power corresponding to its coefficient in the balanced equation. The value of K eq is constant only at a specified temperature.

32 Equilibrium Expressions and Constants Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 K eq > 1: More products than reactants at equilibrium. K eq < 1: More reactants than products at equilibrium.

33 Constants for homogeneous equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 How would you write the equilibrium constant expression for this reaction in which hydrogen and iodine react to form hydrogen iodide? This reaction is a homogeneous equilibrium, which means that all the reactants and products are in the same physical state.

34 Constants for homogeneous equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 All participants are gases. To begin writing the equilibrium constant expression, place the product concentration in the numerator and the reactant concentrations in the denominator. The expression becomes equal to K eq when you add the coefficients from the balanced chemical equation as exponents.

35 Constants for homogeneous equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 K eq for this homogeneous equilibrium at 731 K is Note that 49.7 has no units. In writing equilibrium constant expressions, its customary to omit units.

36 Equilibrium Constant Expressions for Homogeneous Equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Write the equilibrium constant expression for the reaction in which ammonia gas is produced from hydrogen and nitrogen. The form of the equilibrium constant expression is

37 Equilibrium Constant Expressions for Homogeneous Equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22

38 Equilibrium Constant Expressions for Homogeneous Equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Place the product concentration in the numerator and the reactant concentrations in the denominator. Raise the concentration of each reactant and product to a power equal to its coefficient in the balanced chemical equation and set the ratio equal to K eq.

39 Constants for heterogeneous equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 When the reactants and products of a reaction are present in more than one physical state, the equilibrium is called a heterogeneous equilibrium. When ethanol is placed in a closed flask, a liquid-vapor equilibrium is established.

40 Constants for heterogeneous equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 To write the equilibrium constant expression for this process, you would form a ratio of the product to the reactant. At a given temperature, the ratio would have a constant value K.

41 Constants for heterogeneous equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Note that the term in the denominator is the concentration of liquid ethanol. Because liquid ethanol is a pure substance, its concentration is constant at a given temperature. Thats because the concentration of a pure substance is its density in moles per liter.

42 Constants for heterogeneous equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 At any given temperature, density does not change. No matter how much or how little C 2 H 5 OH is present, its concentration remains constant. Therefore, the term in the denominator is a constant and can be combined with K.

43 Constants for heterogeneous equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 The equilibrium constant expression for this phase change is

44 Constants for heterogeneous equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Solids also are pure substances with unchanging concentrations, so equilibria involving solids can be simplified in the same way. For example, notice the experiment involving the sublimation of iodine crystals. The equilibrium depends only on the concentration of gaseous iodine in the system.

45 Equilibrium Constant Expressions for Heterogeneous Equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Write the equilibrium constant expression for the decomposition of baking soda (sodium hydrogen carbonate).

46 Equilibrium Constant Expressions for Heterogeneous Equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 You are given a heterogeneous equilibrium involving gases and solids. The general form of the equilibrium constant expression for this reaction is Because the reactant and one of the products are solids with constant concentrations, they can be omitted from the equilibrium constant expression.

47 Equilibrium Constant Expressions for Heterogeneous Equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Known Unknown equilibrium constant expression = ?

48 Equilibrium Constant Expressions for Heterogeneous Equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Write a ratio with the concentrations of the products in the numerator and the concentration of the reactant in the denominator. Leave out [NaHCO 3 ] and [Na 2 CO 3 ] because they are solids.

49 Equilibrium Constant Expressions for Heterogeneous Equilibria Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Because the coefficients of [CO 2 ] and [H 2 O] are 1, the expression is complete.

50 Determining the Value of Equilibrium Constants Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 When equilibrium is established, the concentration of each substance is determined experimentally. Although an equilibrium system has only one value for K eq at a particular temperature, it has an unlimited number of equilibrium positions. Equilibrium positions depend upon the initial concentrations of the reactants and products.

