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Chapter 2

Q. 2.5

Calculating K_{a} from p K_{ a }

According to the data in Table 2.3, acetic acid has p K_{ a } = 4.76. What is its K_{ a } ?


TABLE 2.3 relative Strengths of Some Common acids and their Conjugate Bases
Acid Name p K_{ a } Conjugate

CH_{3}CH_{2}OH Ethanol 16.00 CH_{3}CH_{2}O^{-} Ethoxide ion  

H_{2}O Water 15.74 HO^{-} Hydroxide ion
HCN Hydrocyanic acid 9.31 CN^{-} Cyanide ion
H_2PO_4^{-} Dihydrogen phosphate ion 7.21 {HPO_4}^{2-} Hydrogen phosphate ion
CH_{3}CO_2H Acetic acid 4.76 CH_{3}CO_2^{-} Acetate ion
H_{3}PO_4 Phosphoric acid 2.16 {H_{2}PO_{4}}^- Dihydrogen phosphate ion
HNO_{3} Nitric acid -1.3 {NO_{3}}^- Nitrate ion
HCI Hydrochloric acid -7.0 Cl^{-} Chloride ion

S t r a t e g y
Since p K_{ a } is the negative logarithm of K_{ a } , it’s necessary to use a calculator with an ANTILOG or INV LOG function. Enter the value of the p K_{ a } (4.76), change the sign (4.76), and then find the antilog (1.74 \times 10^{-5}) .


Verified Solution

K_{ a }= antilog – 4.76 (1.74 \times 10^{-5}) .