The initial pH of the solution containing barium hydroxide is to be determined. Concept introduction: A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Barium hydroxide is a strong base while hydrochloric acid is a strong acid. When these two chemical species react with each other, they form their ionic forms and a neutral solution.
The initial pH of the solution containing barium hydroxide is to be determined. Concept introduction: A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Barium hydroxide is a strong base while hydrochloric acid is a strong acid. When these two chemical species react with each other, they form their ionic forms and a neutral solution.
The initial pH of the solution containing barium hydroxide is to be determined.
Concept introduction:
A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Barium hydroxide is a strong base while hydrochloric acid is a strong acid. When these two chemical species react with each other, they form their ionic forms and a neutral solution.
(a)
Expert Solution
Answer to Problem 64E
The pH of the initial barium hydroxide solution is 13.30 .
Explanation of Solution
Initial pH of the analyte solution; Ba(OH)2 is a strong base. Therefore, the reaction of this base with water is Ba(OH)2+H2O→Ba+2+2OH−+H2O . Then the initial pOH of the solution can be calculated using the concentration of the base directly. There, the stoichiometry of the above reaction should be considered (one mole of barium hydroxide gives two moles of hydroxide ion).
Using the relationship between pH and pOH, we can determine the initial pH of the analyte solution.
14=pH+pOH14=pH+0.70pH=13.30
(b)
Interpretation Introduction
Interpretation:
The pH of the barium hydroxide solution when 20.0mL of acid is added to it is to be determined.
Concept introduction:
A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Barium hydroxide is a strong base while hydrochloric acid is a strong acid. When these two chemical species react with each other, they form their ionic forms and a neutral solution.
(b)
Expert Solution
Answer to Problem 64E
The pH of the resulting solution is 12.91 .
Explanation of Solution
When 20.0mL of the acid is added to the base, some amount of the base reacts with the acid. Therefore, we can determine the pH of the resulting solution by calculating the remaining amount of base in the solution.
Total amount of base =0.100mol/L×80.0×10−3L=8×10−3mol
Amount of the acid added to the base =0.400mol/L×20.0×10−3L=8×10−3mol
The reaction between barium hydroxide and hydrochloric acid is as follows:
Ba(OH)2+2HCl→BaCl2+2H2O
This means two moles of the acid react with one mole of the base. Therefore, only 4×10−3mol of the base is reacted with the added acid. Then 4×10−3mol of base remains in the solution. One mole of barium hydroxide produces two moles of hydroxide ions.
moles of Ba(OH)2=4×10−3molmoles of OH−=4×10−3mol×2=8×10−3molConcentation of OH−=8×10−3mol(80.0+20.0)×10−3L=0.08mol/L
Using this value, the pH of the solution can be determined.
The pH of the barium hydroxide solution when 30.0mL of acid is added to it is to be determined.
Concept introduction:
A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Barium hydroxide is a strong base while hydrochloric acid is a strong acid. When these two chemical species react with each other, they form their ionic forms and a neutral solution.
(c)
Expert Solution
Answer to Problem 64E
The pH of the resulting solution is 12.56 .
Explanation of Solution
When 30.0mL of the acid is added to the base, some amount of the base reacts with the acid. Therefore, we can determine the pH of the resulting solution by calculating the remaining amount of base in the solution.
Total amount of base =0.100mol/L×80.0×10−3L=8×10−3mol
Amount of the acid added to the base =0.400mol/L×30.0×10−3L=12×10−3mol
The reaction between barium hydroxide and hydrochloric acid is as follows:
Ba(OH)2+2HCl→BaCl2+2H2O
This means two moles of the acid react with one mole of the base. Therefore, only 12×10−3mol2=6×10−3mol of the base is reacted with the added acid. Then (8−6)×10−3mol=2×10−3mol of base remains in the solution. One mole of barium hydroxide produces two moles of hydroxide ions.
moles of Ba(OH)2=2×10−3molmoles of OH−=2×10−3mol×2=4×10−3molConcentation of OH−=4×10−3mol(80.0+30.0)×10−3L=0.036mol/L
Using this value, the pH of the solution can be determined.
The pH of the barium hydroxide solution when 40.0mL of acid is added to it is to be determined.
Concept introduction:
A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Barium hydroxide is a strong base while hydrochloric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.
(d)
Expert Solution
Answer to Problem 64E
The pH of the resulting solution is 7.00 .
Explanation of Solution
When 20.0mL of the acid is added to the base, some amount of the base reacts with the acid. Therefore, we can determine the pH of the resulting solution by calculating the remaining amount of base in the solution.
Total amount of base =0.100mol/L×80.0×10−3L=8×10−3mol
Amount of the acid added to the base =0.400mol/L×40.0×10−3L=16×10−3mol
The reaction between barium hydroxide and hydrochloric acid is as follows:
Ba(OH)2+2HCl→BaCl2+2H2O
This means two moles of the acid react with one mole of the base. Therefore, 16×10−3mol2=8×10−3mol of the base is reacted with the added acid. Then there is no base remains in the solution. At this pH, the solution becomes neutral and the pH is 7.00 .
(e)
Interpretation Introduction
Interpretation:
The pH of the barium hydroxide solution when 80.0mL of acid is added to it is to be determined.
Concept introduction:
A strong acid or a strong base can completely dissociate into its ions when they are in an aqueous solution. Barium hydroxide is a strong base while hydrochloric acid is a strong acid. When these two chemical species are reacted with each other, they form their ionic forms and a neutral solution.
(e)
Expert Solution
Answer to Problem 64E
The pH of the resulting solution is 1.30 .
Explanation of Solution
When 20.0mL of the acid is added to the base, some amount of the base reacts with the acid. Therefore, we can determine the pH of the resulting solution by calculating the remaining amount of base in the solution.
Total amount of base =0.100mol/L×80.0×10−3L=8×10−3mol
Amount of the acid added to the base =0.400mol/L×80.0×10−3L=32×10−3mol
The reaction between barium hydroxide and hydrochloric acid is as follows:
Ba(OH)2+2HCl→BaCl2+2H2O
This means two moles of the acid react with one mole of the base. Therefore, 32×10−3mol2=16×10−3mol of the base is reacted with the added acid. Then there is no base remains in the solution and there is excess of acid in the solution.
The amount of acid in the solution =16×10−3mol−8×10−3mol=8×10−3mol
Concentration of the acid =8×10−3mol(80.0+80.0)×10−3L=0.05mol/L
One mole of acid produces one mole of hydrogen ion. Therefore,
pH=−log[H+]=−log[0.05]=1.30
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IX) By writing the appropriate electron configurations and orbital box diagrams briefly EXPLAIN in your own words
each one of the following questions:
a) The bond length of the Br2 molecule is 2.28 Å, while the bond length of the compound KBr is 3.34 Å. The radius of
K✶ is 1.52 Å. Determine the atomic radius in Å of the bromine atom and of the bromide ion.
Br =
Br
b) Explain why there is a large difference in the atomic sizes or radius of the two (Br and Br).
T
When 15.00 mL of 3.00 M NaOH was mixed in a calorimeter with 12.80 mL of 3.00 M HCl, both initially at room temperature (22.00 C), the temperature increased to 29.30 C. The resultant salt solution had a mass of 27.80 g and a specific heat capacity of 3.74 J/Kg. What is heat capacity of the calorimeter (in J/C)? Note: The molar enthalpy of neutralization per mole of HCl is -55.84 kJ/mol.
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Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell