Answer:
0.287 M
Ba(OH)₂ + 2 HClO₄ ⇒ Ba(ClO₄)₂ + 2 H₂O
1.62 M
Explanation:
Step 1: Calculate the concentration of Ba(OH)₂
We will use the following expression.
[Ba(OH)₂] = mass Ba(OH)₂/ molar mass Ba(OH)₂ × liters of solution
[Ba(OH)₂] = 2.29 g/ 171.34 g/mol × 0.0466 L = 0.287 M
Step 2: Write the balanced neutralization equation
Ba(OH)₂ + 2 HClO₄ ⇒ Ba(ClO₄)₂ + 2 H₂O
Step 3: Calculate the reacting moles of Ba(OH)₂
25.5 mL of 0.287 M Ba(OH)₂ react.
0.0255 L × 0.287 mol/L = 7.32 × 10⁻³ mol
Step 4: Calculate the reacting moles of HClO₄
7.32 × 10⁻³ mol Ba(OH)₂ 2 mol HClO₄/1 mol Ba(OH)₂ = 0.0146 mol HClO₄
Step 5: Calculate the concentration of HClO₄
0.0146 moles of HClO₄ are in 8.99 mL of solution.
[HClO₄] = 0.0146 mol/0.00899 L = 1.62 M