What is the concentration of H in 0.50 M hydroiodic acid [Solved]

please present steps.

HI is a halogen acid. Apart from HF, all halogen acids are sturdy acids. They dissociate utterly in water answer.
HI H I-
[HI] = [H ]
Subsequently, in a 0.50M answer of HI, [H ] = 0.50M

Each chemist ought to know that aside from HF the HX acids (X = Cl, Br,I) are v sturdy acids and the acid energy follows the order HI>HBr>HCl.
HI H2O H3O^ I^-
Ka = [H3O^ ][I-]/[HI]
pKa = -logKa = pKa = -9.5 therefore Ka is 3.16 10^10 (i.e. 10)
Therefore we will assume full dissociation:
HI H2O H3O^ I^-
0.50M of HI will give 0.5 [H^ ] pH = -log(0.5) = 0.30

Whats the focus of H in 0.50 M hydroiodic acid
http://en.wikipedia.org/wiki/Hydrogen_iodide
From the web site above, pKa = -9.5
Ka = 10^-9.5 = 3.16 * 10^-10
Ka = [H ] * [I-] [HI]
focus of H = [H ]
1 liter of a 0.50 M Hl answer comprises 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
Remedy quadratic equation for x
I exploit the web site beneath to resolve 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 far 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 ]

This post is last updated on hrtanswers.com at Date : 1st of September – 2022