CHAPTER 4: ANSWERS TO ASSIGNED PROBLEMS

1 1 CHAPTER 4: ANSWERS TO ASSIGNED PROBLEMS Hauser- General Chemistry I revised 10/14/08 You are presented with three white solids, A, B, and C, which are glucose (a sugar substance), NaOH, and AgBr. Solid A dissolves in water to form a conducting solution. B is not soluble in water. C dissolves in water to form a nonconducting solution. Identify A, B, and C. AgBr is an ionic solid that is NOT soluble in water. Since A and C dissolve in water, B is AgBr. C dissolves in water but does not conduct a current, so it cannot be ionic. C is glucose. Substance A dissolves in water and is comprised of soluble ions. Substance A is NaOH. Which of the following ions will always be a spectator ion in a precipitation reaction? Explain briefly. (a) Cl- NO. Can form ppts with Ag+, Hg22+ and Pb2+. (b) NO3- YES. All nitrates are soluble. (c) NH4+ YES. All ammoniums are soluble. (d) S2- NO. Sulfides usually form ppts. (e) SO42- NO. Sulfates usually form ppts. Using solubility guidelines, predict whether each of the following compounds is soluble or insoluble in water: (a) NiCl2 SOLUBLE most chlorides are soluble (b) Ag2S INSOLUBLE most sulfides not soluble (c) Cs3PO4 SOLUBLE while most phosphates are not soluble, Grp IA Cs+ is (d) SrCO3 INSOLUBLE most carbonates not soluble (e) PbSO4 INSOLUBLE most sulfates soluble, but not with Pb+2 Will precipitation occur when the following solutions are mixed?

2 If so, write a balanced chemical equation for the reaction. (a) Na2CO3 and AgNO3 PPT. Na2CO3 (aq) + 2 AgNO3 (aq) 2 NaNO3 (aq) + Ag2CO3 (s) or (b) NaNO3 and NiSO4 NO PPT. (c) FeSO4 and Pb(NO3)2 PPT. FeSO4 (aq) + Pb(NO3)2 (aq) Fe(NO3)2 (aq) + PbSO4 (s) or 2 Write balanced net ionic equations for the reactions that occur in each of the following cases. Identify the spectator ion or ions in each reaction. (a) Cr2(SO4)3 (aq) + (NH4)2CO3 (aq) dissociate compounds into ions Cr3+ (aq)+ SO42- (aq) + NH4+ (aq) + CO32- (aq) swap partners and write neutral formulas Cr3+ (aq)+ SO42- (aq) + NH4+ (aq) + CO32- (aq) (NH4)2SO4 + Cr2(CO3)3 balance 2 Cr3+ (aq)+ 3 SO42- (aq) + 6 NH4+ (aq) + 3CO32- (aq) 3 (NH4)2SO4 + Cr2(CO3)3 check solubility rules 2 Cr3+ (aq)+ 3 SO42- (aq) + 6 NH4+ (aq) + 3 CO32- (aq) 3 (NH4)2SO4 (aq) + Cr2(CO3)3 (s) write net ionic to show stable product formation 2 Cr3+ (aq) + 3 CO32- (aq) Cr2(CO3)3 (s) spectator ions were SO42- (aq) and NH4+ (aq) (b) Ba(NO3)2 (aq) + K2SO4 (aq) Ba2+ (aq) + SO4 2- (aq) BaSO4 (s) spectator ions were NO3 and K+ (c) Fe(NO3)2 (aq) + KOH (aq) Fe 2+ (aq) + 2 OH- (aq) Fe(OH)2 (s) spectator ions were NO3- and K+ Complete and balance the following molecular equations, and then write the net ionic equation for each.

3 (a) HBr (aq) + Ca(OH)2 (aq) 2 HBr (aq) + Ca(OH)2 (aq) CaBr2 (aq) + 2 H2O (l) NET H+ (aq) + OH- (aq) H2O (l) (b) Cu(OH)2 (s) + HClO4 (aq) Cu(OH)2 (s) + 2 HClO4 (aq) Cu(ClO4)2 (aq) + 2 H2O (l) NET H+ (aq) + OH- (aq) H2O (l) (USE LOWEST RATIO) 3 (c) Al(OH)3 (s) + HNO3 (aq) Al(OH)3 (s) + 3 HNO3 (aq) Al(NO3)3 (aq) + 3 H2O (l) NET H+ (aq) + OH- (aq) H2O (l) Write a balanced molecular equation and a net ionic equation for the reaction that occurs when (a) solid CaCO3 reacts with an aqueous solution of nitric acid CaCO3 (s) + 2 HNO3 (aq) Ca(NO3)2 (aq) + H2O (l) + CO2 (g) acceptable answer for NET IONIC NET CO3 2- (aq) + 2 H+ (aq) H2O (l) + CO2 (g) (STABLE PRODUCTS) [officially 2H+ (aq) + CaCO3 (s) H2O (l) + CO2 (g) + Ca2+ (aq) since not all aqueous] (b) solid iron(II) sulfide reacts with an aqueous solution of hydrobromic acid FeS (s) + HBr (aq) FeBr2 (aq) + H2S (g) acceptable answer for NET IONIC S 2- (aq) + 2 H+ (aq) H2S (g) [officially 2H+ (aq) + FeS (s) H2S (g) + Fe2+ (aq) since not all aqueous] Define oxidation and reduction in terms of (a) electron transfer OXIDATION IS LOSS of e- REDUCTION IS GAIN of e- "OIL RIG" (b) oxidation numbers OXIDATION IS INCREASE IN OX # REDUCTION IS DECREASE IN OX # Determine the oxidation number for the indicated element in each of the following substances: (a) S in SO2 S = +4 (b) C in COCl2 (note: NOT Co) C = +4 (assign O, then Cl, then C) (c) Mn in MnO4- Mn = +7 (take charge of ion into account) (d) Br in HBrO Br = +1 (assign O, then Br) Which element is oxidized and which is reduced in the following reactions?