51 Calculating the Value of Equilibrium Constants Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Calculate the value of K eq for the equilibrium constant expression given concentration data at one equilibrium position: [NH 3 ] = mol/L, [N 2 ] = mol/L, [H 2 ] = mol/L.

52 Calculating the Value of Equilibrium Constants Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 You have been given the equilibrium constant expression and the concentration of each reactant and product. You must calculate the equilibrium constant. Because the reactant, H 2, has the largest concentration and is raised to the third power in the denominator, K eq is likely to be less than 1.

53 Calculating the Value of Equilibrium Constants Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Known [NH 3 ] = mol/L [N 2 ] = mol/L [H 2 ] = mol/L Unknown K eq = ?

54 Calculating the Value of Equilibrium Constants Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Substitute the known values into the equilibrium constant expression and calculate its value.

55 Factors Affecting Chemical Equilibrium Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Le Châteliers principle states that if a stress is applied to a system at equilibrium, the system shifts in the direction that relieves the stress. If an additional amount of reactant (NO or Br 2 ) is added to the system, the equilibrium will shift to the right, that is, more product (NOBr) will be formed. For example, consider the equilibrium system

56 Factors Affecting Chemical Equilibrium Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Conversely, adding more NOBr to the system will result in a shift to the left, forming more NO and Br 2. The removal of a reactant or product also results in a shift in the equilibrium. Removing a reactant causes the equilibrium to shift to the left, forming more reactants. Removing the product causes a shift to the right, forming more product.

57 Changes in volume Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Le Châteliers principle also applies to changes in the volume of the reaction vessel containing the equilibrium system. The equilibrium will shift to relieve the stress of increased pressure. Suppose the volume of the reaction vessel for the system is decreased, resulting in an increase in pressure.

58 Changes in volume Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 In this case, the shift will be to the right because three moles of reactant gas combine to form only two moles of product gas. Thus, a shift toward the product will reduce the pressure of the system. If the volume of the reaction vessel was increased, the equilibrium would shift to the left, and more of the reactants would be formed.

59 Changes in volume Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Note that changing the volume of the reaction vessel causes no shift in the equilibrium when the number of moles of product gas equals the number of moles of reactant gas; an example is the equilibrium

60 Changes in temperature Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Even though equilibrium may shift to the right or left in response to a change in concentration or volume, the value of the equilibrium constant remains the same. A change in temperature, however, alters both the equilibrium position and the value of K eq.

61 Changes in temperature Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 For example, consider the thermochemical equation for the reversible formation of hydrogen chloride gas from its elements. The forward reaction releases heat, so you can consider heat as a product in the forward reaction and a reactant in the reverse reaction.

62 Changes in temperature Chemical Equilibrium: Basic Concepts Topic 22 Topic 22 Raising the temperature of this system requires the addition of heat, which shifts the equilibrium to the left and reduces the concentration of hydrogen chloride. Thus, the value of K eq decreases. Lowering the temperature of the system means that heat is removed, so the equilibrium relieves the stress by shifting to the right, increasing both the concentration of hydrogen chloride and K eq.

63 Basic Assessment Questions Question 1 Write equilibrium constant expressions for the following homogeneous equilibria. Topic 22 Topic 22

64 Answer 1a Question 1a Topic 22 Topic 22 Basic Assessment Questions

65 Answer 1b Question 1b Topic 22 Topic 22 Basic Assessment Questions

66 Question 2 Write equilibrium constant expressions for the following heterogeneous equilibria. Topic 22 Topic 22

67 Answer 2a Question 2a Topic 22 Topic 22 Basic Assessment Questions

68 Answer 2b Question 2b Topic 22 Topic 22 Basic Assessment Questions

69 Question 3 Hydrogen and carbon disulfide react to form methane and hydrogen sulfide according to this equation. Topic 22 Topic 22 Calculate K eq if the equilibrium concentrations are [H 2 ] = mol/L, [CS 2 ] = mol/L, [CH 4 ] = mol/L, and [H 2 S] = mol/L.

