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Untitled - Kelly Walsh High School

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242 CHEMISTRY FOR THE UTTERLY CONFUSED<br />

have been added. Usually you will find the moles by multiplying the concentration<br />

by the volume of the solution. In a few cases, you may find the number of<br />

moles from the mass and the molar mass.<br />

In the second region of the titration curve, the substance added is the limiting<br />

reagent. As a limiting reagent, it is no longer present to affect the pH. The other<br />

reactant is in excess, and this excess will have an affect upon the pH. The moles<br />

of excess reactant divided by the total volume of the solution (the initial volume<br />

plus the volume of the solution added) gives the concentration. The solution<br />

also contains the products of the reaction. If the excess reactant is either a weak<br />

acid or a weak base, the concentration of the products will be important. A<br />

buffer solution will be present any time you have an excess of a weak acid or<br />

weak base along with its conjugate.<br />

The next point in the titration curve is the equivalence point. At this point,<br />

both the material added and the material originally present are limiting. At<br />

this point, neither of the reactants will be present and therefore will not<br />

affect the pH. If the titration involves a strong acid and a strong base, the pH<br />

at the equivalence point is 7. If the titration involves a weak base, only the<br />

conjugate acid is present to affect the pH. This will require a K a calculation.<br />

If the titration involves a weak acid, only the conjugate base is present to<br />

affect the pH. This will require a K b calculation. The calculation of the conjugate<br />

acid or base will be the moles produced divided by the total volume of<br />

the solution.<br />

The final region of the titration curve is after the equivalence point. In this<br />

region, the material originally present in the container is limiting. The excess<br />

reagent, the material added, will affect the pH. If this excess reactant is a weak<br />

acid or a weak base, this will be a buffer solution.<br />

Let’s use the following example to see how to deal with each of these regions.<br />

This will be a titration of the weak acid with a strong base.<br />

A 100.0 mL sample of 0.150 M nitrous acid (pK a 3.35) was titrated with 0.300 M<br />

sodium hydroxide, NaOH. Determine the pH of the solution after the following<br />

quantities of base have been added to the acid solution: a. 0.00 mL, b. 25.00 mL,<br />

c. 50.00 mL, and d. 55.00 mL.<br />

a. 0.00 mL. Since no NaOH has been added, this is the initial pH section of the<br />

titration curve.<br />

The only substance present is nitrous acid, HNO 2, a weak acid. Since this is a<br />

weak acid, this must be a K a problem. As a K a problem, we can set up a simple<br />

equilibrium problem:

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