A 15.0 ml solution of Ba(OH)₂ is neutralized with 30.3 ml of 0.200 m HCl. the neutralized of the original Ba(OH)₂ solution is 0.2M
The reaction is given as follows :
Ba(OH)₂ + 2HCl ----> BaCl₂ + 2HCl
concentration of HCl , molarity = 0.200 M
volume = 30.3 mL = 0.030 L
moles of HCl = molarity × volume
= 0.200 × 0.030
= 0.006 mol
moles of Ba(OH)₂ = 1/2 moles of HCl
moles of Ba(OH)₂ = 1 /2 × 0.006 = 0.003 mol
molarity of Ba(OH)₂ = moles / volume
= 0.003 / 0.015
= 0.2 M
The concentration of Ba(OH)₂ = 0.2 M
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