What would be the pH of a solution of hy-

poiodous acid (HOI) prepared by dissolving
144 grams of the acid in 200 mL of pure water
(H2O)? The Ka of hypoiodous acid is 2¡¿10−11

1. 7
2. 10
3. 13
4. 5
5. 1

HOI ==> H^+ + OI^-

Ka = (H^+)(OI^-)/(HOI)
At equilibrium,
(H^+) = x
(OI^-) = x
(HOI) = moles/L. moles = 144 g/molar mass HOI and L = 0.2. So at equilibrium (HOI) = M-x
Solve for x and convert to pH.

said it twice for added effect......

To determine the pH of a solution of hypoiodous acid (HOI), we need to consider the dissociation of the acid and the equilibrium expression. The dissociation of HOI can be represented as follows:

HOI ⇌ H+ + OI-

Based on the given information, the Ka (acid dissociation constant) of HOI is 2 x 10^-11. The Ka expression for this reaction is:

Ka = [H+][OI-] / [HOI]

Since HOI is a monoprotic acid, the concentration of [H+] will be equal to the concentration of [HOI] at equilibrium. Thus, we can assume that [H+] = [HOI] and simplify the expression:

Ka = [H+]^2 / [HOI]

Now, we can substitute the given values into the equation.

1. Convert 144 grams of HOI to moles using its molar mass:
Moles of HOI = mass / molar mass = 144 g / (127.913 g/mol) = 1.125 mol

2. Convert 200 mL of water to liters:
Volume of water = 200 mL = 0.2 L

3. Calculate the initial concentration of HOI:
Initial concentration of HOI = moles / volume = 1.125 mol / 0.2 L = 5.625 M

4. Substitute the values into the Ka expression:
2 x 10^-11 = [H+]^2 / 5.625

5. Rearrange the equation to solve for [H+]:
[H+]^2 = (2 x 10^-11) * 5.625
[H+]^2 = 1.125 x 10^-10
[H+] = √(1.125 x 10^-10)

6. Calculate the pH using the formula:
pH = -log[H+]

Now, you can solve the equation to find the value of [H+] and then calculate the pH. Once you have the pH, you can select the corresponding option from the given choices.