What is the correct name and bonding of BF 3? 1. boron trifluoride, covalent compound 2. boron trifluoride, ionic compound 3. boron fluoride, covalent compound 4. boron fluoride, ionic compound What is the correct name and bonding of BF 3? Covalent molecule since all atoms are nonmetal. Does not contain polyatomic ions. 1 boron atom and 3 fluorine atoms, use prefixes, except mono of first element. Boron trifluoride 1
You have a solution of.1 M F and.1 M CO 3 2-. You add.1 M magnesium nitrate dropwise into the solution. K sp for MgF 2 is 7.4 x 1 9 and for MgCO 3 is 3.5 x 1 8. Which of the following will precipitate first? 1. magnesium nitrate 2. magnesium carbonate 3. magnesium fluoride 4. cannot be determined from the information given 5. none of these You have a solution of.1 M F and.1 M CO 3 2-. You add.1 M magnesium nitrate dropwise into the solution. K sp for MgF 2 is 7.4 x 1 9 and for MgCO 3 is 3.5 x 1 8. Which of the following will precipitate first? K sp (MgF 2 ) = [Mg 2+ ][F - ] 2 7.4 x 1-9 = (x)(2x) 2 x=1.2 x 1-3 M K sp (MgCO 3 ) = [Mg 2+ ][CO 3 2- ] 3.5 x 1-8 = (x)(x) x=1.9 x 1-4 M x= [Mg 2+ ]=solubility of Mg salt Smaller x is less soluble - will precipitate first MgCO 3 2
A comparision of 1. M HOCl and.1 M HOCl would be expected to show. 1. that the percent ionization in both solutions is the same 2. that the percent ionization of 1. M HOCl is greater than that of.1 M HOCl 3. that the percent ionization of.1 M HOCl is greater than that of 1. M HOCl 4. that a comparison of the percent ionization is not possible for the two solutions A comparision of 1. M HOCl and.1 M HOCl would be expected to show. K a is independent of molarity/initial concentration. Less concentrated solution (more dilute) will have higher number of molecules of products to reach K a. Higher number of molecules of products means higher percentage dissociation. 3
Which of the following K a values belongs to the weakest acid? 1. 1.4 x 1-4 2. 1.8 x 1-5 3. 4. x 1-4 4. 6.2 x 1-1 5. 6.4 x 1-5 Which of the following K a values belongs to the weakest acid? Smaller K a means less dissociation weaker acid. 4
It is desired to prepare a system that functions as a buffer at a ph of 5.6. Of the following systems, the best choice would be. 1. HC 2 H 3 O 2 (pk a = 4.7), C 2 H 3 O 2 (pk b = 9.3) 2. HOBr (pk a = 8.6), OBr (pk b = 5.4) 3. H 2 PO 4 (pk a = 7.2), HPO 2 4 (pk b = 6.8) 4. NH 2 OH + 2 (pk a = 5.9), NH 2 OH (pk b = 8.1) It is desired to prepare a system that functions as a buffer at a ph of 5.6. Of the following systems, the best choice would be. ph = pk a + log([a - ]/[HA]) Best buffer has [A - ] = [HA] Choose acid/salt combination with pk a closest to the desired ph NH 2 OH 2 + (pk a = 5.9), NH 2 OH (pk b = 8.1) 5
Addition of NaNO 2 to an aqueous solution of HNO 2 will cause. 1. both [NO 2 ] and the ph to increase 2. both [NO 2 ]and the ph to decrease 3. both [HNO 2 ] and the ph to decrease 4. [HNO 2 ] to decrease and the ph to increase Addition of NaNO 2 to an aqueous solution of HNO 2 will cause. At equilibrium HNO 2 (aq) H + (aq)+ NO -! 2 (aq) Addition of NO - 2 from NaNO 2 will increase [NO 2- ] and cause the equilbrium to shift to the left, increasing [HNO 2 ], and decreasing [H + ], which increases ph 6
