The acid with larger Ka value should be detected and reason of larger value should be explained. Concept Introduction: The value of K a , gives the dissociation constant for an acid. Hence, stronger the value of acid, more will be dissociation and so larger will be the value of K a .
The acid with larger Ka value should be detected and reason of larger value should be explained. Concept Introduction: The value of K a , gives the dissociation constant for an acid. Hence, stronger the value of acid, more will be dissociation and so larger will be the value of K a .
Solution Summary: The author explains the titration curve for acid A and B, where the amount of base solution is stoichiometrically enough to neutralize the sample.
The acid with larger Ka value should be detected and reason of larger value should be explained.
Concept Introduction:
The value of Ka, gives the dissociation constant for an acid. Hence, stronger the value of acid, more will be dissociation and so larger will be the value of Ka.
(b)
Interpretation Introduction
Interpretation:
pH should be estimated for each acid and difference between their values should be explained.
Concept Introduction:
Equivalence point is the point of titration curve where, the amount of titrant added is stoichiometrically enough to neutralize the amount of sample. It is detected at halfway of the steep curve.
The pH at this point will be the equivalent pH.
(c)
Interpretation Introduction
Interpretation:
If volume of acids A and B are taken equal then, acid with higher initial concentration should be determined.
Concept Introduction:
pH gives the concentration of H+ ions, which can explain the concentration of the acid.pH=−log[H+]
Q3: Write in the starting alkyl bromide used to form the following products. Include any
reactants, reagents, and solvents over the reaction arrow. If more than one step is
required, denote separate steps by using 1), 2), 3), etc.
H
OH
racemic
OH
OH
5
racemic
Draw the Lewis structure of the SO3-O(CH3)2 complex shown in the bottom right of slide 2in lecture 3-3 (“Me” means a CH3 group) – include all valence electron pairs and formal charges.From this structure, should the complex be a stable molecule? Explain.
Predict all organic product(s), including stereoisomers when applicable.
Chapter 12 Solutions
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