Answers to selected problems, Chapter 4

Review questions

1. (a) CH₃OH, methanol, is a nonelectrolyte because that's the way organic alcohols are.
    (b) KCl is a strong electrolyte because it is ionic. (See Chapter 3 for reasons why it is ionic.)
    (c) HI is a strong electrolyte because of I's high electronegativity. Except for HF, all HX (X = a halogen) are this way. But the bonds on HX are covalent, not ionic. So these are molecules that yield strong electrolytes. (See the table of common strong acids on page 139.)
    (d) HCOOH, formic acid, is a weak electrolyte like other organic acids. (See the discussion in the middle of page 138.)
    (e) NaOH is a strong electrolyte because it is an ionic compound.
    (f) HNO₂ is a weak electrolyte because it is a weak acid. Best just to learn this one. You can also note that it isn't in the group of strong acids listed in Table 4.1, on page 139.
    (g) HBr is a strong electrolyte for the reason given for HI in (c).
    (h) CH₂OHCHOHH₂OH, propane with a hydroxyl group on each carbon, is a nonelectrolyte for the same reason that methanol is (a).

4. Which of these solutions has the highest total concentration of ions?
    (a) 0.012 M Al₂(SO₄)₃ has a total concentration of ions = 5x0.012 M = 0.060 M because each mole of the formula gives 2 moles of Al³⁺ and 3 moles of (SO₄)^2-. The reason is that Al₂(SO₄)₃ has an implied coefficient of unity, i.e., 1 Al₂(SO₄)₃. The coefficient represents the number of moles of the entire formula unit, and the subscripts do the same for the components of the formula.
    (b) 0.030 M KCl. You can view this formula units as 1 K₁Cl₁. Thus, for one mole of KCl there is one more of K⁺ and one of Cl^-. The total concentration of ions is then 2x0.03 M, or 0.06 M.
    (c) 0.022 M CaCl₂. Since each mole of CaCl₂ gives one more of Ca²⁺ and two of Cl^-, the total concentration of ions here is 3x0.022 M = 0.066 M.
    (d) 0.025 M K₂SO₄. The molar concentration of total ions is three times the molar concentration of the formula unit (2 K's and 1 SO₄), or 3x0.025 M = 0.075 M. This is the highest of the four cases.

14. Why does only one of these systems react?
    (a) ZnCl₂(aq) + MgSO₄(aq) ® ??
    (b) ZnCl₂(aq) + KOH ® ??
    System (a) doesn't react because both of its products, ZnSO₄ and MgCl₂, are soluble. (See rules for solubility in this chapter.)
    System (b) reacts because one of the products, Zn(OH)₂, is insoluble.

21. Oxidation numbers of underlined atoms.
    (a) Cr = 0 (by definition) because it is an element;
    (b) ClO₂^- = +3 because Cl + 2(-2) = -1. (O is usually -2)
    (c) K₂Se = -2 because 2(+1) + Se = 0. (K is always +1)
    (d) TeF₆ = +6 because Te + 6(-1) = 0. (F is always -1)
    (e) PH₄⁺ =  -3 because P + 4(+1) = -3 (H is +1)
    (f) CaRuO₃ = +4 because 2 + Ru + 3(-2) = 0.
    (g) SrTiO₃ = +4 because 2 + Ti  + 3(-2) = 0.
    (h) P₂O₇^4- = +5 because 2P + 7(-2) = -4.
    (i) S₄O₆^2- = +2.5 because 4S + 6(-2) = -2.
    (j) NH₂OH = -1 because N + 2(+1) + (-1) = 0.

26. Even though Mg and Al react with acids to produce hydrogen, one of these equations is wrong. Why?
    (a) Mg(s) + 2 H⁺(aq) ® Mg²⁺(aq) + H₂(g)
    (b) Al(s) + 2 H⁺(aq) ® Al³⁺(aq) + H₂(g)
    Equation (b) is wrong because its charges aren't balanced (2+ on the left and 3+ on the right). Equation is OK because it has 2+ on both sides.

