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## 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 base Name $CH_{3}CH_{2}OH$ 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 $(1.74 \times 10^{-5})$.