The indicator phenol red (pk a =7.9) can be used to detect the equivalence point for the titration of a strong base with a strong acid, but not for the titration of a weak base with a strong acid. This is because. 1. the weak acid product is not strong enough to cause a color change in the indicator 2. the color change in the indicator occurs before the equivalence point is reached 3. the color change in the indicator occurs after the equivalence point is reached 4. the indicator changes color twice during the course of the titration The indicator phenol red (pk a =7.9) can be used to detect the equivalence point for the titration of a strong base with a strong acid, but not for the titration of a weak base with a strong acid. This is because. The indicator will change color at its pk a. At the equivalence point of a titration of a weak base and strong acid, the ph is determined by the pk a of the conjugate acid of te weak base. If this acid is weak, or the pk a is too small, the ph will be too high to induce a color change. 7
Which of the following statements is true? 1. When two opposing processes are proceeding at identical rates, the system is at equilibrium. 2. Catalysts are an effective means of changing the position of an equilibrium. 3. The concentration of the products equals that of the reactants, and is constant at equilibrium. 4. An endothermic reaction shifts toward reactants when heat is added to the reaction. 5. None of the above statements is true. Which of the following statements is true? At equilibrium the rates of the forward and reverse reactions are equal, resulting in a macroscopically static and microscopically dynamic system. Catalysts affect rate, but not the position of the equilibrium. Concentrations of products and reactants are constant, but not necessarily equal. An endothermic reaction requires heat to proceed, so an increase in temperature would shift the reaction towards the products. 8
According to the Bronsted -Lowry definition, an acid is. 1. a substance that increases the hydroxide ion concentration in a solution. 2. a substance that increases the hydrogen ion concentration in a solution. 3. a substance that can accept a proton from another species in solution. 4. a substance that can donate a proton to another species. 5. an electron pair acceptor. According to the Bronsted -Lowry definition, an acid is. Bronsted-Lowry defined an acid as a proton donor and a base as a proton acceptor. Effects on the [H + ] or [OH - ] (compared to water on it s own) may occur, but are not the definition. 9
In the following reaction, which species is the reducing agent? 3Cu (s) + 6H + (aq) + 2HNO 3 (aq) 3Cu 2+ (aq) + 2NO (g) + 4H 2 O (l) 1. H + 2. Cu 2+ 3. N in NO 4. Cu 5. N in HNO 3 In the following reaction, which species is the reducing agent? 3Cu (s) + 6H + (aq) + 2HNO 3 (aq) 3Cu 2+ (aq) + 2NO (g) + 4H 2 O (l) The reducing agent is the reactant that is oxidized. Oxidation: Cu (s) Cu 2+ (aq) +2e - Reducing agent Cu (s) 1
What is the percent H by mass of NH 4 HCO 3? 1. 5.1% 2. 6.3% 3. 9.5% 4. 11.9% 5. none of these What is the percent H by mass of NH 4 HCO 3? Atomic mass of NH 4 HCO 3 = 14+(5 x 1)+12+(3 x 16) = 79 amu Mass of H per molecule= 5 x 1 = 5 amu % mass H = (5/79) x 1% = 6.3% 11
