What volume of a 0.179 M barium hydroxide solution is required to neutralize 25.6 mL of a 0.377 M perchloric acid solution?
To determine the volume of the barium hydroxide solution needed to neutralize the perchloric acid solution, we can use the concept of stoichiometry and the balanced chemical equation for the reaction between barium hydroxide (Ba(OH)2) and perchloric acid (HClO4). The balanced equation is as follows:
2 Ba(OH)2 + HClO4 → Ba(ClO4)2 + 2 H2O
From the balanced equation, we can see that 2 moles of barium hydroxide react with 1 mole of perchloric acid to produce 1 mole of barium perchlorate and 2 moles of water.
Step 1: Calculate the number of moles of perchloric acid:
Given that the concentration of the perchloric acid solution is 0.377 M and the volume used is 25.6 mL (or 0.0256 L), we can calculate the number of moles using the formula:
Moles = Concentration (M) × Volume (L)
Moles of HClO4 = 0.377 M × 0.0256 L = 0.0096512 mol
Step 2: Use the stoichiometry of the balanced equation to find the number of moles of barium hydroxide:
From the balanced equation, we know that 2 moles of barium hydroxide react with 1 mole of perchloric acid. Thus, the number of moles of barium hydroxide required will be half of the moles of perchloric acid.
Moles of Ba(OH)2 = 0.0096512 mol / 2 = 0.0048256 mol
Step 3: Calculate the volume of the barium hydroxide solution:
To determine the volume, we need to divide the number of moles of barium hydroxide by its concentration. The concentration of the barium hydroxide solution is given as 0.179 M.
Volume = Moles / Concentration
Volume of Ba(OH)2 = 0.0048256 mol / 0.179 M ≈ 0.0269 L
Since the concentration of the barium hydroxide solution is given in moles per liter (M), the volume obtained will be in liters. To convert it to milliliters (mL), we can multiply it by 1000.
Volume of Ba(OH)2 = 0.0269 L × 1000 mL/L ≈ 26.9 mL
Therefore, approximately 26.9 mL of the 0.179 M barium hydroxide solution is required to neutralize 25.6 mL of the 0.377 M perchloric acid solution.
2HClO4 + Ba(OH)2 ==> Ba(ClO4)2 + 2H2O
mols HClO4 = M x L = ?
mols Ba(OH)2 = 1/2 mols HClO4 (from the coefficients in the balanced equation)
Then mols Ba(OH)2 = M x L. You know mols and M, solve for L.