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Calculating the pH at equivalence of a titration A chemist titrates 120.0 mL of a 0.7513...


Calculating the pH at equivalence of a titration A chemist titrates 120.0 mL of a 0.7513 M lidocaine (C4H2, NONH) solution wi
Calculating the pH at equivalence of a titration A chemist titrates 120.0 mL of a 0.7513 M lidocaine (C4H2, NONH) solution with 0.4926 M HNO, solution at 25°C. Calculate the pH at equivalence. The PK, of lidocaine is 7.94 Round your answer to 2 decimal places Note for advanced students: you may assume the total volume of the solution equals the initial volume plus the volume of HNO, solution added. DH-
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base = 120 x 0.7513 = 90.156

volume of salt at equivalence point   = 90.156 / 0.4926 = 183. 0 m
salt concentration = 90.156 / 183 +120 = 0.2975 M

pH = 7 - 1/ 2[pKb + log C]

     = 7 - 1/2 [7.94 + log 0.2975 ]

     = 3.29

pH = 3.29

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