2007Exam1Q4Answer - 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

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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.
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This note was uploaded on 11/07/2009 for the course CH Organic taught by Professor Iverson during the Spring '09 term at University of Texas at Austin.

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2007Exam1Q4Answer - 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

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