Answer:
The correct answer is D) 7.06.
Step-by-step explanation:
The first step in solving this problem is to write the balanced chemical equation for the reaction that occurs when HClO and KClO are mixed:
HClO + KClO → K+ + ClO- + HClO
The reaction is a neutralization reaction in which HClO acts as an acid and KClO acts as a base.
Next, we need to calculate the moles of acid (HClO) and base (KClO) that are present in the solution after mixing. We can use the formula:
moles = concentration × volume
For HClO, we have:
moles of HClO = (0.30 mol/L) × (0.200 L) = 0.060 mol
For KClO, we have:
moles of KClO = (0.20 mol/L) × (0.100 L) = 0.020 mol
The HClO will react with the KClO to form ClO- and H3O+ ions:
HClO + ClO- → H2O + ClO2-
So, we can set up an ICE (initial, change, equilibrium) table to determine the concentration of H3O+ ions at equilibrium:
HClO ClO- H3O+
Initial 0.060 M 0.020 M 0 M
Change -x -x +x
Equilibrium 0.060-x 0.020-x x
The value of x represents the concentration of H3O+ ions at equilibrium. We can use the equilibrium concentrations to set up an expression for the acid dissociation constant (Ka) of HClO:
Ka = [H3O+][ClO-] / [HClO]
Substituting in the equilibrium concentrations, we get:
Ka = x^2 / (0.060 - x)
The value of x can be calculated using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / 2a
where a = 1, b = -Ka, and c = Ka × 0.060.
After substituting these values into the quadratic formula, we get:
x = 7.63 × 10^-5 M
The pH of the solution is given by:
pH = -log[H3O+]
Substituting the value of [H3O+] into this equation, we get:
pH = -log(7.63 × 10^-5) ≈ 4.12
However, we need to consider the dilution factor when mixing the two solutions. The total volume of the solution after mixing is:
V = 200.0 mL + 100.0 mL = 300.0 mL = 0.300 L
The concentration of H3O+ ions in the final solution is:
[H3O+] = x / V = 7.63 × 10^-5 M / 0.300 L ≈ 0.00025 M
Taking the negative logarithm of this concentration gives:
pH = -log(0.00025) ≈ 3.60
Therefore, the pH of the solution is approximately 3.60. This corresponds to answer choice E).