Suspension of Mg

Suspension of Mg - Milk of magnesia is a suspension of...

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Milk of magnesia is a suspension of Mg(OH) 2 in water. It can be made by adding a base to a solution containing Mg +2 . Suppose that 40.0 mL of 0.200 M NaOH solution is added to 25.0 mL of 0.300 M MgCl 2 solution. What mass of Mg(OH) 2 will formed, and what will be the concentrations of the ions in the solution after the reaction is complete? 40.0 mL 25.0mL ?? grams 0.200 M 0.300 M 2 NaOH (aq) + MgCl 2 (aq) Mg(OH) 2 (s) + 2 NaCl (aq) 40.0 mL 1.00 L 0.200 moles NaOH 1 mole Mg(OH) 2 58.3 g Mg(OH) 2 = 0.233 g 1000 mL 1.00 L 2 moles NaOH 1 moles Mg(OH) 2 25.0 mL 1.00 L 0.300 moles MgCl 2 1 mole Mg(OH) 2 58.3 g Mg(OH) 2 = 0.437 g 1000 mL 1.00 L 1 moles MgCl 2 1 moles Mg(OH) 2 Therefore, all calculations must be based on NaOH. 2 NaOH (aq) + MgCl 2 (aq) Mg(OH) 2 (s) + 2 NaCl (aq) start 0.008 moles 0.0075 moles 0 moles 0 moles reacted -0.008 moles -0.004 moles +0.004 moles +0.008 moles end 0 moles NaOH .0035 moles MgCl 2 .004 moles Mg(OH) 2 .008 moles NaCl Next step is to calculate the concentration of the remaining ions. first recognize what we have remaining in solution
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This note was uploaded on 04/28/2008 for the course CHEM 111 taught by Professor Zombeck during the Winter '07 term at Saginaw Valley.

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Suspension of Mg - Milk of magnesia is a suspension of...

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