Chapter 4 Suggested end-of-chapter problems with solutions

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1 Chapter 4 Suggested end-of-chapter problems with solutions a. 5.6 g NaHCO 1 mol NaHCO g NaHCO = mol NaHCO M = mol 50.0 m 1000 m = M NaHCO b g K Cr O 7 1 mol K 94.0 g K Cr O Cr O 7 7 = mol K Cr O 7 M = mol = M K Cr O 7 c g Cu 1 mol Cu 6.55 g Cu = mol Cu = mol Cu + M = mol Cu 00.0 m 1000 m = M Cu +

2 75.0 m 0.79 g m 1 mol = 1. mol C H 5 OH; molarity = g 1. mol 0.50 = 5. M C H 5 OH g (NH 4 ) SO 4 1 mol 1.15 g = mol (NH 4 ) SO 4 Molarity = mol m 1000 m = M (NH 4 ) SO 4 Moles of (NH 4 ) SO 4 in final solution: mol Molarity of final solution = = mol mol ( ) m (NH 4 ) SO 4 (s) NH 4 + (aq) + SO 4 (aq); NH m = 0.16 M (NH 4 ) SO 4 M = (0.16) = 0.7 M; M = 0.16 M SO 4

3 total mol HNO Molarity = ; total volume = = total volume Total mol HNO = mol HNO mol HNO Total mol HNO = mol mol = mol HNO Molarity = mol HNO = M HNO As expected, the molarity of HNO is between M and 0.00 M.. The solubility rules referenced in the following answers are outlined in Table 4.1 of the text. a. Soluble: Most nitrate salts are soluble (Rule 1). b. Soluble: Most chloride salts are soluble except for Ag +, Pb +, and Hg + (Rule ). c. Soluble: Most sulfate salts are soluble except for BaSO 4, PbSO 4, Hg SO 4, and CaSO 4 (Rule 4.) d. Insoluble: Most hydroxide salts are only slightly soluble (Rule 5). Note: We will interpret the phrase slightly soluble as meaning insoluble and the phrase marginally soluble as meaning soluble. So the marginally soluble hydroxides Ba(OH), Sr(OH), and Ca(OH) will be assumed soluble unless noted otherwise. e. Insoluble: Most sulfide salts are only slightly soluble (Rule 6). Again, slightly soluble is interpreted as insoluble in problems like these. f. Insoluble: Rule 5 (see answer d). g. Insoluble: Most phosphate salts are only slightly soluble (Rule 6).

4 . In these reactions, soluble ionic compounds are mixed together. To predict the precipitate, switch the anions and cations in the two reactant compounds to predict possible products; then use the solubility rules in Table 4.1 to predict if any of these possible products are insoluble (are the precipitate). Note that the phrase slightly soluble in Table 4.1 is interpreted to mean insoluble, and the phrase marginally soluble is interpreted to mean soluble. a. Possible products = FeCl and K SO 4 ; both salts are soluble, so no precipitate forms. b. Possible products = Al(OH) and Ba(NO ) ; precipitate = Al(OH) (s) c. Possible products = CaSO 4 and NaCl; precipitate = CaSO 4 (s) d. Possible products = KNO and NiS; precipitate = NiS(s) a. (NH 4 ) SO 4 (aq) + Ba(NO ) (aq) NH 4 NO (aq) + BaSO 4 (s) Ba + (aq) + SO 4 (aq) BaSO 4 (s) b. Pb(NO ) (aq) + NaCl(aq) PbCl (s) + NaNO (aq) Pb + (aq) + Cl (aq) PbCl (s) c. Potassium phosphate and sodium nitrate are both soluble in water. No reaction occurs. d. No reaction occurs because all possible products are soluble. e. CuCl (aq) + NaOH(aq) Cu(OH) (s) + NaCl(aq) Cu + (aq) + OH (aq) Cu(OH) (s)