Problems

27. Determine the molarity of each of the following:
    (a) Li⁺ in 0.0385 M LiNO₃ = 0.0385 M because there is one Li per formula unit (1 Li₁NO₃).
    (b) Cl^- in 0.035 M CaCl₂ = 0.070 M because there are two moles of Cl for each mole of CaCl₂. (Note the significant figures: 0.070 vs. 0.035. They each have two significant figures because the 2 that multiplies 0.035 is an exact number, not something experimental.)
    (c) Al³⁺ in 0.0112 M Al₂(SO₄)₃ = 0.0224 M because there are two moles of Al³⁺ in each more of Al₂(SO₄)₃.
    (d) Na⁺ in 0.12 M Na₂HPO₄ = 0.24 M because there are two Na's for each Na₂HPO₄.

33. The total concentration of chloride ion in seawater, in mg/L, by using data from Exercise 4.1A.
    Seawater is essentially 0.438 M NaCl and 0.0512 M MgCl₂, plus a few minor solutes. This problem contains two steps, first calculating the total molar concentration of chloride and then converting it to a mass concentration.
    (1) The contributions to chloride are 0.438 M + 2(0.0512 M) = 0.438 + 0.1024 = 0.5404 M (limited to the third decimal place in subsequent calculations, though, as required by the 0.438 term).
    (2) Converting to mass: (0.5404 mol Cl/L)(35.45 g Cl/mole Cl) = 19.16 g Cl/L = 19.2x10³ mg Cl/L (rounded to three places) = 1.92x10⁴ mgCl/L (proper scientific notation).

39. Ionization of acids and bases. ("<=>" will substitute for the double arrows of the book.)
    (a) HI(aq) ® H⁺(aq) + I^-(aq) (strong acid)
    (b) KOH(aq) ® K⁺(aq) + OH^-(aq) (strong base)
    (c) HNO₂(aq) <=> H⁺(aq) + NO₂^-(aq) (weak acid)
    (d) H₂PO₄^-(aq) <=> H⁺(aq) + PO₄^2-(aq) (weak acid)
    (e) CH₃NH₂(aq) + H₂O <=> CH₃NH₃⁺(aq) + OH^-(aq) (Weak base)
    (f)CH₃CH₂COOH(aq) <=>  CH₃CH₂COO^-(aq) + H⁺(aq) (Weak acid)

43. Equation for dissolving calcium carbonate in vinegar (dilute acetic acid):
    To write the proper equation, we must remember what happens during the dissolution reaction: the acid end of acetic acid attacks the calcium carbonate to  form carbonic acid and liberate ionic Ca and the acetate counterion in the process. The qualitative version of this reaction is: 
    CaCO₃(s) + CH₃COOH(aq) ® Ca²⁺(aq) + CH₃COO^-(aq) + H₂CO₃(aq)
    The balanced version is :
    CaCO₃ + 2 CH₃COOH ® Ca²⁺ + 2 CH₃COO^- + H₂CO₃ 
    Additionally, the carbonic acid can decompose into carbon dioxide and water:
    H₂CO₃ ® CO₂ + H₂O

51. Without doing any detailed calculations, determine whether Alka-Seltzer (NaHCO₃) or Tums (CaCO₃) can neutralize more stomach acid [HCl(aq)] per unit mass.
    The approach here is to see which of the two compounds has the greater acid-neutralizing power per unit mass. First we quickly calculate the formula masses for each compound, which you should be able to do in you head by now: 84 u for NaHCO₃ and 100 for CaCO₃. The two values are roughly the same. Second, we note that one mole of NaHCO₃ can react with one mole of HCl (you can balance the neutralization reaction to be sure), whereas one mole of CaCO₃ can neutralize two moles of HCl. Thus Alka-Seltzer neutralizes 1 mole of HCl per mole of itself, and Tums neutralizes two moles. So Tums wins by a mile. Expressed mathematically, 2 moles/100 g > 1 mole/ 84 g.

