Any help would be greatly appreciated. 1.What is the hydroxide ion concentration

Question

Any help would be greatly appreciated. 1.What is the hydroxide ion concentration of a solutionprepared by dissolving 0.731 g of Mg(OH)2 in 427 mL ofwater?

2.How many milliliters of 0.254 MH2SO4 would be required to exactly neutralize450 mL of 0.242 M NaOH? Thank you very much! Any help would be greatly appreciated. 1.What is the hydroxide ion concentration of a solutionprepared by dissolving 0.731 g of Mg(OH)2 in 427 mL ofwater?

2.How many milliliters of 0.254 MH2SO4 would be required to exactly neutralize450 mL of 0.242 M NaOH? Thank you very much! Thank you very much!

Explanation / Answer

1. There are three steps to this problem,    a) Calulate the moles of Mg(OH)2(the molar mass is 58.3g/mol)          Mg(OH)2= mass / molar mass = (0.731g) / (58.3g/mol) = 0.0125 mol    b) Calculate the concentration of theMg(OH)2 in solution          [Mg(OH)2]= Mg(OH)2 / volume = (0.0125 mol / 0.427 L)= 0.0294 M    c) Determine the concentration ofOH- .          Takeinto account the fact that for every 1 mole of Mg(OH)2there is 2 moles of OH- : Mg(OH)2 ----->    Mg+2 + OH-           [OH-]= 2 x [Mg(OH)2] = (2 x 0.0294M)= 0.0587M 2. Since NaOH completely dissociates intoNa+  and OH-, toneutalize we need to react the all moles of OH-with a source of H+ ions. the source of these ions isthe H2SO4 which is a strong acid, meaningthat the first proton readily comes off: H2SO4----->   HSO4- + H+ To neutralize we must have the same number of moles ofOH- as H+, therefore the following can besaid [OH-] x VOH- = [H+] *VH+ To neutralize we must have the same number of moles ofOH- as H+, therefore the following can besaid [OH-] x VOH- = [H+] *VH+                Since  H2SO4 readily dissociates the follwing can be written [OH-] x VOH- =[H2SO4] * VH2SO4                Rearrangingfor VH2SO4 we get, VH2SO4 = ( [OH-] xVOH- ) / [H2SO4] = ( 0.242 Mx  0.450L) / 0.254 M = 0.429 L = 429 mL

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