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