a. List the following alcohols in order from strongest acid to weakest acid: CCl3CH2OH Ka = 5.75 1013 Ka = 1.29 1013 Ka = 4.90 1013 CH2ClCH2OH CHCl2CH2OH b. Explain the relative acidities.
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Textbook Solutions for Organic Chemistry
Question
a. Indicate whether a carboxylic acid \((\mathrm{RCOOH})\) with a \(p K_{a}\) value of \(4.5\) will have more charged molecules or more neutral molecules in a solution with the following \(p H\):
1. \(p H=1\) 3. \(p H=5\) 5. \(p H=10\)
2. \(p H=3\) 4. \(p H=7\) 6. \(p H=13\)
b. Answer the same question for a protonated amine with a \(p K_{a}\) value of \(9\).
c. Answer the same question for an alcohol \((R O H)\) with a \(p K_{a}\) value of \(15\).
Solution
Solution 38P
Step 1 of 3:
Here, we are asked to indicate if the given compound with a given pKa value will have more charged molecules or more neutral molecules in a solution for given pH values.
If the pH value of the solution is lesser than the pKa value, then it exists in its acidic form and will have more neutral molecules.
If the pH value of the solution is greater than the pKa value, then it exists in its basic form will have more charged molecules.
a.
Given Compound: carboxylic acid (RCOOH) with a pKa of 4.5.
A carboxylic acid is neutral in its acidic form (RCOOH) and charged in its basic form (RCOO-).
Hence, a Carboxylic acid at pH = 1 and pH =3 with a pKa of 4.5 has more neutral molecules(RCOOH). A Carboxylic acid at pH = 5,7,10 and 13 with a pKa of 4.5 has more charged molecules(RCOO-).
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