Explanation Given: The density of the solution is 1 .002 g/mol . Formula used: The formula for density is, Density = Mass Volume K a = Concentration of products Concentration of reactants Moles = Given mass Molar mass Δ T b = K b × m Where, Δ T b is change in boiling point m is the molality K b is the ebullioscopic constant Calculation: Solution of 1 mL of NaHSO 4 contains 1 .002 g . Thus, 1000 mL of NaHSO 4 = 1 .002 g × 1000 mL 1 mL = 1002 g Molar mass of NaHSO 4 is, = 22 .98 g/mol+1 .01 g/mol + 32 .065 g/mol+4 ( 16 g/mol ) = 120 .055 g/mol Formula for the number of moles is, Moles = Given mass Molar mass Rearrange the above equation in terms of mass, Given mass = Moles × Molar mass Substitute the value of moles and molar mass in the above equation, Given mass = 0 .10 mol × 120 .055 g/mol = 12 .0055g Formula for calculating the mass of is, Mass of solvent = Mass of solution − Mass of solute Substitute the value of mass of solution and mass of solvent in the above equation. Mass of solvent = 1002 g − 12 .0055 g = 989 .9 g The dominant equilibrium reaction that takes place in the given case is, HSO 4 − ( a q ) ⇌ H + ( a q ) + SO 4 2 − ( a q ) The change in the concentration of HSO 4 − is assumed to be x . The ICE table is formed for the given reaction. HSO 4 − ( a q ) → H + ( a q ) + SO 4 2 − ( a q ) Initial : 0.10 0 0 Change : − x + x + x Equilibrium: 0.10 − x x x The equilibrium concentration of HSO 4 − is ( 0.10 − x ) M . The equilibrium concentration of H + is x M . The equilibrium concentration of SO 4 2 − is x M . At equilibrium, the equilibrium constant expression is expressed by the formula, K a = Concentration of products Concentration of reactants Where, K a is the acid dissociation constant. The equilibrium constant expression for the given reaction is, K a = [ H + ] [ SO 4 2 − ] [ HSO 4 − ] The value of K a is 1 .2 × 10 − 2 . Substitute the value of K a ,[ HSO 4 − ],and [ SO 4 2 − ] in the above equation. 1 .2 × 10 − 2 = [ x ] 2 [ 0 .10 − x ] The value of K a is not small, so x cannot be neglected in comparison to 0.075 M . 1 .2 × 10 − 2 ( 0

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Chapter 16, Problem 122IE

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To determine: The boiling point of a 0.10 M solution of NaHSO4 .

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Chapter 16, Problem 121IE

Chapter 16, Problem 123IE

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Chapter 16 Solutions

CHEMISTRY THE CENTRAL SCIENCE 14TH EDI

Ch. 16.2 - Practice Exercise 1 Consider the following...Ch. 16.2 - Prob. 16.1.2PE Ch. 16.2 - Prob. 16.2.1PE Ch. 16.2 - Practice Exercise 2 When lithium oxide (Li2O) is...Ch. 16.2 - Based on information in Figure 16.4, place the...Ch. 16.2 - Practice Exercise 2 For each reaction, use Figure...Ch. 16.3 - Practice Exercise 1 In a certain acidic solution...Ch. 16.3 - Practice Exercise 2 Indicate whether solutions...Ch. 16.3 - Prob. 16.5.1PE Ch. 16.3 - Prob. 16.5.2PE

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The boiling point of a 0 .10 M solution of NaHSO 4 .

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To determine: The boiling point of a 0.10 M solution of NaHSO4 .

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Chapter 16 Solutions