Question

19.20 grams of barium hydroxide, Ba(OH)₂ is dissolved in 178.0 ml of water (you may assume no volume change). How many milliliters of 0.500 M HCL solution would be needed to completley neutralize 34.12 ml of the barium hydroxide solution?

Answer 1

Answer – Given, mass of Ba(OH)₂ = 19.520 g , volume of water = 178.0 mL

Molarity of HCl = 0.500 M , volume of barium hydroxide solution = 34.12 mL

We need to calculate the moles of barium hydroxide first –

Moles of barium hydroxide, Ba(OH)₂ = mass / molar mas

                                                             = 19.20 g / 171.34 g.mol^-1

                                                             = 0.112 moles

Now we need to calculate the initial concentration of Ba(OH)₂

[Ba(OH)₂]= 0.112 moles / 0.178 L

                 = 0.630 M

For the complete reaction 34.12 ml of the barium hydroxide solution used , means

Used moles of Ba(OH)₂ = 0.630 M * 0.03412 L

                                       = 0.0215 moles

Reaction –

Ba(OH)₂ + 2 HCl -----> BaCl₂ + 2H₂O

From the above balanced equation –

1 moles of Ba(OH)₂ = 2 moles of HCl

So, 0.0215 moles of Ba(OH)₂ = ?

= 0.0430 moles of HCl

So, volume of HCl = moles of HCl / moalrity of HCl

                               = 0.0430 moles / 0.500 M

                                = 0.0859 L

                              = 85.9 mL

So, 85.9 milliliters of 0.500 M HCl solution would be needed to completely neutralize 34.12 ml of the barium hydroxide solution