If it requires 37.5 mL of a 0.245 M HBr solution to neutralize 18.0 mL Mg(OH)2, what is the concentration of the Mg(OH) 2 solution? 2 HBr + Mg(OH) 2 yields Mg(Br) 2 + 2 H2O
2 HBr + Mg(OH) 2 yields Mg(Br) 2 + 2 H2O M1V1 = M2V2 (0.245 M)(37.5 mL) = (M2)(18.0 mL) 9.1875 = (M2)(18.0 mL) 9.1875/(18.0 mL) = (M2)(18.0 mL)/(18.0 mL) 0.51041 this is not the complete answer as the mol to mol ratio must be considered! For every 2 mols of HBr there is 1 mole of Mg(OH)2; ratio = 2:1 0.51041 / 2 = 0.25520 M2 = 0.25520 0.25520 M of Mg(OH)2 solution
