Hydrochloric acid (12.0 ml of 0.233 m) is added to 296.0 ml of 0.0579 m ba (oh) 2 solution. what is the concentration of the excess h + or oh - ions left in this solution?

The balanced equation for the reaction is as follows;

Ba (OH) ₂ + 2HCl - - - > BaCl₂ + 2H₂O

stoichiometry of base to acid is 1:2

Ba (OH) ₂ is a strong acid and HCl is a strong base therefore complete ionization takes place

the number of HCl moles reacted - 0.233 mol/L x 0.0120 L = 0.00280 mol

number of Ba (OH) ₂ moles reacted - 0.0579 mol/L x 0.2960 L = 0.0171 mol

we have to first find the limiting reactant

if HCl is the limiting reactant

if 2 mol of HCl reacts with 1 mol of Ba (OH) ₂

then 0.00280 mol of HCl reacts with - 0.00280 / 2 = 0.00140 mol of Ba (OH) ₂

and 0.0171 mol of Ba (OH) ₂ is present, therefore Ba (OH) ₂ is in excess and HCl is the limiting reactant

number of excess Ba (OH) ₂ moles = 0.0171 - 0.00140 = 0.0157 mol

total volume - 12.0 mL + 296.0 mL = 308.0 mL

[Ba (OH) ₂] = 0.0157 mol / 0.3080 L = 0.0510 M

since Ba (OH) ₂ is a strong acid

Ba (OH) ₂ - - > Ba²⁺ + 2OH⁻

therefore [OH⁻] = 2[Ba (OH) ₂]

[OH⁻] = 2 x 0.0510 = 0.102 M

hydroxide ion concentration is 0.102 M