The volume of [tex]Ba(OH)_2[/tex] required will be 83.33 mL
Stoichiometric reactions
The equation of the reaction is as follows:
[tex]Ba(OH)_2 + 2HBr --- > 2H_2O + BaBr_2[/tex]
Mole ratio of reactants = 1:2
Mole of 50 mL, 0.05 M HBr = 0.05 x 0.05 = 0.0025 moles
Equivalent mole of [tex]Ba(OH)_2[/tex] = 0.0025/2 = 0.00125 moles
Volume of 0.015 M, 0.00125 moles [tex]Ba(OH)_2[/tex] = 0.00125/0.015 = 0.0833 L
0.0833 L = 83.33 mL
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