70 Basic Assessment Questions Answer Topic 22 Topic 22

71 Basic Assessment Questions Question 4 How would decreasing the volume of the reaction vessel affect these equilibria? Topic 22 Topic 22

72 shift to the right Answer 4a Question 4a Topic 22 Topic 22 Basic Assessment Questions

73 shift to the left Answer 4b Question 4b Topic 22 Topic 22 Basic Assessment Questions

74 Chemical Equilibrium: Additional Concepts Topic 22 Topic 22 Additional Concepts

75 Chemical Equilibrium: Additional Concepts Using Equilibrium Constants When K eq is known, the equilibrium concentration of a substance can be calculated if you know the concentrations of all other reactants and products. The following example problem shows you how to determine an equilibrium concentration. Topic 22 Topic 22

76 Chemical Equilibrium: Additional Concepts Calculating Equilibrium Concentrations At 350 C, K eq = 66.9 for the formation of hydrogen iodide from its elements. What is the concentration of HI if [H 2 ] = mol/L and [I 2 ] = mol/L? Topic 22 Topic 22

77 Chemical Equilibrium: Additional Concepts Calculating Equilibrium Concentrations Write the equilibrium constant expression. Multiply both sides of the equation by [H 2 ] [I 2 ]. Topic 22 Topic 22

78 Chemical Equilibrium: Additional Concepts Calculating Equilibrium Concentrations Substitute the known quantities into the equation and solve for [HI]. An extra digit is retained here for accuracy, but the final answer will be rounded to three digits. Topic 22 Topic 22 The equilibrium concentration of HI is mol/L.

79 Chemical Equilibrium: Additional Concepts Solubility equilibria The solubility product constant (K sp ) is an equilibrium constant for the dissolving of a sparingly soluble ionic compound in water. Topic 22 Topic 22

80 Chemical Equilibrium: Additional Concepts Solubility equilibria Topic 22 Topic 22 The solubility product constant expression is the product of the concentrations of the ions with each concentration raised to a power equal to the coefficient of the ion in the chemical equation.

81 Chemical Equilibrium: Additional Concepts Solubility equilibria For example, copper (II) hydroxide dissolves in water according to this equation. The coefficient of Cu 2+ is 1, and the coefficient of OH – is 2, so the following is the solubility product constant expression. Topic 22 Topic 22

82 Chemical Equilibrium: Additional Concepts Solubility equilibria Tabulated K sp values may be used to calculate the molar solubility of a sparingly soluble ionic compound and also to calculate ion concentrations in a saturated solution. Topic 22 Topic 22

83 Chemical Equilibrium: Additional Concepts Calculating Molar Solubility and Ion Concentration from K sp The K sp for lead(II) fluoride (PbF 2 ) is 3.3 x 10 –8 at 25 C. Use this K s value to calculate the following. Topic 22 Topic 22 a.The solubility in mol/L of PbF 2 b.The fluoride ion concentration in a saturated solution of PbF 2.

84 Chemical Equilibrium: Additional Concepts Calculating Molar Solubility and Ion Concentration from K sp Topic 22 Topic 22 a. Write the balanced equation for the solubility equilibrium, and write the K sp expression. The moles of Pb 2+ ions in solution equal the moles of PbF 2 that dissolved.

85 Chemical Equilibrium: Additional Concepts Calculating Molar Solubility and Ion Concentration from K sp Topic 22 Topic 22 Therefore, let [Pb 2+ ] equal s, where s represents the solubility of PbF 2. Because there are two F – ions for every Pb 2+ ion, [F – ] = 2s.

86 Chemical Equilibrium: Additional Concepts Calculating Molar Solubility and Ion Concentration from K sp Topic 22 Topic 22 Substitute these terms into the K sp expression and solve for s. Here three digits are retained for accuracy, but the final answer will be rounded to two digits.

87 Chemical Equilibrium: Additional Concepts Calculating Molar Solubility and Ion Concentration from K sp Topic 22 Topic 22 The molar solubility of PbF 2 in water at 25°C is 2.0 x 10 –3 mol/L.

