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Applications of Equilibrium
Buffers are solutions which resist changes in pH. They consist of solutions of a weak acid and its conjugate base in approximately equal concentrations. The solutions resists changes in pH because the acid can react to neutralize small amounts of base which are added, and the conjugate base can neutralize small amounts of acid which are added. Each conjugate acid/base pair has a characteristic pH at which it makes the most effective buffer. It is important to remember that even though the acid and its conjugate base (usually in the form of a salt) are both present initially, it is the equilibrium between them that must be considered. That is, the chemical reaction involves the acid on one side of the double arrows and the conjugate base on the other. Buffers & Acid/Base Titrations
a) Determine the pH of a solution which contains 0.50M HClO and 0.60M NaClO. Ka for HClO = 3.5 x 10-8.
b) Suppose 5.00 mLs of 0.010M HNO3 is added to 50.0 mLs of the above buffer. Calculate the change in pH.
c) Calculate the change in pH when 5.00 mLs of 0.0100M HNO3 is added to 50.0 mLs of water.To Make a Buffer with a Specific pH 1. Choose an acid/base pair which has a Ka near the desired [H3O+], or with a pKa near the desired pH.
2. Calculate the relative concentration of base and acid needed to obtain the desired pH. For buffers, the equilibrium calculation can be simplified using the equation:
pH = pKa + log {[base]/[acid]}
- Using a table of Ka values how would you prepare a buffer with a pH of 8.75?
Strong Acid with a Strong Base: This type of titration results in the formation of a "neutral" salt. Since the products of the titration will not effect the pH, the pH at the equivalence point will be 7. The profile of the titration curve is very steep and nearly vertical at the equivalence point, so an indicator which changes color anywhere from a pH of 5-9 will provide accurate results. Acid/Base Titrations
Weak Acid with a Strong Base: This type of titration results in the formation of the conjugate base of the weak acid (such as acetate ion from acetic acid). Thus the pH at the equivalence point will be greater than 7. For accurate results. the indicator much change color at a pH of greater than 7. As in the previous case, a buffer is formed mid way through the titration, and the profile of the titration curve reflects this resistance to changes in pH.
Strong Acid with a Weak Base: This type of titration will result in the formation of the conjugate acid of the weak base (such as NH4+ from NH3). Thus, the pH at the equivalence point will be less than 7. The indicator used must change color at a pH less than 7, or inaccuracy will result. In addition, the profile of the titration curve will show that mid-way through the titration, the pH doesn't change much. This is due to the formation of a buffer. At the mid-point of the titration, the concentration of the weak base will equal that of its conjugate acid.
- Problem: 25.0 mLs of 0.10M benzoic acid (Ka=6.4 x 10-5) is titrated with 0.10M NaOH. Calculate the pH a) initially; b) after 12.5 mLs of NaOH added; c) at the equivalence point; and d) after 30.0 mLs of NaOH have been added
Most salts (ionic compounds) which are described as insoluble are actually sparingly soluble. The solubility, though small, can be expressed in several ways. Often it is expressed in grams of compound per 100g of water. It can also be expressed in terms of molar solubility (M). Since solubility is governed by an equilibrium reaction, it can be described using an equilibrium constant called Ksp. All these methods of describing solubility are related. Solubility Equilibria
The reactions involving solubility are all written the same way (just like weak acid dissociation reactions). The solid is always on the left side of the equation, and the aqueous ions are always on the right. Solubility is highly temperature dependent, and most values of Ksp are obtained at 25oC.
example: Ksp for PbI2 is 1.4 x 10–8. This is the equilibrium constant for the following reaction.PbI2(s)<--> Pb+2(aq) + 2 I–(aq) • Ksp for calcium phosphate = 1.3 x 10-32. Calculate the molar solubility of calcium phosphate.Predicting If Precipitates Will Form If the ions which may form a precipitate come from two different sources, (the cation comes from one soluble salt, the anion from another), calculate Q. If Q>K, the reaction goes to the left, and a precipitate will form.
• Will a precipitate form if 10.0 mLs of 0.010M AgNO3(aq) is added to 20.0 mLs of 0.10M Na2SO4(aq)? [Ksp for Ag2SO4=1.2 x 10–5]pH and Solubility Since most anions are weak bases, solubility is highly pH dependent. Many of these salts become more soluble as the pH decreases, because the anion gets protonated by hydronium ions, and is effectively removed from the solubility equilibrium.
• Calculate the solubility of Fe(OH)2 in water, and in buffers of pH=3.00 and pH= 11.00 [Ksp=1.8 x 10–15]
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Copyright ©1998 Beverly J. Volicer and Steven F. Tello, UMass Lowell. You may freely edit these pages for use in a non-profit, educational setting. Please include this copyright notice on all pages.
