Question

The pH of a solution prepared by the addition of 100mL 0.002M HCL to 100mL distilled water is closest to:

a.1.0
b.1.5
c.2.0
d.3.0

Answer 1

The pH of a solution resulting from adding 100 mL 0.002M HCl to 100 mL distilled water is 3.0, since the concentration is halved in the final mixture and HCl fully dissociates.

Step-by-step explanation:

The student asked about the pH of a solution prepared by the addition of 100 mL 0.002M HCl to 100 mL distilled water. To calculate this, we first need to find the new concentration of HCl in the final solution. Since the volumes are equal, the concentration of HCl in the final 200 mL solution will be half the original concentration, which is 0.001M (0.002M / 2). The pH of a solution is given by the negative logarithm (base 10) of the hydrogen ion concentration (pH = -log[H+]). For a 0.001M HCl solution, which is a strong acid and fully dissociates, the pH is therefore:

pH = -log(0.001) = 3.0

Thus, the correct answer is (d) 3.0.

Answer 2

d.3.0

Step-by-step explanation:

Step 1: Calculate the final volume of the solution

The final volume is equal to the sum of the volumes of the initial HCl solution and the volume of distilled water.

V₂ = 100 mL + 100 mL = 200 mL

Step 2: Calculate the final concentration of HCl

We will use the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁/V₂ = 0.002 M × 100 mL/200 mL = 0.001 M

Step 3: Calculate the pH of the final HCl solution

Since HCl is a strong acid, [H⁺] = HCl. We will use the definition of pH.

pH = -log [H⁺] = -log 0.001 = 3