A student was required to prepare 250.0 mL of a dichloroacetic acid/sodium dichl

Question

A student was required to prepare 250.0 mL of a dichloroacetic acid/sodium dichloroacetate buffer in which the concentration of the weak acid component was 0.078 M and the concentration of the conjugate base was 0.02 M. The student was supplied with 0.508 M dichloroacetic acid and 1.0M NaOH to perform this task. What volume (in L) of the acid would the student need to prepare this buffer solution? Hint: assume that all of the conjugate base comes directly from the reaction of NaOH with the weak acid (in other words, there is negligible dissociation of the weak acid).

Explanation / Answer

Total no of mol of buffer = weak acid(dichloroacetic acid) + conjugate base

no of mol of weak acid present in buffer = 250/1000*0.078 = 0.0195 mol

no of mol of conjugate base present in buffer = 250/1000*0.02 = 0.005 mol

total weak acid required = 0.0195+0.005 = 0.0245 mol

volume of dichloroacetic acid needed = n/M

                              = 0.0245/0.508

                              = 0.0482 L

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