57. Predict which of these combinations leads to a reaction and, if so, write the net ionic equation (no spectator ions).
    (a) MgO(s) + HI(aq) ® ??
        They react--the basic oxide MgO neutralizes the strong acid HI to make the salt plus water:
        MgO(s) + 2 H⁺(aq) ® Mg²⁺(aq) + H₂O(l)  
    (b) HCOOH(aq) + NH₃(aq) ® ??
        The weak acid formic acid reacts with the weak base ammonia:
        HCOOH(aq) + NH₃(aq) ® NH₄⁺(aq) + HCOO^-(aq)
    (c) CH₃COOH(aq) + H₂SO₄(aq) ® ??
        No reaction because one acid (formic) can't react with another (sulfuric).
    (d) CuSO₄(aq) + Na₂CO₃(aq) ® ??
        They react because one of the products, copper(II) carbonate, is insoluble:
        Cu²⁺(aq) + CO₃^2-(aq) ® Cu₂CO₃(s) (Na⁺ and SO₄^2- are spectator ions.)
    (e) KBr(aq) + Zn(NO₃)₂ ® ??
        No reaction because both products, potassium nitrate and zinc bromide, are soluble.

61. Write the net ionic reaction for the reaction of copper(II) nitrate and potassium carbonate. 
    From the ions involved you see that combining these solutions will create the insoluble copper(II) carbonate. The net ionic reaction is very simple:
    Cu²⁺(aq) + CO₃^2-(aq) ® CuCO₃(s)

65. Is the substance oxidixed, reduced, or neither?
    (a) Blue CrCl₂(aq) changes to green CrCl₃(aq) when exposed to air.
        Cr's oxidation state changes from +2 (because of the two Cl's) to +3 (because of the three Cl's), so it is oxidized.
    (b) Yellow K₂CrO₄(aq) changes to orange K₂Cr₂O₇(aq) when acidified.
        This one is a little trickier. In K₂CrO₄, the oxidation states of K and O are =1 and -2, as usual. You can find the oxidation state of Cr by summing all the oxidation numbers to zero, as we did above: 2(+1) + Cr + 4(-2) = 0 ® Cr = +6. For K₂Cr₂O₇, you do the same: 2(+1) + 2 Cr + 7(-2) = 0 ® Cr = 12/6 = 6 = the same as in the first compound. How can this be? It is because the difference between the two compounds of Cr is just CrO₃, where Cr has an oxidation number of +6, the same as in the first compound, Therefore, adding CrO₃ to K₂CrO₄ won't change Cr's oxidation state.
    (c) Nitrogen pentoxide reacts with water to form nitric acid. 
        "Nitrogen pentoxide" is officially dinitrogen pentoxide, N₂O₅, as shown on page 51. The oxidation number of nitrogen it is is +5. (Work it out for yourself.) Nitric acid is of course HNO₃. The oxidation number of its nitrogen is also +5. So when nitrogen pentoxide reacts with water to form nitric acid, the nitrogen is neither oxidized nor reduced.

69. Balancing redox equations.

    (a) Cu²⁺ + I₂ ® Cu⁺ + I^- 
        Cu decreases from +2 to +1, while I also decreases from 0 to -1. This reaction cannot be balanced as it stands.

    (b) Ag + H⁺ + NO₃^- ® Ag⁺ + H₂O + NO
        Ag increases from 0 to +1, and N decreases from +5 to +2. This requires three Ag's for each N. Insert the 3.
        3 Ag + H⁺ + NO₃^- ® 3 Ag⁺ + H₂O + NO
        This unbalances the electrical charges, with net 0 on the left side and +3 on the right. Balance the charges by making 4 H⁺'s on the left:
        3 Ag + 4 H⁺ + NO₃^- ® 3 Ag⁺ + H₂O + NO
        Now balance the H's by making 2 H₂O on the right:
        3 Ag + 4 H⁺ + NO₃^- ® 3 Ag⁺ + 2 H₂O + NO
        Now check to see that the O's balance. They do, with three on each side. The equation is balanced with respect to oxidation numbers, charge, and elements.

