Horvath Practice Problems 65

Horvath Practice Problems 65 - and 0.0014 mol C2 H3 O 2 =...

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and 0 . 0014 mol C 2 H 3 O 2 0 . 050 L =0 . 028 MC 2 H 3 O 2 . Finally we now need to setup an ICE chart, [ HC 2 H 3 O 2 ] £ H + ¤£ C 2 H 3 O 2 ¤ I 0 . 052 0 0 . 028 C xx x E 0 . 052 0 . 028 + x . Using K a =1 . 7 × 10 5 = ( x )(0 . 028 + x ) (0 . 052 x ) and solving the quadratic equation we get x = £ H + ¤ =3 . 15 × 10 5 .So , pH = log ¡ 3 . 15 × 10 5 ¢ =4 . 50 So a change in pH of only 0.05 units. We now calculate the pH for the situation of adding the HCl to pure water. From above we see that 0.0001 mol of H + are produced. These moles are now in a total volume of 40 mL so, £ H + ¤ = 0 . 0001 mol H + 0 . 040 mL . 0025 M Hence a pH of pH =
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This document was uploaded on 11/01/2011 for the course CHM 2046 at University of Florida.

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