3.a solution has 64 grams of HI dissolved in 250 L of water. what is its [H+]

4. What is the pH of the solution in question #3?
5. A solution has 189 grams of HNO3 dissolved in 1000 L of water. What is pH?

First off, remember HI is a strong acid; i.e., it ionizes 100%. Therfore, (H^+) = (HI).

mols HI = grams/molar mass
M HI = mols HI/L solution
(H^+) = (HI)
pH = -log(H^+)

5 is done the same way since HNO3 is a strong acid also. I should point out it is stated incorrectly. The way it is stated apparently doesn't take into account that the solution volume is not the same as the added H2O. By the way, is that 1000 L or 1000 mL.

1000L

What is the pH of the solution in question 3

To find the answers to these questions, we need to use the concepts of molarity, dissociation, and pH.

1. [H+]:
To find the [H+], we need to calculate the molarity of HI. Molarity (M) is defined as the number of moles of solute per liter of solution.

First, we need to find the number of moles of HI. The molar mass of HI is 127.91 g/mol:
moles of HI = mass of HI / molar mass of HI
moles of HI = 64 g / 127.91 g/mol ≈ 0.5001 mol

Next, we need to calculate the molarity:
Molarity (M) = moles of solute / volume of solution (in liters)
Molarity (M) = 0.5001 mol / 250 L = 0.002001 M

Therefore, the [H+] is equal to the molarity of HI, which is 0.002001 M.

2. pH:
The pH is a measure of the acidity or alkalinity of a solution. It is defined as the negative logarithm (base 10) of the concentration of H+ ions in a solution.

To find the pH, we can use the equation:
pH = -log10([H+])

For the solution in question 3:
pH = -log10(0.002001) ≈ 2.7

Therefore, the pH of the solution in question 3 is approximately 2.7.

3. pH of HNO3 solution:
The molar mass of HNO3 is 63.01 g/mol.

First, we need to find the number of moles of HNO3:
moles of HNO3 = mass of HNO3 / molar mass of HNO3
moles of HNO3 = 189 g / 63.01 g/mol ≈ 3.0004 mol

Next, we calculate the molarity of HNO3:
Molarity (M) = moles of solute / volume of solution (in liters)
Molarity (M) = 3.0004 mol / 1000 L = 0.0030004 M

To find the pH, we use the same equation as before:
pH = -log10([H+])

For the HNO3 solution:
pH = -log10(0.0030004) ≈ 2.5

Therefore, the pH of the solution in question 5 is approximately 2.5.