88 Chemical Equilibrium: Additional Concepts Calculating Molar Solubility and Ion Concentration from K sp Topic 22 Topic 22 b. As shown above, the fluoride ion concentration in a saturated solution of PbF 2 at 25 C is as follows.

89 Chemical Equilibrium: Additional Concepts Predicting precipitates Topic 22 Topic 22 The solubility product constant expression can also be used to predict whether a precipitate will form when two solutions of ionic compounds are mixed. The molar concentrations of the ions in a solution are used to calculate the ion product, Q sp.

90 Chemical Equilibrium: Additional Concepts Predicting precipitates Topic 22 Topic 22 If Q sp < K sp, a precipitate will form, reducing the ion concentrations until the system reaches equilibrium and the solution is saturated. If Q sp < K sp, no precipitate forms.

91 Chemical Equilibrium: Additional Concepts Predicting precipitates Topic 22 Topic 22 The following example problem demonstrates how to use Q sp and K sp to determine whether a precipitate will form. Predict whether a precipitate will form if 200 mL of 0.030M CaCl 2 is added to 200 mL of 0.080M NaOH. Predicting a Precipitate

92 Chemical Equilibrium: Additional Concepts Predicting a Precipitate Topic 22 Topic 22 A double-replacement reaction might occur according to this equation. You know that NaCl is a soluble compound and will not form a precipitate. However, Ca(OH) 2 is sparingly soluble with K sp = 5.0 x 10 –6, so it might precipitate if the concentrations of its ions are high enough.

93 Chemical Equilibrium: Additional Concepts Predicting a Precipitate Topic 22 Topic 22 Write the equation for the dissolving of Ca(OH) 2. The ion product expression is as follows. Q sp is a trial value that will be compared to K sp.

94 Chemical Equilibrium: Additional Concepts Predicting a Precipitate Topic 22 Topic 22 Next, find the concentrations of the Ca 2+ and OH – ions. Divide the initial concentrations in half because the volume doubles on mixing.

95 Chemical Equilibrium: Additional Concepts Predicting a Precipitate Topic 22 Topic 22 Calculate Q sp. Calculate Q sp with K sp. The concentrations of Ca 2+ and OH – are high enough to cause a precipitate of Ca(OH) 2 to form.

96 Chemical Equilibrium: Additional Concepts Common ion effect Topic 22 Topic 22 The solubility of a substance is reduced when the substance is dissolved in a solution containing a common ion. This is called the common ion effect. For example, PbI 2 is less soluble in an aqueous solution of NaI than in pure water because the common ion I –, already present in the NaI solution, reduces the maximum possible concentration of Pb 2+ and thus reduces the solubility of PbI 2.

97 Additional Assessment Questions At a certain temperature, K eq = for the following reaction. Question 1 Topic 22 Topic 22 Calculate the concentration of [H 2 ] in an equilibrium mixture with [CH 4 ] = mol/L and [C 2 H 2 ] = mol/L.

98 Additional Assessment Questions mol/L Answer Topic 22 Topic 22

99 Additional Assessment Questions Use the data in this table to calculate the solubility in mol/L of these ionic compounds at 298 K. Topic 22 Topic 22 Question 2

100 MgCO x 10 –3 mol/L Answer 2a Question 2a Additional Assessment Questions Topic 22 Topic 22

101 Question 2b AlPO x 10 –11 mol/L Answer 2b Additional Assessment Questions Topic 22 Topic 22

102 Additional Assessment Questions Will a precipitate form when 125 mL of 0.010M K 2 SO 4 is mixed with 250 mol of M CaBr 2 ? (Hint: Note that the volumes of the two solutions are not equal, and the volume after mixing is 375 mL.) Question 3 Topic 22 Topic 22

103 Additional Assessment Questions For CaSO 4 : Q sp (3.3 x 10 –5 ) < K sp (4.9 x 10 –5 ); no precipitate forms Answer Topic 22 Topic 22

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