What is the pH of a solution prepared by dissolving 0.13 grams of HI (molar mass HI = 127.9 g/mol) in 300 mL of solution?

Question

Quizplus · Accepted Answer

Step 1: Calculate the number of moles of HI:
0.13 g HI x (1 mol HI/127.9 g HI) = 0.00102 mol HI

Step 2: Calculate the molarity of the solution:
Molarity = (moles of solute)/(volume of solution in L)
Volume of solution = 300 mL = 0.3 L
Molarity = 0.00102 mol/0.3 L = 0.0034 M

Step 3: HI is a strong acid, so it will completely dissociate in water:
HI → H+ + I-

Step 4: Calculate the concentration of H+ ions:
[H+] = 0.0034 M

Step 5: Calculate the pH:
pH = -log[H+]
pH = -log(0.0034)
pH = 2.47

Therefore, the answer is B.

What Is the PH of a Solution Prepared by Dissolving