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 All of the NaOH is consumed in the reaction, ie none leftover .0035 moles of MgCl

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