Solutions to CHEM 301 Review Exercises

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1 Solutions to CHEM 301 Review Eercises naming 1. a) calcium phosphate b) chromium (III) oide c) chlorine dioide. a) NaOCl b) HgSO 4 significant figures 3. [H + ] M has three significant figures, therefore the ph is reported to three places after the decimal point. So ph is reported. (The whole number to the left of the decimal point reports the order of magnitude in powers of ten). 4. p sp log sp log ( ) 9.7 (note: there is only one significant figure in this result) units of concentration 5. a) We must convert mmol of P into mass (mg) of PO 4 3. Since each mol of PO 4 3 contains one P, there must be 3. mmol of PO 4 3 per liter. So all we need to do is multiply the number of mols of PO 4 3 by its molar mass (94.97 g/mol) mmol P 1mol PO g PO mg ppm PO 3 4 L 1mol P 1mol PO 1g 4 b) Recall, 1 ppb in freshwater is equivalent to the ratio 1 μg/kg which in dilute solution (freshwater d ~ 1.00 g/ml) is equal to 1 μg/l mol Cu g Cu 10 ug ppb Cu L mol Cu 1g 6 (in freshwater) 6. a) Converting mass (mg) to moles involves dividing by the molar mass of the species involved. In this case we are dealing with molecular oygen (O ), so the molar mass is 3.00 g/mol or 3000 mg/mol. 9.8 mg O L 1mol O M O 3000 mg O 1

2 b) 150 μg As 1mol As 1g kg M As 6 1kg 74.9 g As 10 μg 1L Where the last term has been introduced to correct for the density of seawater. 7. To carry out a charge balance, one must calculate the equivalents of charge (both positive and negative). Therefore, one must convert the given ion concentrations to molar concentrations and then multiply by the magnitude of the ion charge in each case. Note: equiv charge # mol of ion (n ion ) # unit charge/mol (Z) Σ (+ ve) 6.08 meq/l and Σ ( ve).60 meq/l. Since in solution the charge must balance, the calculated discrepancy means one of two things. Either one (or more) of the analysis is in error or one (or more) ion/s have been neglected. (Since, humic substances (e.g., DOC) carry a negative charge, they are sometimes associated with small anion deficiencies). 8. Equivalents of acid refer to the number of moles of protons (rather than the number of moles of unit charge, as above). Since each mole of sulfuric acid (H SO 4 ) supplies moles of H +, a M solution is equivalent to moles of H + /L (N or equivalents/l). Thus, the number of equivalents of acid neutralized in this reaction is: 1L.6 ml 1000 ml equiv L 1000 mequiv 1equiv mequiv of H equilibrium epressions 9. Very often, we will be required to evaluate an equilibrium constant of a comple reaction by combining the known equilibrium epressions of known simpler reactions. In this case, we will rearrange and combine the given epressions such that they sum to equal the reaction of interest. CaCO 3 (s) Ca + (aq) + CO 3 (aq) sp CO (g) CO (aq) H CO 3 (aq) + H 3 O + (aq) H O + HCO 3 (aq) 1/ a CO (aq) + H O HCO 3 (aq) + H + (aq) a1 CaCO 3 (s) + H O(l) + CO (g) Ca + (aq) + HCO 3 (aq) Thus, eq sp H a a1 [Ca + ][HCO P CO 3 ]

3 Since all of the equilibrium constants on the right hand side of this epression are known, we could calculate the value of eq for the reaction of interest. This value could be used then, for eample to calculate the equilibrium concentration of HCO 3 if [Ca + ] and the P CO were measured or known. 10. a) At ph 8.0, the activity (concentration) of [H + ] M. Since, + [H3O ][OCl ] eq [HOCl] can be rearranged + [HOCl] [H3O ] [OCl ] eq Since eq and [H 3 O + ] are known, this ratio is given by / Hence, [HOCl]/[OCl ] 0.4 (in other words, [HOCl] 0.4 [OCl ]) The total OCl [HOCl] + [OCl ], so the percent of HOCl to the total is given by: [HOCl] 0.4[OCl ] [HOCl] + [OCl ] 0.4[OCl ] + [OCl ] 1.4 b) Substituting the instantaneous concentrations into the equilibrium epression to calculate Q yields, Q (10 7 ) (10 4 )/(10 3 ) 10 8 Since Q < eq, the reaction will proceed to right to increase to concentration of the products relative to that of the reactants. Consequently, the [HOCl] will decrease as the reaction approaches equilibrium. 9% acids/bases 11. Using the rules for predicting the ph of aqueous solutions. a) acidic (Cu + metal cations other than alkali and alkali earth metals are slightly acidic, Cl is neutral) b) basic (Na + is an alkali metal, therefore neutral and PO 4 3 is the conjugate base of HPO 4, which is a weak acid. Therefore the solution is basic) c) acidic (ammonium ion is acidic, chloride is neutral) d) basic (Mg + is an alkali earth metal, therefore neutral and S is the conjugate base of a weak acid HS. Therefore the solution is basic) 1. The relevant equilibria are potentially: + NH 4 + H O NH 3 + H 3 O + a Cl + H O HCl + OH H O H 3 O + + OH w

