1. Tonic water, quinine, and malaria Quinine (C2oH24N202, MW 324 g/mol) has two

Question

1. Tonic water, quinine, and malaria Quinine (C2oH24N202, MW 324 g/mol) has two amine groups with pKb 5.10 and pKb2 9.70. Quinine was originally extracted from the bark of the cinchona tree, which grows in the rain forest on the eastern slopes of the Andes. The Spanish first heard about the medicinal properties of the "fever tree" from the natives in the early 17h century. According to tradition, this bark was used to cure Countess Anna del Chinchon, wife of the viceroy of Peru, in 1638. Because of this, botanists later named the tree for her. | For 300 years, quinine was the only effective treatment for malaria. Since quinine is also found in tonic water, there is an urban legend that the British Army prescribed gin and tonics to troops stationed in India as a malaria prophylactic. Alas, this is only a legend, as one would have to consume more than 10 gin and tonics a day to pe effective. Calculate the concentrations of all ions in a saturated solution of quinine which is made from 1.00 g of quinine dissolved in 1.90 liters of water. Explicitly state and test any assumptions.

Explanation / Answer

[quinine] = 1 g/324 g/mol x 1.90 L = 0.0016 M

quinine + H2O <==> quinineH+ + OH-

let x amount reacted

Kb1 = [quinineH+][OH-]/[quinine]

7.9 x 10^-6 = x^2/0.0016

x = [OH-] = 1.12 x 10^-4 M

[quinine] = 0.0016 - 1.12 x 10^-4 = 0.0015 M

[H3O+] = 8.93 x 10^-11 M

[quinineH+] = 1.12 x 10^-4 M

then,

quinineH+ + H2O <=> quinineH2^2+ + OH-

let x amount reacted

Kb2 = [quinineH2^2+][OH-]/[quinineH+]

2 x 10^-10 = x^2/1.12 x 10^-4

[quinineH2^2+] = x = 1.50 x 10^-7 M

So,

concentrations of,

[OH-] = 1.12 x 10^-4 M

[quinine] = 0.0015 M

[H3O+] = 8.93 x 10^-11 M

[quinineH+] = 1.12 x 10^-4 M

[quinineH2^2+] = 1.50 x 10^-7 M

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