This preview shows page 1. Sign up to view the full content.
Unformatted text preview: Chem 1B Exam 3 Name October 27, 2010 Lab M T(AM) T(PM) Instructions: For all mathematical problems, show your work. Give the basic formula (if there is one), rearrange it if necessary, plug in data with units, round properly to the correct number of significant figures, and place a box around the answer. For essay questions, be complete, but concise. Complete sentences with proper spelling and punctuation are expected. Useful information: Kw = 1.0 × 10-14 Spectrochemical series: CO > CN1- > en > NH3 > H2O > OH1- > F1- > Cl1- > Br1- > I11. (10 pts.) How many moles of cyanic acid, HCNO, must be added to 1.00 L of 0.385 M sodium cyanate, NaCNO, to form a buffer solution of pH 3.25? Ka for cyanic acid is 3.46 × 10-4. HCNO(aq) + H2O(l ) ∏ H3O+(aq) + CNO1-(aq) [CNO1-] pH = pKa + log[HCNO] pH = pKa + log[CNO1-] - log[HCNO] log[HCNO] = pKa + log[CHO1-] - pH log[HCNO] = -log(3.46 x 10-4) + log(0.385) – 3.25 = -0.20 [HCNO] = 0.63 M 0.63 mol HCNO must be added. 2. (10 pts.) To 125 mL of 0.250 M propanoic acid, HC3H5O2, is added 75.0 mL of 0.125 M sodium hydroxide solution. Calculate the pH of the resulting solution. Ka for propanoic acid is 1.3 × 10-5. HC3H5O2(aq) + H2O(l ) ∏ H3O+(aq) + C3H5O21-(aq) Initial moles HC3H5O2 = 0.125 L × 0.250 M = 0.0313 mol Added moles OH1- = moles HC3H5O2 consumed = moles C3H5O21- generated = 0.0750 L × 0.125 M = 0.00938 mol [C3H5O21-] pH = pKa + log[HC3H5O2] 0.00938/V pH = -log(1.3 × 10-5) + log(0.0313 - 0.00938)/V pH = 4.89 + log(0.429) = 4.52 3. A buffer solution is prepared containing 0.10 M benzoic acid, HC7H5O2 and 0.10 M sodium benzoate, NaC7H5O2. (a) (3 pts.) Write the chemical equation that describes what will happen if a small amount of NaOH(aq) is added to this buffer. HC7H5O2(aq) + OH1-(l ) ∏ H2O(aq) + C7H5O21-(aq) (b) (3 pts.) Write the chemical equation that describes what will happen if a small amount of HCl(aq) is added to this buffer. C7H5O21-(aq) + H3O+(l ) ∏ H2O(aq) + HC7H5O2(aq) 4. (2 pts.) Acid–base indicators show color changes (a) (b) (c) (d) (e) at exactly the pKin of the indicator. at exactly pH = 7. generally over the range pKin ± 1. between pH of 6 and 8. always at a pH above 7. 5. (2 pts.) Which of the following is not a buffer system? A solution containing roughly equal concentrations of (a) (b) (c) (d) (e) fluoride ion and hydrofluoric acid. phosphate ion and hydrogenphosphate ion. bromide ion and hydrobromic acid. carbonate ion and bicarbonate ion. All of these are buffers. 6. (2 pts.) A phosphate buffer (H2PO41-/HPO42-) has a pH of 8.3. Which of the following changes will cause the pH to increase? (a) (b) (c) (d) (e) Dissolving a small amount of Na2HPO4. Dissolving a small amount of NaH2PO4. Adding a small amount of dilute hydrochloric acid. Adding a small amount of dilute phosphoric acid. Dissolving a small amount of sodium chloride. 