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 <span>Mg(OH)2; ratio = 2:1
</span>0.51041 / 2 = 0.25520
M2 = 0.25520
0.25520 M of<span> Mg(OH)2 solution</span>
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