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

• 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

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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