    (c) H₂O₂ + MnO₄^- + H⁺ ® Mn²⁺ + H₂O + O₂
        Oxidation numbers: the O of H₂O₂ is -1, as opposed to the usual -2 for O. The O of MnO₄^- is -2, as usual. The O of H₂O is also -2, and the O of O₂ is zero (because that O is in elemental form). So the easiest thing is to consider that the O₂ of H₂O₂ becomes the O₂ and the O of MnO₄^- goes to H₂O. Mn decreases from +7 in MnO₄^- to +2 in Mn²⁺. That's a decrease of five units for each Mn, which must be balanced by the increase in O from -1 in H₂O₂ to 0 in O₂. Since those O's come in pairs, the net oxidation numbers decrease by 2. To balance the oxidation numbers, we need 5 O₂'s for 2 Mn's:
        5 H₂O₂ + 2 MnO₄^- + H⁺ ® 2 Mn²⁺ + H₂O + 5 O₂    
        Balance the O's in MnO4- and H2O by taking 8 H₂O:
        5 H₂O₂ + 2 MnO₄^- + H⁺ ® 2 Mn²⁺ + 8 H₂O + 5 O₂       
        Then balance the charges by taking 6 H⁺ on the left side:
        5 H₂O₂ + 2 MnO₄^- + 6 H⁺ ® 2 Mn²⁺ + 8 H₂O + 5 O₂       
        This also balances the H's. All is balanced. 

    (d) PbO + V³⁺ + H₂O ® PbO₂ + VO²⁺ + H⁺
        Redox: Pb increases from +2 to +4 (+2). V increases from +3 to +4 (+1). This cannot be a real reaction because Pb and V increase their oxidation state, but nothing decreases to compensate.

    (e) IO₄^- + I^- + H⁺ ® I₂ + H₂O
        This is the opposite of a disproportionation reaction because some I increases its oxidation number, while some decreases.
        The I in IO₄^- decreases from +7 to 0 in I₂, while the I in I^- increases from -1 to 0 in I₂. That requires 7 I^- for each IO₄^-. Insert the coefficients.
        IO₄^- + 7 I^- + H⁺ ® I₂ + H₂O
        This requires 8 atoms of I, or 4 I₂:
        IO₄^- + 7 I^- + H⁺ ® 4 I₂ + H₂O
        Now you can balance the charge because you have coefficients for two of the three charged species, which incidentally are all on the left side of the equation:
        IO₄^- + 7 I^- + 8 H⁺ ® 4 I₂ + H₂O
        Having just set the hydrogens on the left side, they can be balanced on the right:
        IO₄^- + 7 I^- + 8 H⁺ ® 4 I₂ + 4 H₂O
        The only thing left to balance is the oxygens. They can be checked. There are indeed four atoms on the left and four on the right. So the equation is balanced with respect to redox, I, charge, H, and O.

77. Using the activity series of page 161 to predict whether particular combinations of metals and ions react, and balancing the reactions that are found. The key is to understand that elements near the top of Figure 4.15 tend to be oxidized (lose electrons) relative to those near the bottom, which tend to be reduced (gain electrons). Thus if a metal in a higher position is given as an ion and one lower on the arrow is given in the elemental form, they will be in their preferred configurations relative to one another and will not react further. If the combination is presented in the opposite way, with the "higher" metal as an element and the lower metal as an ion, they will react to return to their preferred states. 
    (a) Zn(s) + H⁺(aq) ® ??
        Since Zn is above H in Figure 4.15, Zn will reduce H⁺:
        Zn(s) + 2 H⁺(aq) ® Zn²⁺(aq) + H₂(g)
    (b) Cu(s) + Zn²⁺(aq) ® ??
        Since Zn, the oxidized metal here, is above Cu in the figure, they are in their preferred states and will not react further.
    (c) Fe(s) + Ag⁺(aq) ® ??
        Since Fe is well above Ag in the figure but is not oxidized, they will react to oxidize Zn and reduce Ag⁺:
        Fe(s) + 2 Ag⁺(aq) ® Fe²⁺(aq) + 2 Ag(s)
    (d) Au(s) + H⁺ ® ??
        Since Au falls below H in the activity series and is also the reduced one of the pair, it will not react further with H⁺.

