What is the concentration of H+ in 0.50 M hydroiodic acid?
What is the concentration of H⁺ ions in a 0.50 M solution of hydroiodic acid (HI)? Please provide a detailed explanation of the steps involved in your calculation.
3 Answers
Jan 05, 2025
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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