4 Using the first and last of these, we can set up an equilibrium table as follows. Where , since ph 5.13, the [H + ] M + H O NH 3 + H 3 O + a I i C E i i NH 4 + Using the epression for a (NH 4 + ) a [NH 3 ] [H 3 O + ]/[NH 4 + ] So, ( ) ( )/(i ) Solving for i yields, i M (this is the initial concentration of NH 4 +, which corresponds to the number of moles of NH 4 Cl dissolved per liter. In grams, this is 0.5 g/l) 13. The potentially relevant equilbria are: CN + H O HCN + OH b (CN ) { w / a (HCN)} + + H O OH + H + H O H 3 O + + OH w Using the first and last of these, we can set up an equilibrium table as follows. Where M, since ph 10.76, the poh 3.4 and [OH ] M, CN + H O HCN + OH b I M C M E M ~ b [HCN] [OH ]/[CN ] ( ) /( ) (From reference tables of a values a (HCN) , so b (CN ) w / a ) 14. a) Sodium hypochlorite is basic since OCl is the conjugate base of a weak acid. b) b w / a (HOCl) / c) First, we must convert the 5% (wt/wt) concentration into one we can use in chemical calculations, molarity (or more precisely activities). For a 5% soln there is 5 g NaOCl per 100 g of solution. Assuming the density of the solution is close to 1000 g/l, we have; 5 g NaOCl 1mol NaOCl 1000 g 0.67 M 100 g soln g NaOCl 1L The potentially relevant equilibria are: OCl + H O HOCl + OH Na + + H O NaOH + H + 4

5 H O H 3 O + + OH w Using the first and last of these, we can set up an equilibrium table as follows. OCl + H O HOCl + OH b I 0.67 M C E b [HOCl] [OH ]/[OCl ] () ( )/(0.67) Epand and solve using quadratic formula OR If >> 10 7 and << 0.67, Then the above epression simplifies to / so, and assumptions check Therefore, [OH ] M and poh 3.31 Therefore, ph 14 poh 10.7 solubility 15. a) AgCl(s) b) CdS(s) c) Cu(OH)(s), BaSO4(s) 16. a) CaF (s) Ca + + F let quantity dissolved be s s s sp [Ca + ][F ] (s) (s) s Therefore, s M (this is the molar solubility) Since CaF has a molar mass of 78 g/mol, g of CaF will dissolve per liter of H O) b) less c) less oidation states 17. a) (+V), (III), (III), (0) b) (II), (+IV), (+II), (+VI) on sulfur c) (+VII), (+V), (+III), (+I) d) (+IV), (+IV), (III and I), (III and +I) 18. a) reduction b) oidation c) reduction b) both (disproportionation) 5

6 19. oidized reduced a) I IO 4 b) Cl NO 3 c) Cu NO 3 0. a) oidation b) reduction c) oidation 1. See reference tables of standard half reaction reduction potentials. Half reactions (written as reductions) with a large positive E o value are energetically favorable. Therefore, species on the lefthand side of the half reactions with large positive reduction potentials are the most powerful oidizing agents. Increasing order of oidizing ability is given by O < ClO < Cl < O 3. The potential reaction is: Cu(s) + Ag + Cu + + Ag(s) The relevant half reactions are: Cu + + e Cu(s) E o 0.51 V Ag + + e Ag(s) E o V Reversing the first of these, and combining yields the reaction above with E o 0.78 V Since E o > 0, ΔG o < 0 and therefore the reaction will proceed as written at standard state conditions. Since the concentrations of the relevant ions was not given, we could use the Nernst equation To check the effect of other ion concentrations. E 0.78 V 0.057/1 ln {[Cu + ]/[Ag + ]} (at room temp) Where n 1, and Q [Cu + ]/[Ag + ] So even if the [Cu + ]/[Ag + ] 10 4, E V and the reaction still proceeds as written. 3. This is a tricky one. First of all, the aqueous iron salts should be written out in their dissociated forms. Br (l) + Fe + (aq) + 4 Br (aq) Fe 3+ (aq) + 6 Br (aq) Canceling the bromide ions from both sides yields Br (l) + Fe + (aq) Fe 3+ (aq) + Br (aq) The two half reactions are: Br (l) + e Br (aq) E o 1.066V Fe 3+ (aq) + e Fe + (aq) E o 0.771V ΔG o nfe o (96480 C/mol) (0.95 V) 56.9 kj/mol ep {ΔG o /RT} e where R J/mol and T 98 6