Titrations

Titrations - 61 VII TITRATIONS A titration involves adding...

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VII TITRATIONS A titration involves adding a solution that is acidic to another solution that is basic. One can either start with a fixed volume of an acid and add the base (normally contained in a burette) or a fixed volume of base and add an acid. The objective in this section is to predict a titration curve. For beginning with a fixed volume of an acid, the titration curve is the pH of the solution versus the volume of base added. One can titrate a strong acid with a strong base, a weak acid with a strong base or a weak base with a strong acid. The case of titrating a weak acid with a weak base is not considered here. The titration a strong acid with a strong base is not very interesting in the sense that the conjugate acid and base are inert ions. Therefore this section will treat the case of titrating a weak acid with a strong base (sodium hydroxide). The treatment of a weak base with a strong acid is very similar. To begin with the titration of a weak monoprotic acid with a strong base is considered. In particular the weak acid is acetic acid (CH 3 COOH, HAc). Experimentally one knows what the acid is so that K a =1.8x10 -5 is known. The molarity ([HAc] o ) and volume (V HAc ) of the acid are also known. Therefore the number of moles of the acid to be titrated is given by n HA co = [HA c] o V HA c . (7.1) If one started with 25mL of a 0.10M solution of HAc then the number of moles of HAc would be 0.0025mol. These rather small numbers can be avoided by introducing another set of units for molarity. Let 1m o l = 1000mm o l (7.2) where mmol is a millimole and 1L = 1000mL . (7.3) Then the molarity of the acid [HAc] is given by [HAc] = n HAc (mol) V(L) = n HAc (mol) 1000mmol 1mol V(L) 1000mL 1L = n HAc (mmol) V(mL) . (7.4) Now the above 0.10M solution can be interpreted as either 0.10mol of HAc pre Liter of solution or as 0.10mmol of HAc per milliliter of solution. However for 25mL of solution there would now be 2.5mmol of HAc rather than 0.0025mol of HAc. It is easier to deal with 2.5 than 0.0025. The numerical value of molarity is the same in either set of units. The molarity of the base [NaOH] o is also known so that the number of mmol of base added is given by n N aOH (mm o l) = [N aOH ]V N aOH (mL ) . (7.5) 61
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If 10.mL of 0.10M of NaOH is added to the 25mL of HAc then n NaOH would be 1.0mmol and the total volume would be V tot = 25mL+10.mL=35mL. In a titration of a weak acid with a strong base the solution will start out as an acidic solution and the reaction needed to determine equilibrium concentrations is shown below. HAc (aq) ←→ H (aq) + + Ac (aq) - init. [HAc] o 10 -7 M [Ac - ] o equil. [HAc] eq = [HAc] o - x [H + ] eq = x + 10 - 7 [Ac - ] eq = x Also Q a = 10 -7 [A - ] o [HA] o (7.6) and as long as Q a <K a the solution is acidic and the above reaction is used to determine equilibrium concentrations. When Q a >K a , the solution becomes basic and one must change to the reaction shown below. Ac
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This note was uploaded on 04/29/2008 for the course CHE 1302 taught by Professor Young during the Spring '08 term at Baylor.

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Titrations - 61 VII TITRATIONS A titration involves adding...

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