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How do I make a Vitamin C standard solution?

How do you find the pH of this solution?

  • Answer:

    Moles of acid =0.5g / 175 = 0.00286 moles. Moles of H+ ions is , 0.00286 x 2 = 0.00571 = 5.71x10^3 moles H+ ions pH = -log of 5,71x10^-3 3.00 - log of .571 = 0.76 pH = 2.24

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Moles of acid = 0.5g / (176.12g/mol) = 0.0028 mol Molarity = 0.0028mol / 0.2L = .014M https://en.wikipedia.org/wiki/Ascorbic_acid give you Ka1 = 7.9*10^-5, Ka2 = 1.6*10^12 Using ICE table, Ka1 = 7.9*10^-5 = [H+]*[HC6H6O6-] / [H2C6H6O6] = x^2 / (0.014M - x), Quadratic formula, x= 0.001 = [H+] = [HC6H6O6-] (unecessary because the H+ contribution is so little) Ka2 = 1.6*10^-12 = [H+]*[C6H6O6^2-] / [HC6H6O6-] Ka2 = x^2 / (0.001M -x) *you can ignore the x from 0.001M - x because the Ka2 value is so small compare to 0.001 x = sqrt(0.001M * 1.6*10^-12) = 4*10^-8 pH = -log(1 + 4*10^-8) = 3

Vu

Moles of acid =0.5g / 175 = 0.00286 moles. Moles of H+ ions is , 0.00286 x 2 = 0.00571 = 5.71x10^3 moles H+ ions pH = -log of 5,71x10^-3 3.00 - log of .571 = 0.76 pH = 2.24

John W

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