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. Given the \(K_{a}\) values, estimate the \(p K_{a}\) value of each of the following acids without using a calculator (that is, is it between 3 and4, between 9 and 10, and so on?):
1. nitrous acid \(\left(H N O_{2}\right), K_{a}=4.0 \times 10^{-4}\) 3. bicarbonate \(\left(\mathrm{HCO} 3^{-}\right), K_{a}=6.3 \times 10^{-11}\)
2. nitric acid \(\left(H N O_{3}\right), K_{a}=22\) 4. hydrogen cyanide \((H C N), K_{a}=7.9 \times 10^{-10}\)
5. formic acid \((H C O O H), K_{a}=2.0 \times 10^{-4}\)
b. Determine the exact \(p K_{a}\) values, using a calculator.
c. Which is the strongest acid?
Solution
Solution:
Here we will have to calculate the pKa value.
Step 1
a.
1.Nitrous acid (HNO2), Ka = 4.0 10-4
It is known that pKa = -log Ka
The Ka of HNO2 is =4.0 10-4
Substituting the value of Ka , pKa = -log (4.0 10-4)
=-log 4 + 4
Thus pKa will be between 3 and 4.
2. Nitric acid (HNO3), Ka = 22
Substituting the value of Ka, pKa = -log (22)
= -1.34
Thus pKa will be between -2 and -1.
3.Bicarbonate (HCO3-), Ka = 6.310-11
Substituting the value of Ka , pKa = -log (6.310-11)
= -log 6.3 +11
Thus pKa will be between 10 to 11.
4.Hydrogen cyanide (HCN), Ka = 7.910-10
Substituting the value of Ka, pKa = -log (7.910-10)
= -log 7.9 +10
Thus pKa will be between 9 and 10.
5.Formic acid (HCOOH), Ka = 2.010-4
Substituting the value of Ka, pKa = -log (2.010-4)
= -log 2.0 + 4
Thus pKa will be between 3 and 4.
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