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Calculate the analytical and equilibrium molar concentrations of the solute species in an aqueous solution that contains 285 mg of trichloroacetic acid, Cl,ccoOH (163.4 g/mol), in 10.0 mL (the acid is 73% ionized in water).

Calculate the analytical and equilibrium molar concentrations of the solute species in an aqueous solution that contains 285 mg of trichloroacetic acid, Cl,ccoOH (163.4 g/mol), in 10.0 mL (the acid is 73% ionized in water).

Principles of Modern Chemistry
Principles of Modern Chemistry
8th Edition
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Chapter16: Solubility And Precipitation Equilibria
Section: Chapter Questions
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I want this example to find a value mole fraction and molality. Please help me. And thank you so much
1 % YA . N Fundamentals+of+a... -> into a mixture of H', HSO, , SO, ions. 1 here are essentially no H,SO, molecules in this solution. The equilibrium concentrations of the ions are 1.01, 0.99, and 0.01 M, respectively. Equilibrium molar concentrations are usually symbolized by placing square brack- ets around the chemical formula for the species. So, for our solution of H,SO, with an analytical concentration of qH so, =1.0 M, we write [H,SO] = 0.00 M [H*] = 1.01 M [HSO,] = 0.99 M [SO,?-] = 0.01 M EXAMPLE 4-4 Calculate the analytical and equilibrium molar concentrations of the solute species in an aqueous solution that contains 285 mg of trichloroacetic acid, Cl,CCOOH (163.4 g/mol), in 10.0 mL (the acid is 73% ionized in water). Solution As in Example 4-3, we calculate the number of moles of Cl,CCOOH, which we desig- nate as HA, and divide by the volume of the solution, 10.0 mL, or 0.0100 L. Therefore, 1g HA 1 mol HA 163.4 g HA amount HA = nHA = 285 mg HA X 1000 mg HA = 1.744 X 10-3 mol HA The molar analytical concentration, CHA, is then 1.744 X 10 mol HA 1000 mŁ mol HA CHA = 0.174 = 0.174 M 10.0 mŁ 1 L L In this solution, 73% of the HA dissociates, giving H* and A: HA=H* + A The equilibrium concentration of HA is then 27% of CHA- Thus, (HA] = CHA X (100 – 73)/100 = 0.174 X 0.27 = 0.047 mol/L = 0.047 M The equilibrium concentration of A is equal to 73% of the analytical concentration of HA, that is, 73 mol A mol HA [A] = X 0.174 = 0.127 M 100 mol HA (continued) Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied., scanned, or duplicated, in whole or in part. Due to electronic rights, some thind party coment may be suppressed deemed that any suppressed content does not materially affect the overall leaming experience. Cengage Learning reserves the right to remove additional content at any time i CНАPТER 4 Calculations Used in Analytical Chemistry Because 1 mole of H* is formed for each mole of A", we can [H*] [A"] = 0.127 M and ...
Transcribed Image Text:1 % YA . N Fundamentals+of+a... -> into a mixture of H', HSO, , SO, ions. 1 here are essentially no H,SO, molecules in this solution. The equilibrium concentrations of the ions are 1.01, 0.99, and 0.01 M, respectively. Equilibrium molar concentrations are usually symbolized by placing square brack- ets around the chemical formula for the species. So, for our solution of H,SO, with an analytical concentration of qH so, =1.0 M, we write [H,SO] = 0.00 M [H*] = 1.01 M [HSO,] = 0.99 M [SO,?-] = 0.01 M EXAMPLE 4-4 Calculate the analytical and equilibrium molar concentrations of the solute species in an aqueous solution that contains 285 mg of trichloroacetic acid, Cl,CCOOH (163.4 g/mol), in 10.0 mL (the acid is 73% ionized in water). Solution As in Example 4-3, we calculate the number of moles of Cl,CCOOH, which we desig- nate as HA, and divide by the volume of the solution, 10.0 mL, or 0.0100 L. Therefore, 1g HA 1 mol HA 163.4 g HA amount HA = nHA = 285 mg HA X 1000 mg HA = 1.744 X 10-3 mol HA The molar analytical concentration, CHA, is then 1.744 X 10 mol HA 1000 mŁ mol HA CHA = 0.174 = 0.174 M 10.0 mŁ 1 L L In this solution, 73% of the HA dissociates, giving H* and A: HA=H* + A The equilibrium concentration of HA is then 27% of CHA- Thus, (HA] = CHA X (100 – 73)/100 = 0.174 X 0.27 = 0.047 mol/L = 0.047 M The equilibrium concentration of A is equal to 73% of the analytical concentration of HA, that is, 73 mol A mol HA [A] = X 0.174 = 0.127 M 100 mol HA (continued) Copyright 2013 Cengage Learning. All Rights Reserved. May not be copied., scanned, or duplicated, in whole or in part. Due to electronic rights, some thind party coment may be suppressed deemed that any suppressed content does not materially affect the overall leaming experience. Cengage Learning reserves the right to remove additional content at any time i CНАPТER 4 Calculations Used in Analytical Chemistry Because 1 mole of H* is formed for each mole of A", we can [H*] [A"] = 0.127 M and ...
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