5 Because no precipitates formed upon addition of NaCl or Na SO 4, we can conclude that Hg + and Ba + are not present in the sample because Hg Cl and BaSO 4 are insoluble salts. However, Mn + may be present since Mn + does not form a precipitate with either NaCl or Na SO 4. A precipitate formed with NaOH; the solution must contain Mn + because it forms a precipitate with OH [Mn(OH) (s)]. The reaction is AgNO (aq) + NaBr(aq) AgBr(s) + NaNO (aq). Assuming AgNO is limiting: m AgNO m mol AgNO AgNO 1 mol AgBr mol AgNO g AgBr mol AgBr =.8 g AgBr Assuming NaBr is limiting: 0.0 m NaBr m 1.00 mol NaBr NaBr 1 mol AgBr mol NaBr g AgBr mol AgBr =.76 g AgBr The AgNO reagent produces the smaller quantity of AgBr, so AgNO is limiting and.8 g AgBr can form.

6 M SO 4 (aq) + CaCl (aq) CaSO 4 (s) + MCl(aq) 1.6 g CaSO 4 1 mol CaSO g CaSO 4 1 mol M SO4 = mol M SO 4 mol CaSO 4 From the problem, 1.4 g M SO 4 was reacted, so: molar mass = g M SO 4 mol M SO 4 = 14 g/mol 14 u = (atomic mass M) (16.00), atomic mass M = u From periodic table, M = Na (sodium). All the bases in this problem are ionic compounds containing OH -. The acids are either strong or weak electrolytes. The best way to determine if an acid is a strong or weak electrolyte is to memorize all the strong electrolytes (strong acids). Any other acid you encounter that is not a strong acid will be a weak electrolyte (a weak acid), and the formula should be left unaltered in the complete ionic and net ionic equations. The strong acids to recognize are HCl, HBr, HI, HNO, HClO 4, and H SO 4. For the following answers, the order of the equations are formula, complete ionic, and net ionic. a. HClO 4 (aq) + Mg(OH) (s) H O(l) + Mg(ClO 4 ) (aq) H + (aq) + ClO 4 (aq) + Mg(OH) (s) H O(l) + Mg + (aq) + ClO 4 (aq) H + (aq) + Mg(OH) (s) H O(l) + Mg + (aq) b. HCN(aq) + NaOH(aq) H O(l) + NaCN(aq) HCN(aq) + Na + (aq) + OH (aq) H O(l) + Na + (aq) + CN (aq) HCN(aq) + OH (aq) H O(l) + CN (aq) c. HCl(aq) + NaOH(aq) H O(l) + NaCl(aq) H + (aq) + Cl (aq) + Na + (aq) + OH (aq) H O(l) + Na + (aq) + Cl (aq) H + (aq) + OH (aq) H O(l)

7 a. Perchloric acid plus potassium hydroxide is a possibility. HClO 4 (aq) + KOH(aq) H O(l) + KClO 4 (aq) b. Nitric acid plus cesium hydroxide is a possibility. HNO (aq) + CsOH(aq) H O(l) + CsNO (aq) c. Hydroiodic acid plus calcium hydroxide is a possibility. HI(aq) + Ca(OH) (aq) H O(l) + CaI (aq) If we begin with m of 0.00 M NaOH, then: mol = 1.00 a. NaOH(aq) + HCl(aq) NaCl(aq) + H O(l) mol NaOH 1 mol HCl mol NaOH 10 mol NaOH is to be neutralized mol = or 100. m b. HNO (aq) + NaOH(aq) H O(l) + NaNO (aq) mol NaOH 1 mol HNO mol NaOH c. HC H O (aq) + NaOH(aq) H O(l) + NaC H O (aq) mol NaOH 1 mol HCHO mol NaOH 1 = or 66.7 m mol HNO 1 = mol HC H O

