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