4 (a) N2 (g) + 3 H2 (g) 2 NH3 (g) assign OX #'s first 0 0 -3 +1 N2 (g) + 3 H2 (g) 2 NH3 (g) ignore coefficients for this application H atom (0 to +1) oxidized N atom (0 to -3) reduced 4 (b) 3 Fe(NO3)2 (aq) + 2 Al (s) 3 Fe (s) + 2 Al(NO3)3 (aq) +2 +5 -2 0 0 +3 +5 -2 3 Fe(NO3)2 (aq) + 2 Al (s) 3 Fe (s) + 2 Al(NO3)3 (aq) Al atom (0 to +3) oxidized Fe atom (+2 to 0) reduced ADDITIONAL EXERCISE #2 Examine the following reactions and identify whether or not a REDOX (oxidation-reduction) reaction has occurred. If REDOX has occurred, report the following: - the atom oxidized - the atom reduced - the oxidizing agent - the reducing agent +4 -1 0 +2 -1 0 A) SiCl4 + 2 Mg 2 MgCl2 + Si Mg atom ( 0 to +2) oxidized, Mg is reducing agent Si atom (+4 to 0) reduced, SiCl4 is oxidizing agent +4 -1 +1 -2 +1 -1 +4 -2 B) SiCl4 + 2 H2O 4 HCl + SiO2 NO REDOX ADDITIONAL EXERCISE #3 Use the Activity Series (Table ) to predict the outcome (if any) of each of the following reactions: ANY METAL ON LIST CAN BE OXIDIZED BY THE IONS OF AN ELEMENT BELOW IT.

5 A) Cr (s) + 3 AgNO3 Cr(NO3)3 (aq) + Ag (s) because Ag ion is below Cr on list b) Fe (s) + Mn+2 NR because Mn ion is above Fe on list (a) Calculate the molarity of a solution that contains mol NH4Cl in exactly 500. mL of solution. M = mol solute / L sol'n = mol / L = mol/L or M (Did you convert mL to L?) (b) How many moles of HNO3 are present in mL of a M solution of nitric acid? convert mL to L while maintaining SF = mL ( 1 L / 1000 mL) = L M = mol solute / L sol'n so rearrange to mol solute = M L = M ( L); convert M to its units of mol/L; L cancels ANSWER: mol 5 (c) How many milliliters of M KOH solution are needed to provide mol of KOH? M = mol solute / L sol'n so rearrange to L = mol solute / M L = mol / ( mol/L) = L; convert to 183 mL Calculate (b) the molar concentration of a solution containing g of Ca(NO3)2 in L M = mol solute / L sol'n convert grams to moles Ca 1 X = N 2 X = O 6 X = sum = g / mol ( g) ( 1 mol Ca(NO3)2 / g) ---------------------------------------- ----------------- = = mol / L (4 SF) L (a) Which will have the highest concentration of potassium ion: M KCl, M K2 CrO4, or M K3PO4?

6 Beware of dissociation! M KCl 1 X mol/L = M K+ M K2 CrO4 2 X mol/L = M K+ M K3PO4 3 X mol/L = M K+ M K2 CrO4 has the highest potassium ion concentration (a) You have a stock solution of M NH3. How many milliliters of this solution should you dilute to make mL of M NH3? CiVi = CfVf Ci = M Vf = mL Cf = M SOLVE THE ALGEBRA! Vi = CfVf / Ci Vi = ( M) ( mL) / M = mL = mL (3 SF) Check work: You have more volume and lesser concentration at end. 6 (a) What volume of M HClO4 solution is needed to neutralize mL of M NaOH? Start your calculation by writing a balanced chemical equation: 1 NaOH + 1 HCLO4 H2O + NaCLO4 Start the math with the item you know the most about (both volume and molarity here). Don't forget to convert mL to L so you can use molarity concepts. FOLLOW THE LABELS! RELATE MOLES OF ACID TO MOLES OF BASE. mL (1 L / 1000 mL) = L ( mol NaOH) ( 1mol HCLO4) (1 L HCLO4) L NaOH _____ _____ _____ (1 L NaOH) (1 mol NaOH) mol HCLO4 = L or mL (b) What volume of M HCl is needed to neutralize g of Mg(OH)2?

7 BEWARE! "DIBASIC" COMPOUND! Start your calculation by writing a balanced chemical equation: 1 Mg(OH)2 + 2 HCL 2H2O + MgCL2 You will be converting grams to moles of Mg(OH)2: Mg 1 X = O 2 X = H 2 X = sum g /mol (1 mol Mg(OH)2) (2 mol HCL) ( 1 L HCL) (1000 mL) g Mg(OH)2 _____ _____ _____ _____ = ( g Mg(OH)2) (1 mol Mg(OH)2) mol HCL (1L) = L = L or 769 mL (3 SF)