When the pH of a solution is 1, the concentration of hydronium ions is M, which is the same as 0.1 M. Assume that the volume of this solution is 500 mL and that the solution is not buffered. What would be the new pH of this solution after 500 mL of pure water is added? You will need a calculator with a log function to answer this question.

Question

When the pH of a solution is 1, the concentration of hydronium ions is  M, which is the same as 0.1 M. Assume that the volume of this solution is 500 mL and that the solution is not buffered. What would be the new pH of this solution after 500 mL of pure water is added? You will need a calculator with a log function to answer this question.

Quizplus · Accepted Answer

To solve this problem, we will use the equation for calculating pH:

pH = -log [H3O+]

Initially, the pH of the solution is 1. This means that:

pH = -log [H3O+] = -log 0.1 = 1

Now, we add 500 mL of pure water to the solution. This will dilute the concentration of H3O+:

[H3O+]final = [H3O+]initial x (Vinitial / Vfinal)

where Vinitial is the initial volume of the solution (500 mL), Vfinal is the final volume of the solution (500 mL + 500 mL = 1000 mL), and [H3O+]initial is the initial concentration of H3O+ (0.1 M).

[H3O+]final = 0.1 x (500 / 1000) = 0.05 M

Now we can calculate the new pH of the solution:

pH = -log [H3O+] = -log 0.05 = 1.3

Therefore, the answer is B, and the new pH of the solution is 1.3.

When the PH of a Solution Is 1, the Concentration