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What is the pH of a 0.28 M solution of ascorbic acid (Vitamin C)? (The values for Ka1 and Ka2 for ascorbic acid are 8.0×10−5 and 1.6×10−12, respectively.) What is the of a 0.28 solution of ascorbic acid (Vitamin C)? (The values for and for ascorbic acid are and , respectively.) 2.04 2.32 2.82 4.65 6.17

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Olajidey

Answer:

pH = 2.32

Explanation:

H2A + H2O -------> H3O+ + HA-    

Ka2 is very less so i am not considering that dissociation.

now Ka = 8.0×10−5

            = [H3O+] [HA-] / [H2A]

lets concentration of H3O+ = X then above equation will be

8.0×10−5 = [X] [X] / [0.28 -X]

8.0×10−5 = X2 /  [0.28 -X]

X2 + 8.0×10−5 X - 2.24 x 10−5

solve the quardratic equation

X =0.004693 M

pH = -log[H+}

    = -log [0.004693]

    = 2.3285

    ≅2.32

pH = 2.32

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