Question

A) A barium hydroxide solution is prepared by dissolving 1.34 g of Ba(OH)2 in water to make 61.1 mL of solution. What is the concentration of the solution in units of molarity? b) If 15.1 mL of the barium hydroxide solution was needed to neutralize a 5.92 mL aliquot of the perchloric acid solution, what is the concentration of the acid?

Answer 1

A. Molarity of the Ba(OH)₂ solution is 0.128 mol/L or 0.128 M

B. Molarity of the HClO₄ solution is 0.653 mol/L or 0.653 M.

Step-by-step explanation:

A) Given: Given mass of Ba(OH)₂: w= 1.34 g, Molar mass of Ba(OH)₂: m= 171.34 g/mol, Volume of the solution: V = 61.1 mL

The molarity of Ba(OH)₂:
`M = (n*1000 )/(V_(solution))= (w*1000 )/(m* V_(solution))= (1.34 g*1000 )/(171.34 g/mol* 61.1 mL) = 0.128 mol/L`

Therefore, the molarity of the Ba(OH)₂ solution is 0.128 mol/L or 0.128 M

B) The neutralization reaction:

Ba(OH)₂ (aq) + 2 HClO₄(aq) → Ba(ClO₄)₂ (aq) + 2 H₂O(l)

Volume of Ba(OH)₂: V₁ = 15.1 mL ; volume of HClO₄: V₂ = 5.92 mL

Concentration of Ba(OH)₂: M₁ =
`0.128 mol/L = (n_(1)*1000 )/(V_(1))=(n_(1)*1000 )/(15.1 mL)`

`n_(1)= 0.0019328 mole`

Now, concentration of HClO₄: M₂= ?

As 1 mole of barium hydroxide neutralizes 2 moles of perchloric acid.

∴ number of moles of HClO₄: n₂= 2 × number of moles of Ba(OH)₂ = 2 \times 0.0019328 mole = 0.00386 moles

Now, the molarity of HClO₄:
`M = (n_(2)*1000 )/(V_(2)) = (0.00386 mol*1000 )/(5.92 mL) = 0.653 mol/L`

Therefore, the molarity of the HClO₄ solution is 0.653 mol/L or 0.653 M.