Chapter 17 Homework Problem Solutions

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1 Chapter 17 Homework Problem Solutions D 2 O D + + OD, K w = [D + ] [OD ] = Since [D + ] = [OD ], we can rewrite the above expression to give: = ([D + ]) 2, [D + ] = M = [OD ] pd = log[d + ] = log( ) = 7.52 pod = log[od ] = log( ) = 7.52 pk w = pd + pod = At 25 C, K w = = [H + ] [OH ]. Let x = [H + ], for each of the following: (a) x(0.0068) = [H + ] = ( ) (0.0068) = M ph = log[h + ] = log( ) = poh = 14 ph = = 2.17 (b) x( ) = [H + ] = ( ) ( ) = M ph = log[h + ] = log( ) = 9.81 poh = 14 ph = = 4.19 (c) x( ) = [H + ] = ( ) ( ) = M ph = log[h + ] = log( ) = 6.20 poh = 14 ph = = 7.80 (d) x( ) = [H + ] = ( ) ( ) = M ph = log[h + ] = log( ) = 2.91 poh = 14 ph = = [H + ] = 10 ph = = M

2 If the concentration of H + is doubled: [H + ] = M ph = log[h + ] = log( ) = HNO 3 is a strong acid so [H + ] = [HNO 3 ] = M ph = log[h + ] = log( ) = 3.19 poh = ph = = [OH ] = 10 poh = = M poh = ph = =2.40 [OH ] = 10 poh = = M [ ] mol OH 1 mol Ca(OH) 2 Ca(OH) 2 = 1 L solution 2 mol OH = M Ca(OH) 2 ph = poh = = 3.4 [OH ] = 10 poh = = M mol OH 1 mol Ca(OH) Ca(OH) 2 2 = 1 L solution 2 mol OH 4 = M Ca(OH) 2 [ ] [H + ] = 10 ph = = M % ionization = 100% = 0.43% (a) At 25 C, K a K b = K w K b = K w /K a = = (b) K b for acetic acid is 1.0 x /1.8 x 10-5 = 5.6 x Therefore, since K b for the lactate ion is smaller than that for the acetate ion, the lactate ion is a weaker base.

3 17.66 ph = 3.22 [H + ] = = M Percentage ionization = moles ionized per liter moles available per liter 100% Percentage ionization = HA H + + A M 0.20 M 100% = 0.30% [HA] [H + ] [A ] I 0.20 C E K a = H A = = [ HA] [ 0.20] K b = 10 pkb = = Cod + H 2 O HCod + + OH b + HCod OH K = = [ Cod] 6 [Cod] [HCod + ] [OH ] I E x + x + x Substituting these values into the mass action expression gives: ( x)( x) 6 K b = = x

4 If we assume that x << we get; x 2 = , x = M = [OH ] poh = log[oh ] = log( ) = 3.75 ph = poh = = CN + H 2 O HCN + OH [ ] 14 HCN OH Kw K b = = = = CN K 10 a [CN ] [HCN] [OH ] I E x + x + x Assume that x << ( x)( x) 5 4 K b = = x = = [OH ] Wait!!!! is not << So, use the method of successive approximations. ( x)( x) 5 4 K b = = x = (x)(x) K b = = x = x = = [OH ] poh = 3.51 ph = NaCN will be basic in solution since CN is a basic ion and Na + is a neutral ion.

5 CN + H 2 O HCN + OH For HCN, K a = , we need K b for CN ; K b = K w /K a = ( ) ( ) = [ HCN] OH 5 K b = = CN [CN ] [HCN] [OH ] I 0.20 E 0.20 x + x + x Substituting these values into the mass action expression gives: ( x)( x) 5 K b = = x Assuming that x << 0.20 we can solve for x and determine; x = M = [OH ] poh = log[oh ] = log( ) = 2.70 ph = poh = = Concentration of HCN is equal to that of hydroxide ion: M OCl + H 2 O HOCl + OH [ ] 14 HOCl OH Kw K b = = = = OCl K 8 a g NaOCl 1 mol NaOCl 1.0 g 1000 ml [OCl ] = 100 g solution g NaOCl 1 ml 1 L = 0.68 M [OCl ] [HOCl] [OH ]