8 Ba(OH) (aq) + HCl(aq) BaCl (aq) + H O(l); H + (aq) + OH (aq) H O(l) mol HCl = mol Ba(OH) 10 mol HCl = mol H + = 1.4 = mol Ba(OH) mol Cl 10 mol Ba mol OH The net ionic equation requires a 1 : 1 mole ratio between OH and H +. The actual mole OH to mole H + ratio is greater than 1 : 1, so OH is in excess. Because mol OH will be neutralized by the H +, we have ( ) 10 = mol OH in excess. M = OH mol OH excess total volume mol OH = M OH KHP is a monoprotic acid: NaOH(aq) + KHP(aq) H O(l) + NaKP(aq) Mass KHP = NaOH mol NaOH NaOH 1 mol KHP mol NaOH 04. g KHP mol KHP = g KHP

9 Apply the rules in Table 4.. a. KMnO 4 is composed of K + and MnO 4 ions. Assign oxygen an oxidation state of, which gives manganese a +7 oxidation state because the sum of oxidation states for all atoms in MnO 4 must equal the 1 charge on MnO 4. K, +1; O, ; Mn, +7. b. Assign O a oxidation state, which gives nickel a +4 oxidation state. Ni, +4; O,. c. Na 4 Fe(OH) 6 is composed of Na + cations and Fe(OH) 6 4 anions. Fe(OH) 6 4 is composed of an iron cation and 6 OH anions. For an overall anion charge of 4, iron must have a + oxidation state. As is usually the case in compounds, assign O a oxidation state and H a +1 oxidation state. Na, +1; Fe, +; O, ; H, +1. d. (NH 4 ) HPO 4 is made of NH 4 + cations and HPO 4 anions. Assign +1 as the oxidation state of H and as the oxidation state of O. In NH 4 +, x + 4(+1) = +1, x = = oxidation state of N. In HPO 4, +1 + y + 4( ) =, y = +5 = oxidation state of P. e. O, ; P, + f. O, ; x + 4( ) = 0, x = +8/ = oxidation state of Fe; this is the average oxidation state of the three iron ions in Fe O 4. In the actual formula unit, there are two Fe + ions and one Fe + ion. g. O, ; F, 1; Xe, +6 h. F, 1; S, +4 i. O, ; C, + j. H, +1; O, ; C, 0

10 To determine if the reaction is an oxidation-reduction reaction, assign oxidation states. If the oxidation states change for some elements, then the reaction is a redox reaction. If the oxidation states do not change, then the reaction is not a redox reaction. In redox reactions, the species oxidized (called the reducing agent) shows an increase in oxidation states, and the species reduced (called the oxidizing agent) shows a decrease in oxidation states. Redox? Oxidizing Reducing Substance Substance Agent Agent Oxidized Reduced a. Yes Ag + Cu Cu Ag + b. No c. No d. Yes SiCl 4 Mg Mg SiCl 4 (Si) e. No In b, c, and e, no oxidation numbers change.

11 a. The first step is to assign oxidation states to all atoms (see numbers above the atoms) C H 6 + O CO + H O Each carbon atom changes from to +4, an increase of 7. Each oxygen atom changes from 0 to, a decrease of. We need 7/ O atoms for every C atom in order to balance electron gain with electron loss. C H 6 + 7/ O CO + H O Balancing the remainder of the equation by inspection: or C H 6 (g) + 7/ O (g) CO (g) + H O(g) C H 6 (g) + 7 O (g) 4 CO (g) + 6 H O(g) b. The oxidation state of magnesium changes from 0 to +, an increase of. The oxidation state of hydrogen changes from +1 to 0, a decrease of 1. We need H atoms for every Mg atom in order to balance the electrons transferred. The balanced equation is: Mg(s) + HCl(aq) Mg + (aq) + Cl (aq) + H (g) c. The oxidation state of nickel increases by (0 to +), and the oxidation state of cobalt decreases by 1 (+ to +). We need Co + ions for every Ni atom in order to balance electron gain with electron loss. The balanced equation is: Ni(s) + Co + (aq) Ni + (aq) + Co + (aq) d. The equation is balanced (mass and charge balanced). Each hydrogen atom gains one electron (+1 0), and each zinc atom loses two electrons (0 +). We need H atoms for every Zn atom in order to balance the electrons transferred. This is the ratio in the given equation: Zn(s) + H SO 4 (aq) ZnSO 4 (aq) + H (g)

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