The balanced equation for the reaction between KOH and HBr is as follows;
KOH + HBr --> KBr + H₂O
stoichiometry of KOH to HBr is 1:1
number of KOH moles reacted - 0.25 mol/L x 0.015 L = 0.00375 mol
according to molar ration
number of KOH moles reacted = number of HBr moles reacted
number of HBr moles reacted - 0.00375 mol
if 12 mL of HBr contains - 0.00375 mol
then 1000 mL of HBr contains - 0.00375 mol / 12 mL x 1000 mL = 0.313 mol
therefore molarity of HBr is 0.313 M
Full electron configuration of barium: 1s2 2s2 2p6 3s2 3p6 3d10 4s2 4p6 4d10 5s2 5p6 6s2
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