S11-HW5-review

S11-HW5-review - HOMEWORK WEEK 5 REVIEW TITRATIONS 16.31 a First since the pH of the solution is less than seven at the beginning of the titration

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HOMEWORK WEEK 5 REVIEW TITRATIONS 16.31 a. First, since the pH of the solution is less than seven at the beginning of the titration, you can conclude that an acid is being titrated with a base. As the pH in much of the region before the equivalence point is quite high, about 5, this indicates that the acid is weak – compare Figures 16.12 and 16.13. b. The pH of the equivalence point is about 8.5. c. You need to pick an indicator that changes color in the pH range of about 7 to 10, i.e. the range for the near-vertical region around the equivalence point. Therefore, thymol blue or phenolphthalein would work fine for this titration (See Fig 15.10) 16.32 a. The balanced chemical equation for the titration of aqueous NaOH with HCl is HCl(aq) + NaOH(aq) NaCl(aq) + H 2 O(l) b. Beaker A contains four Cl - ions and four water molecules. There are no OH - ions present. This represents the equivalence point. Beaker B contains two OH - and two Cl - ions and two water molecules. Since some OH - is still present, this represents the titration at a point before the equivalence point. Beaker C contains six Cl - and two H 3 O + ions and four water molecules. This represents the titration at a point after the equivalence point. Thus, as the titration progresses, we move from Beaker B to A and then to C. Note that the numbers of spectator Cl - ions are consistent with this: we are adding HCl, so it is not surprising that the number of Cl - ions increases throughout the titration. c. The solution in beaker A is at the equivalence point and is neutral. The solution in beaker B contains OH - ions and is basic. The solution in beaker C contains H 3 O + ions and is acidic. 16.86 In a titration problem like this, it is important to work in moles, rather than in molarity, because the volume is changing, as the base is added to the acid. All the H 3 O + (from the HCl) reacts with the OH - from NaOH. Calculate the stoichiometric amounts of OH - and H 3 O + , and subtract the mol of H 3 O + (limiting reagent) from the mol of OH - . Then, divide the remaining OH - by the total volume of 0.040 L + 0.025 L, or 0.065 L, to find the [OH - ]. Then calculate the pOH and pH. Mol OH
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This note was uploaded on 02/20/2011 for the course CHEM 110 taught by Professor Dr.huston during the Spring '06 term at Pittsburgh.

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S11-HW5-review - HOMEWORK WEEK 5 REVIEW TITRATIONS 16.31 a First since the pH of the solution is less than seven at the beginning of the titration

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