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

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Mina Herzog · Accepted Answer

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

Mina Herzog · Answer

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

Mina Herzog · Answer

What is the concentration of H+ in 0.50 M hydroiodic acid?http://en.wikipedia.org/wiki/Hydrogen_iodideFrom the website above, pKa = -9.5Ka = 10^-9.5 = 3.16 * 10^-10Ka = [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 * xx^2 + 3.16 * 10^-10 * x – 1.58 * 10^-10 = 0Solve quadratic equation for xI 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^-10x = √(1.58 * 10^-10) = 1.257 * 10^-5 = [H+]

Lavinia Carroll

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

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