85. The basic approach to this problem is to find the mass of Cl^- in the NaCl and in the MgCl₂ separately, add them, and covert the total to molarity.
    (a) Mass of Cl in NaCl = (mass of NaCl in the 6.85 g of the mixture) x (mass fraction of Cl in NaCl) = (6.85 g mixture)(0.988 g NaCl/g mixture)(35.5 Cl/(23.0 + 35.5 NaCl)) = 4.11 g Cl^-
    (b) Mass of  Cl- in MgCl₂ = (mass of MgCl₂ in the 6.85 g of the mixture) x (mass fraction of Cl in MgCl₂) = (6.85 g mixture)(0.012 g MgCl₂/g mixture)(71.0 Cl/(24.3 + 71.0 MgCl₂)) = 0.061 g Cl^-
    (c) Total Cl^- = 4.11 g + 0.061 g 4.17 g
    (d) Moles of Cl^- = 4.17 g/35.5 g mol-1 = 0.1175 moles
    (e) Molarity of solution = 0.1175 mol Cl^-/0.5000 L = 0.235 M

89. To determine how much sodium carbonate is required to neutralize the sulfuric acid, you must first know the equation of neutralization. The unbalanced reaction is:
    H₂SO₄ + Na₂CO₃ ® Na₂SO₄ + H₂CO₃ (which then decomposes to H₂O and CO₂, but which is irrelevant to this problem)
    Happily for you, this equation is already balanced, with all the (molar) coefficients equal to unity. Thus, one mole of sodium carbonate is required to neutralize one mole of sulfuric acid. As in the previous chapter, you convert the mass of sulfuric acid to moles, then find the number of moles of sodium carbonate required to neutralize it (the same number), and then convert to mass of sodium carbonate. These steps can be summarized as: kg H₂SO₄ ® moles H₂SO₄ ® moles Na₂CO₃ ® kg of Na₂CO₃.
    (a) Moles of H₂SO₄ = (1.5x10³ kg)(10³ g/kg)(0.932)/(98 g/mole) = 1.43x10⁴
    (b) Moles of Na₂CO₃ = 1.43x10⁴ (the same)
    (c) kg of Na₂CO₃ = (moles of Na₂CO₃)(106 g/mole)/(10³ g/kg) = (1.43x10⁴)(106)/(10³) = 1.51x10³ kg = 1510 kg

Note that this procedure can be simplified by combining steps (a) through (c):
    kg of Na₂CO₃ = (moles of H₂SO₄)(106 g/mole)/(10³ g/kg) = (1.5x10³ kg)(10³ g/kg)(0.932)(98 g/mole)^-1(106)/(10³) = (1.5x10³ kg)(0.932)(106 g mole^-1/98 g/mole) = 1510 kg (The two factors of 10³ cancel out.)

95. Balancing more redox equations.
    (a) CrI₃ + H₂O₂ + OH^- ® CrO₄^2- + IO₄^- + H₂O
        Redox: Cr increases from +3 to +6 (+3); I increases from -1 to+7 (+8); O decreases from -1 (in H₂O₂) to -2 (in everything on the right side). While it first appears problematic to have three elements change oxidation numbers rather than the usual two, the problem can be dealt with by noting that both species that increase are in the same formula unit (CrI₃). Thus for each molecule of CrI₃, the Cr increases by 3 and the three I's increase by 8 each, for a total increase of 27 units. This must be balanced by the decrease in oxidation numbers for the 2 O's in H₂O₂, which each decrease by one unit. Thus for each H₂O₂, there are two units of reduction, against the 27 units of oxidation in CrI₃. To balance oxidation and reduction, we can just put a coefficient of 13.5 in front of the H₂O₂. Better to avoid the fractional coefficients by using 2 for CrI₃ and 27 for H₂O₂ instead. That gives:
    2 CrI₃ + 27 H₂O₂ + OH^- ® CrO₄^2- + IO₄^- + H₂O, or 54 units of oxidation and or reduction.
    Next, balance Cr and I:
    2 CrI₃ + 27 H₂O₂ + OH^- ® 2 CrO₄^2- + 6 IO₄^- + H₂O
    Now we can balance the charge because two of the three charged units have been balanced:
    2 CrI₃ + 27 H₂O₂ + 10 OH^- ® 2 CrO₄^2- + 6 IO₄^- + H₂O
    Now we can balance H and check O, or the reverse. We balance H:
     2 CrI₃ + 27 H₂O₂ + 10 OH^- ® 2 CrO₄^2- + 6 IO₄^- + 32 H₂O
    Oxygen checks because there are 64 on each side.

Back to CHM101 Fall 2001
Back to Home Page