What is the concentration of H+ in 0.50 M hydroiodic acid?

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3 Answers

  • HI is a halogen acid. Except for HF, all halogen acids are strong acids. They dissociate completely in water solution.

    HI → H+ + I-

    [HI] = [H+]

    Therefore, in a 0.50M solution of HI, [H+] = 0.50M

  • Every chemist should know that except for HF the HX acids (X = Cl, Br,I) are v strong acids and the acid strength follows the order HI>HBr>HCl.

    HI + H2O ⇋ H3O^+ + I^-

    Ka = [H3O^+][I-]/[HI]

    pKa = -logKa = pKa = -9.5 hence Ka is 3.16 × 10^10 (i.e. +10)

    Hence we can assume complete dissociation:

    HI + H2O → H3O^+ + I^-

    0.50M of HI will give 0.5 [H^+] pH = -log(0.5) = 0.30

  • What is the concentration of H+ in 0.50 M hydroiodic acid?

    http://en.wikipedia.org/wiki/Hydrogen_iodide

    From the website above, pKa = -9.5

    Ka = 10^-9.5 = 3.16 * 10^-10

    Ka = [H+] * [I-] ÷ [HI]

    concentration of H+ = [H+]

    1 liter of a 0.50 M Hl solution contains 0.50 mole of HI.

    As 0.50 mole of HI dissociate in H2O, x mole of H+ and x mole of I- are produced; leaving 0.50 – x mole of HI.

    (x * x) ÷ (0.50 – x) = 3.16 * 10^-10

    x^2 = (0.50 – x) * 3.16 * 10^-10

    x^2 = 1.58 * 10^-10 – 3.16 * 10^-10 * x

    x^2 + 3.16 * 10^-10 * x – 1.58 * 10^-10 = 0

    Solve quadratic equation for x

    I use the website below to solve quadratic equations.

    http://www.math.com/students/calculators/source/qu...

    x = 0.000012569647090969549 ≈ 1.257 * 10^-5 = [H+]

    OR

    (x * x) ÷ (0.50 – x) = 3.16 * 10^-10

    x^2 = (0.50 – x) * 3.16 * 10^-10

    Since 3.16 * 10^-10 is much smaller than 0.50, neglect the – x in (0.05 – x)

    x^2 = 0.50 * 3.16 * 10^-10 = 1.58 * 10^-10

    x = √(1.58 * 10^-10) = 1.257 * 10^-5 = [H+]

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