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This post covers buffered solutions, their importance, and estimating pH using the Henderson-Hasselbalch Equation.
Let's say that we would like to make a buffer solution that is 0.10 M Potassium Dihydrogen Phosphate and 0.15 M Potassium Monohydrogen Phosphate. The Ka of dihydrogenPhosphate ion is 6.3 x 10 exp-8. Using the Henderson-Hasselbalch equation from my last Post we obtain:
pH = pKa + log(A-/HA) = -log(6.3 x 10exp-8) + log(0.15M/0.10M) = 7.2 + 0.2 = 7.4
That's all there is to it! As always, thank you for reading!
A Publication of http://ExcellenceInLearning.biz
This post covers buffered solutions, their importance, and estimating pH using the Henderson-Hasselbalch Equation.
Introduction
A buffered solution is one which resists a change in pH when extra acid or base is added. This buffering of pH (upon the addition of an acid) occurs because of a system of two equilibria; one is the ionization of a weak acid and the other is the capture of hydrogen ion by the salt of the weak base. The addition of a strong base (i.e. Hydroxide Ion) results in the capture of excess hydroxyl ion by the weak acid. A buffer solution is then prepared by combining a particular molar ratio of a weak acid and the salt of its conjugate base.Importance of Buffers
Buffered solutions are found everywhere in nature because they are an essential "ingredient" for life. Living systems require a nearly constant pH level and buffered solutions make this possible. In fact, the blood of living organisms, including our own, is a solution of red blood cells suspended in a buffered serum solution!
Buffer solutions are also necessary for industrial biological processes, such as fermentation; the process whereby certain antibiotics are produced. A fermentation process is used to make bacitracin, an ingredient of common triple-antibiotic ointments. One common buffered solution is prepared from a combination of dihydrogenphosphate ion and the salt of its conjugate base, potassium monohydrogenphosphate. The reactions of the buffered solution are shown below.
Estimating pH of a Buffer Solution
Let's say that we would like to make a buffer solution that is 0.10 M Potassium Dihydrogen Phosphate and 0.15 M Potassium Monohydrogen Phosphate. The Ka of dihydrogenPhosphate ion is 6.3 x 10 exp-8. Using the Henderson-Hasselbalch equation from my last Post we obtain:
pH = pKa + log(A-/HA) = -log(6.3 x 10exp-8) + log(0.15M/0.10M) = 7.2 + 0.2 = 7.4
That's all there is to it! As always, thank you for reading!
A Publication of http://ExcellenceInLearning.biz