Horvath Practice Problems 64

Horvath Practice Problems 64 - 5 = 0 . 038 M So the number...

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Using, K a =1 . 7 × 10 5 = ( x )(0 . 038 + x ) (0 . 063 x ) , and the quadratic formula, we get x = £ H + ¤ =2 . 8 × 10 5 . So, pH = log ¡ 2 . 8 × 10 5 ¢ =4 . 55 . Example, the acetate bu f er (part 2) : Calculate the pH if 10 ml of 0.01 M HCl is added to the above bu f er. Then compare this to the pH if the HCl was added to water. First we f gure out how many moles of H + is coming from the HCl, 0 . 010 L × 0 . 010 mol H + 1 L =0 . 0001 mol H + . We take it that all the H + reacts with C 6 H 4 NO 2 to make HC 2 H 3 O 2 . Thus we must determine the new concentrations of C 6 H 4 NO 2 and HC 2 H 3 O 2 . From above we know the equilibrium concentrations of C 6 H 4 NO 2 and HC 2 H 3 O 2 before the HCl is added. These are [ HC 2 H 3 O 2 ]=0 . 063 1 . 7 × 10 5 =0 . 063 M £ C 2 H 3 O 2 ¤ =0 . 038 + 1 . 7 × 10
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Unformatted text preview: 5 = 0 . 038 M So the number of moles of each present are . 063 mol HC 2 H 3 O 2 1 L . 040 L = 0 . 0025 mol HC 2 H 3 O 2 and . 038 mol HC 2 H 3 O 2 1 L . 040 L = 0 . 0015 mol C 2 H 3 O 2 . After completely reacting with the 0.0005 mol of H + from the HCl we have . 0025 + 0 . 0001 = 0 . 0026 mol HC 2 H 3 O 2 and . 0015 . 0001 = 0 . 0014 mol HC 2 H 3 O 2 . Now we need to get the new concentrations . 0026 mol HC 2 H 3 O 2 . 050 L = 0 . 052 M HC 2 H 3 O 2 THE COMMON ION EFFECT AND BUFFERS 63...
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