A 15.0 mL solution of Ba(OH)₂ is neutralized with 30.5 mL of 0.200 M HCl. What is the concentration of the original Ba(OH)₂ solution?

Answer 1

Strong acids and strong bases come together in perfect ratios to form water and salts in neutralization reactions. In other words, there shouldn't be any leftover acid or base. Use the following equations to find the unknown volumes or concentrations in neutralization problems:

mol H⁺ = mol OH^-

number of H⁺/mol * concentration of acid * volume of acid = number of OH^-/mol * concentration of base * volume of base

In this particular problem, here's the work:

mol H⁺: 30.5 mL HCl * 1 L / 1000 mL * 0.200 mol / 1 L = 0.0061 mol HCl

15.0 mL Ba(OH)₂ * 1 L / 1000 mL * 2 mol OH^- / 1 mol Ba(OH)2 * x mol / L = 0.0061 mol OH^-

x mol / L = 0.0061 mol OH^- / (2 * 15/1000 L) = 0.2033 M Ba(OH)₂

Longer explanation:

How do we measure "leftover" acid or base? You might be tempted to say volume, but that only describes the amount of solution added and not its concentration. Concentration is also tempting, but one drop of highly concentrated acid can't neutralize a swimming pool of concentrated base. Therefore, we combine volume and concentration to get MOLES of acid and base, or the count of H⁺ and OH^- ions floating around in the solution. If the solution is to be neutral, mol H⁺ = mol OH^-.

The 15.0 mL of Ba(OH)₂ has an unknown concentration, so we'll come back to this one.

30.5 mL of HCl has a unknown concentration of 0.200 M, or 0.200 mol / 1 L.

30.5 mL HCl * 1 L / 1000 mL * 0.200 mol / 1 L = 0.0061 mol HCl.

This is virtually the same as 0.0061 mol H⁺ because HCl is a strong acid that dissociates readily in water.

For neutralization, mol H⁺ = mol OH^- = 0.0061 mol.

15.0 mL Ba(OH)₂ needs to contain 0.0061 mol of OH^-. Each mole of Ba(OH)₂ contains two OH^-, so make sure to account for that in the calculations:

15.0 mL Ba(OH)₂ * 1 L / 1000 mL * 2 mol OH^- / 1 mol Ba(OH)2 * x mol / L = 0.0061 mol OH^-

Using algebra, we find that the unknown molarity must've been:

x mol / L = 0.0061 mol OH^- / (2 * 15/1000 L) = 0.2033 M Ba(OH)₂

Check your answer by remembering that we needed mol H⁺ = mol OH^-:

2 * 0.2033 M Ba(OH)₂ * 0.015 L Ba(OH)₂ = 0.200 M HCl * .0305 L HCl