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

14th Edition

ISBN: 9780134809694

Author: Brown

Publisher: PEARSON

See similar textbooks

Concept explainers

Ionic Equilibrium

Chemical equilibrium and ionic equilibrium are two major concepts in chemistry. Ionic equilibrium deals with the equilibrium involved in an ionization process while chemical equilibrium deals with the equilibrium during a chemical change. Ionic equilibri…

Arrhenius Acid

Arrhenius acid act as a good electrolyte as it dissociates to its respective ions in the aqueous solutions. Keeping it similar to the general acid properties, Arrhenius acid also neutralizes bases and turns litmus paper into red.

Bronsted Lowry Base In Inorganic Chemistry

Bronsted-Lowry base in inorganic chemistry is any chemical substance that can accept a proton from the other chemical substance it is reacting with.

Videos

Question

Chapter 16, Problem 122IE

Interpretation Introduction

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

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 16, Problem 121IE

Chapter 16, Problem 123IE

Students have asked these similar questions

All of these compounds would be produced (I think). In my book, I don't see any rules about yield in this case, like explaining that one product would be present in less yield for this reason or that reason. Please explain why some of these produce less yield than others.

5. Fill in the missing molecules in the following reaction pathway. TMSO Heat + CI then HF O₂N (1.0 equiv) AICI 3 OMe

e. O₂N NO2 1. excess H2, Pd/C 2. excess NaNO2, HCI 3. excess CuCN

Chapter 16 Solutions

CHEMISTRY:CENTRAL SCIENCE-W/MOD.ACCESS

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

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.

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

Concept explainers

Videos

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

Want to see the full answer?

Chapter 16 Solutions