Respuesta :
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
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
