The amount of lactic acid,HC 3 H 5 O 3 , produced in a sample ofmuscle tissue wa

Question

The amount of lactic acid,HC3H5O3, produced in a sample ofmuscle tissue was analyzed by reaction with hydroxide ion.Hydroxide ion was produced in the sample mixture by electrolysis.The cathode reaction is given below. 2 H2O(l) + 2 e H2(g) + 2 OH (aq) Hydroxide ion reacts with lactic acid as soon as it is produced.The endpoint of the reaction is detected with an acid-baseindicator. It required 116 s for a currentof 17.4 mA to reach the endpoint. How manygrams of lactic acid (a monoprotic acid) were present in thesample?
2 H2O(l) + 2 e H2(g) + 2 OH (aq)

Explanation / Answer

116 seconds *(17.4 mC/s) * (1 C/1mC )*(1 mol e- /96485 C) *(2 molOH-/ 2mol e-)    = 0.0209193139 mol OH-consumed          (note 2 mol OH-/2 mol e- because of cathode reactionstoichiometry) b/c monoprotic, 1 mol OH- reacts with 1 mol of lactic acid So 0.02092 mol lactic acid produced mass : 0.02092 mol *(90.08 g/mol) = 1.88 grams

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