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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 Mg(Br) 2 + 2 H2O 0.0120 M 0.255 M 0.118 M 0.510 M
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ok lets do it step by step, c=n/V --> n=c*V = 0,245 M * 37,5*10-3 = 9,18 *10^-3 mol n(Mg(OH)2)/n(HBr)=1/2--> n(Mg(OH)2)= 1/2 * n(HBr) = 4,59 * 10^-3 mol c(Mg(OH)2)= 4,59*10^-3 mol /18 *10^-3 dm3 = 0,255M
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