Question

A student mixes 100. mL of 0.25 M HCl(aq) with 200. mL of 0.50 M HClO4(aq) and then dilutes the mixture with distilled water to a total volume of 500. mL. The [H3O+] in the final solution is closest to

(A) 0.0025 M
(B) 0.12 M
(C) 0.25 M
(D) 0.75 M

Answer 1

The answer is B: 0.0025 M

Answer 2

The closest concentration of [H3O+] in the final solution is (C) 0.25 M.

To find the concentration of [H3O+] in the final solution, we need to consider several factors:

1. **Moles of HCl and HClO4:**

* Calculate the moles of each acid initially present:

* HCl: moles = M * V = 0.25 M * 0.1 L = 0.025 mol

* HClO4: moles = M * V = 0.50 M * 0.2 L = 0.1 mol

* Since both acids are dissociated in water, the total moles of H+ ions will be the sum of the individual moles: 0.025 mol + 0.1 mol = 0.125 mol.

2. **Dilution:**

* The total volume after dilution is 500 mL.

* The dilution factor is the final volume divided by the initial volume of the acid mixture: 500 mL / (100 mL + 200 mL) = 1.

3. **[H3O+] calculation:**

* Since the dilution factor is 1, the number of moles of H+ ions remains the same.

* Therefore, the final [H3O+] can be calculated using the final volume:

[H3O+] = moles / V = 0.125 mol / 0.5 L = 0.25 M.

Explanation of other options:

* (A) 0.0025 M: This is much lower than the actual concentration because it only considers the initial concentration of HCl, neglecting the contribution of HClO4.

* (B) 0.12 M: This is the sum of the initial concentrations of both acids, but it doesn't account for the dilution, which reduces the concentration.

* (D) 0.75 M: This is higher than the actual concentration because it assumes all H+ ions come from HCl, neglecting the contribution of HClO4 and the dilution effect.