If 16.0 mL of 8.5 ✕ 10-2 M HClO4 is added to 22.0 mL of 7.6 ✕ 10-4 M KOH, what is the pH of the solution?
To determine the pH of the solution, you need to follow a series of steps. Let's break it down:
Step 1: Write the balanced equation for the reaction between HClO4 and KOH.
HClO4 + KOH -> KClO4 + H2O
Step 2: Calculate the number of moles of HClO4 and KOH.
Moles of HClO4 = volume (in liters) × molarity
Moles of KOH = volume (in liters) × molarity
Given:
Volume of HClO4 = 16.0 mL = 0.016 L
Molarity of HClO4 = 8.5 × 10^-2 M
Volume of KOH = 22.0 mL = 0.022 L
Molarity of KOH = 7.6 × 10^-4 M
Moles of HClO4 = 0.016 L × 8.5 × 10^-2 M = 1.36 × 10^-3 mol
Moles of KOH = 0.022 L × 7.6 × 10^-4 M = 1.672 × 10^-5 mol
Step 3: Determine the limiting reagent.
Comparing the number of moles, HClO4 is in excess as it has a greater number of moles.
Step 4: Calculate the remaining moles of HClO4 after the reaction.
Since HClO4 is in excess, all of it will not react. Thus, all 1.36 × 10^-3 mol of HClO4 remains after the reaction.
Step 5: Calculate the concentration of OH- ions from the reaction.
Using the balanced equation, we can see that KOH is a strong base and completely ionizes in water.
So, the concentration of OH- ions from KOH is equal to the initial concentration of KOH.
Concentration of OH- ions = 7.6 × 10^-4 M
Step 6: Calculate the concentration of H3O+ ions.
Since the reaction involves a strong acid (HClO4) and a strong base (KOH), a neutralization reaction occurs.
From the balanced equation, we can see that the mole ratio between HClO4 and OH- is 1:1.
Therefore, the concentration of H3O+ ions is equal to the concentration of OH- ions.
Concentration of H3O+ ions = 7.6 × 10^-4 M
Step 7: Calculate the pH.
pH = -log[H3O+]
pH = -log(7.6 × 10^-4)
pH ≈ 3.12
So, the pH of the solution is approximately 3.12.