7. (2 pts.) The reaction Cr(NH3)63+(aq) + 3NH2CH2CH2NH2(aq) ∏ Cr(en)33+ (aq) + 6NH3(aq) has a small value for ΔH, yet the reaction has a large free-energy change. Why is this? (a) (b) (c) (d) (e) The rate of reaction is fast. The entropy change is large and positive. While the enthalpy value is small, it is still large enough to drive the reaction, even though the entropy change is negative. The entropy change is large and negative. There is not enough information to answer this question. 8. (10 pts.) The molar solubility of the ionic solid YF3 is 4.2 × 10-6 M. Calculate Ksp for YF3(s). YF3(s) ∏ Y3+(aq) + 3F1-(aq) Ksp = [Y3+][F1-]3 [Y3+] = 4.2 x 10-6 M [F1_] = 3(4.2 x 10-6) M = 1.3 × 10-5 M Ksp = (4.2 x 10-6) × (1.3 × 10-5) = 8.4 × 10-21 9. (10 pts.) The Ksp of Cu(OH)2(s) is 4.8 × 10-20. Calculate the molar solubility of Cu(OH)2(s) in a solution that is buffered at pH 5.00. Cu(OH)2(s) ∏ Cu2+(aq) + 2OH1-(aq) Ksp = [Cu2+][OH1-]2 = 4.8 × 10-20 Let x = molar solubility of Cu(OH)2 [Cu2+] = x [H3O+][OH1-] = Kw 1.0 × 10-14 Kw [OH ] = [H3O+] = 1.0 × 10-5 = 1.0 × 10-9 M
1- 4.8 × 10-20 Ksp [Cu ] = [OH1-]2 = (1.0 × 10-9)2 = 0.048 M
2+ molar solubility = 0.048 M 10. (10 pts.) Mercury(II) ion forms the stable complex ion HgI42-(aq). Write the balanced equation for dissolving HgBr2(s) in an aqueous solution containing iodide ion and compute the value of K for this process. Ksp for HgBr2(s) is 1.3 × 10-19 and K for the formation of HgI42- (aq) is 1.9 × 1030. HgBr2(s) ∏ Hg2+(aq) + 2Br1-(aq) Hg2+(aq) + 4I1-(aq) ∏ HgI42-(aq) HgBr2(s) + 4I1- (aq) ∏ HgI42-(aq) + 2Br1- (aq) K = (1.3 × 10-19) × (1.9 × 1030) = 2.5 × 1011 11. (6 pts.) (a) Write the formula of potassium tetracyanonickelate(II). K2[Ni(CN)4] (b) Name the following: [Co(NH3)4(H2O)Br]Br2 tetraammineaquabromocobalt(III) bromide 12. (6 pts.) Can the square planar complex [Ni(CO)2Cl2] exhibit geometric isomerism? If yes, draw the isomers and assign geometries to each. If no, explain why not.
CO Ni Cl CO Cl Cl Cl Ni CO CO trans 13. cis (4 pts.) Does the complex shown have an enantiomer? Explain.
NH3 OC Ni H3N CO Br Br This complex lacks symmetry. It is chiral and a chiral structure has an enantiomer. 14. (8 pts.) Predict which of the following complex ions is expected to absorb at longer wavelengths in the visible spectrum: [Zr(en)3]2- or [Zr(H2O)6]2-. Explain your reasoning. The ethylenediamine (en) is a strong field ligand while water is a weak field ligand. The splitting of the d orbitals in the aqua complex is less than in the en complex. The smaller energy gap in the aqua complex will absorb longer wavelength (lower energy) radiation. 15. Nitrate ion is a common ligand. The octahedral complex [Ru(NO3)6]4- is diamagnetic. (a) (8 pts.) Draw the crystal field energy level diagram and show the placement of the d electrons for this ion. Ru+2 d6 E (b) (4 pts.) What does the above information indicate about this ligand? Explain your reasoning. The diamagnetic ion has all electrons paired. This results from a relatively large splitting of the d orbitals. Nitrate ion must be a strong field ligand. ...
View Full Document
This note was uploaded on 12/21/2010 for the course CHEM 1B taught by Professor Garyj.decicco during the Fall '10 term at Santa Rosa.
- Fall '10