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# 2007Exam1Q4Answer - Practice Problem using...

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Practice Problem using Henderson-Hassalbalch equation A compound has a pKa of 7.4. To 100 mL of a 1.0 M solution of this compound at pH 8.0 is added 30 mL of 1.0 M hydrochloric acid. The resulting solution is pH a. 6.5 b. 6.8 c. 7.2 d. 7.9 e. 7.5 Unknown Solution pKa = 7.4 pH = 8.0 1.0 M 100 mL HCl Strong acid 1.0 M 30 mL For the Unknown Solution: 1.0 M = [A - ] + [HA] therefore, [HA] = 1.0 M - [A - ] Using the Henderson-Hassalbalch equation: pH = pKa + log([A - ]/[HA]) 8.0 = 7.4 + log([A - ]/{1.0 M - [A - ]}) OR probably easier: 8.0 = 7.4 + log(R) Solving for [A - ] and then [HA] and then dividing the former by the latter one gets R. The alternative is to solve directly for R. R = anti-log(0.6) = 3.9810 = 3.981 M (A - )/ 1.000 M (HA) This means that for every liter of this solution, there are 3.98107 mol of the conjugate base to every 1 mol of the acid. Therefore in the 100 mL original solution, there was 0.398 mol (A - ) and 0.100 mol (HA). Now, 30 mL of 1.0 M HCl is added to the above solution. Remember that the original volume of the solution was 100 mL or 0.1 L, we are now adding an additionally 30 mL or 0.03 L to it. We’ll use this in a bit.

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For now, HCl is a strong acid so it dissociates completely. So we will be adding a certain number of mols of H
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