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Unformatted text preview: Chemistry 2080 Spring 2009
Problem Set #8 - due date: Friday, April 3rd at 2 pm Name; Lab TA Name: Lab Day/Time: Practice problems (not graded). Chapter 17: 32, 46, 71, 74, i) This will require a bit of “library” research, and some calculations. For- each of the titrations listed below, give
a suitable indicator that would allow a reasonably accurate determination of the equivalence point for the titration.
Also, for titraﬁons (b) and (c) only, calculate the pH of the solution at the equivalence point. All solutions are in water, and are at 25°C. (Figure 17.8 in Petrucci et at shows a chart of indicators. Also, Table 16.3 gives values
_ of K3). (a) Titration of 100 mL of 0.10 M NaOH with 0.10 M HCl. (b) Titration of 100 mL of 0. 10 M ethylamine with 0.10 M HCl. (c) Titration of 100 mL of 0. 10 M formic acid with 0.10 M NaOH. (d) Titration of 100 mL of 0.10 M acetic acid with 0.10 M ammonia solution. 2) (3) Suppose that you mix 13.5 g of KH2P04 and 13.0 g of NazHPO4 with enough water to make 500 mL of
solution. What is the pH of this buffer solution at 25°C? (K5l values are on page 682 of Petrucci et at.) (b) How much 0.10 M NaOH solution would you need to add to 100 mL of this buffer solution, to raise its
pH by 0.20 units? 3) There are two major determinants of “acidity” in wine: titratable acidity, and pH. Determination of the titratable
acidity involves adding base (e.g. 0.10 M NaOH) to the wine while measuring the pH of the resulting mixture.
Why is the conventional end—point taken to be a pH of 8.2 rather than 7.0? ...
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- Spring '07
- Chemistry, 10 m, 0.10 M