What is the sum of the coefficients in the balanced reaction between lead(ii) nitrate and potassium iodide? 1. 2 2. 3 3. 4 4. 5 5. 6 What is the sum of the coefficients in the balanced reaction between lead(ii) nitrate and potassium iodide? Pb(NO 3 ) 2 + 2KI PbI 2 + 2KNO 3 Sum of coefficients = 1 + 2 + 1 + 2 = 6 12
The value of K sp for AgI is 1.5 x 1 16. Calculate the solubility, in moles per liter, of AgI in a.5 M NaI solution. 1. 1.7 x 1 8 mol/l 2. 1.2 x 1 8 mol/l 3. 2.8 x 1-9 mol/l 4. 3. x 1 16 mol/l 5. 7.5 x 1 17 mol/l The value of K sp for AgI is 1.5 x 1 16. Calculate the solubility, in moles per liter, of AgI in a.5 M NaI solution. [Ag + ] [I - ] Initial.5 Change +x +x Equilibrium x.5 + x K sp =1.5 x 1-16 = [Ag + ][I - ]=(x)(.5 + x) (x)(.5) = 1.5 x 1-16 x = [Ag + ] = 3 x 1-16 M = solubility of AgI 13
In a.2 M aqueous solution of HC 3 H 3 O 2, [H 3 O + ] is found to be 3.3 x 1 3 M. K a for this acid is therefore. 1. 3.3 x 1 3 2. 1.7 x 1 2 3. 8.3 x 1-2 4. 2.2 x 1 6 5. 5.5 x 1 5 In a.2 M aqueous solution of HC 3 H 3 O 2, [H 3 O + ] is found to be 3.3 x 1 3 M. K a for this acid is therefore. HC 3 H 3 O 2 (aq)! H + (aq)+ C 3 H 3 O - 2 (aq) K a = [H + ][C 3 H 3 O 2- ]/[HC 3 H 3 O 2 ] For every H + formed, one C 3 H 3 O - 2 is also formed K a = (3.3 x 1-3 ) 2 /.2 = 5.5 x 1-5 14
Consider the titration of 1. ml of.25 M aniline (C 6 H 5 NH 2 ; K b = 3.8 1 1 ) with.5 M HCl. Calculate the ph of the solution at the equivalence point. 1. 11.62 2. 8.7 3. 2.68 4..85 5. none of these Consider the titration of 1. ml of.25 M aniline (C 6 H 5 NH 2 ; K b = 3.8 1 1 ) with.5 M HCl. Calculate the ph of the solution at the equivalence point. At equivalence point, equal moles of acid and base are present and ph determined by conjugate acid of the aniline C 6 H 5 NH 2 HCl C 6 H 5 NH + 3 Initial (mmol) 25. 25. Change (mmol) -25. -25. +25. Final (mmol) 25. Final volume = 1. ml + (25mmol/.5M) = 1. ml + 5. ml = 15. ml C 6 H 5 NH 3 + (aq)! C 6 H 5 NH 2 (aq) + H + (aq) [C 6 H 5 NH + 3 ] [C 6 H 5 NH 2 ] [H + ] Initial.167 Change -x +x +x Equilibrium.167 - x x x K a = K w / K b = 1 x 1-14 / 3.8 x 1-1 = 2.6 x 1-5 K a =2.6 x 1-5 = [H + ][C 6 H 5 NH 2 ]/ [C 6 H 5 NH 3 + ] =x 2 /.167 x = [C 6 H 5 NH 2 ]=[H + ]=2 x 1-3 M ph = -log(2 x 1-3 ) = 2.67 15
Consider the reaction 2 BrCl (g)! Br 2 (g) + Cl 2 (g) at T = 25ºC. This reaction has an equilibrium constant K =.172 at this temperature. If P BrCl =.4 atm, P Br2 =.5 atm, and P Cl2 =.5 atm, the reaction. 1. will proceed by forming additional Br 2 (g) and Cl 2 (g). 2. will proceed by forming additional BrCl (g). 3. is at equilibrium. 4. none of the above Consider the reaction 2 BrCl (g) Br 2 (g) + Cl 2 (g) at T = 25ºC. This reaction has an equilibrium constant K =.172 at this temperature. If P BrCl =.4 atm, P Br2 =.5 atm, and P Cl2 =.5 atm, the reaction. K p = P Br2 P Cl2 / (P BrCl ) 2! Q p = (.5 atm)(.5 atm)/(.4 atm) 2 Q p =.156 Q p < K p so reaction will shift to the right (products) 16
Calculate the mass in grams of 3.65 x 1 2 molecules of SO 3. 1. 6.6 x 1-4 g 2. 4.85 x 1-2 g 3. 2.6 g 4. 165 g 5. none of the above Calculate the mass in grams of 3.65 x 1 2 molecules of SO 3. 3.65 x 1 2 molecules x (1 mol / 6.2 x 1 23 molecules = 6.6 x 1-4 moles of SO 3 6.6 x 1-4 moles SO 3 x (8 g / 1 mol SO 3 ) = 4.85 x 1-2 g SO 3 17