Explanation First calculate the molarity of the solution using the equilibrium conditions, then continue with the freezing points of the solution. Molarity of solution Step 1: The equilibrium reaction Let “Ac - “substitute the acetate ion, C H 3 C O 2 − A c − ( a q ) + H 2 O ( l ) ↔ H A c ( a q ) + O H − ( a q ) Step 2: The equilibrium expression K b = [ H A c ] [ O H − ] [ A c − ] Step 3: Substitute known data Let the unknown concentration be “M” and the dissociation factor “x” K b = [ H A c ] [ O H − ] [ A c − ] 5.55 x 10 − 10 = ( x ) ( x ) ( M − x ) Get x via pH p O H = 14 − p H = 14 − 9.07 = 4.93 [ O H − ] = 10 − p O H = 10 − 4.93 [ O H − ] = 0.0000117 M x = [ O H − ] = 0.0000117 Step 4: Get the concentration 5.55 x 10 − 10 = ( x ) ( x ) ( M − x ) 5.55 x 10 − 10 = ( 0.0000117 ) ( 0.0000117 ) ( M − 0.0000117 ) M = ( 0.0000117 ) ( 0.0000117 ) 5.55 x 10 − 10 + 0.0000117 M = 0.2466 Change molarity to molality Step 1: Get mass of solution Assume a basis of 1 Liter, 1000mL solution, so total mass will be: m a s s = ( D e n s i t y ) x ( V o l u m e ) = ( 1

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Chapter 15, Problem 15.152MP

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pH and freezing points

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pH and freezing points

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pH and freezing points