6 I 0.68 E 0.68 x + x + x Assume that x << 0.68 ( x)( x) 7 4 K b = 0.68 = x = = [OH ] poh = 3.33 ph = HC 2 H 3 O 2 H + + C 2 H 3 O 2 + [H ][C2H3O 2 ] 5 K a = = HC2H3O2 [HC 2 H 3 O 2 ] [H + ] [C 2 H 3 O 2 ] I E 0.15 x + x x Substituting these values into the mass action expression gives: (x)( x) 5 K a = = x Assume that x << 0.15 M and x << 0.25 M, then; x x x M = [H + ] ph = log [H + ] = 4.97

7 ph = pk + log a [ anion] A = pk a + log [ acid] [ HA] 4.00 = log( [NaC 2 H 3 O 2 ]/[HC 2 H 3 O 2 ] ) 0.74 = log[nac 2 H 3 O 2 ]/[HC 2 H 3 O 2 ] [NaC 2 H 3 O 2 ]/[HC 2 H 3 O 2 ] = 0.18 [NaC 2 H 3 O 2 ] = 0.18 [HC 2 H 3 O 2 ] = = M Thus to the 1 L of acetic acid solution we add: mol NaC 2 H 3 O g/mol = 2.2 g NaC 2 H 3 O H 3 AsO 4 + H 2 O H 3 O + + H 2 AsO 4 H 2 AsO 4 + H 2 O H 3 O + + HAsO 4 2 HAsO H 2 O H 3 O + + AsO 4 3 K a1 = K a2 = K a3 = [HAsO 4 2 ] = The following assumptions are made: [H + ] total [H + ] first step [H 2 AsO 4 ] total [H 2 AsO 4 ] first step [HAsO 2 4 ] total [HAsO 2 4 ] second step Now solve using K a1 = for the other concentrations: [H 3 AsO 4 ] [H 2 AsO 4 ] [H + ] I 0.25 E 0.25 x + x + x + H2AsO 4 H (x)(x) 3 K a1 = = = H3AsO4 ( 0.25 x) Solve for x by successive approximations or by the quadratic equation: x = M = [H + ] = [H 2 AsO 4 ] [H 3 AsO 4 ] = = M ph = log(0.034) = 1.47 [H 2 AsO 4 ] = M

8 For the concentration of HAsO 4 2 K a2 = 2 HAsO + 4 H H2AsO 4 = K a2 = 2 4 ( ) ( 0.034) HAsO = [HAsO 4 2 ] = The hydrolysis equation is: HSO SO H 2 O HSO 3 + OH 3 OH K b = 2 SO 3 In order to obtain K b we will use the relationship K w = K a K b 14 Kw b K 8 a K = = = [SO 2 3 ] [HSO 3 ] [OH ] I 0.24 E 0.24 x + x + x Since K b is so small, assume that x << 0.24 and we determine x = M = [OH ] poh = log( ) = 3.72, ph = poh = [SO 3 2- ] = 0.24 M [HSO 3 - ] = [OH - ] = 1.9 x 10-4 M [Na + ] = 0.24 M [H 3 O + ] = 1.00 x /1.9 x 10-4 = 5.3 x The second ionization is very small compared to the first ionization so the [HSO 3 - ], [OH - ] and [H 3 O + ] do not change significantly. For the concentration of H 2 SO 3

9 HSO 3 + H 2 O H 2 SO 3 + OH K b = 2 H2SO3 OH HSO 3 K b 2 = a 1 14 Kw = = K K = b 2 4 H2SO = [H 2 SO 3 ] = mol HC2H3O mol HC 2 H 3 O 2 = (25.0 ml HC 2 H 3 O 2 ) 1000 ml HC2H3O2 = mol HC 2 H 3 O 2 mol OH = (40.0 ml OH mol OH ) 1000 ml OH = mol OH excess OH = ( ) ( ) = mol [OH ] moles 1 L ( ml) 1000 ml = = M poh = 1.07 ph = (a) The equation that will be used is the normal Henderson Hasselbach equation, namely: ph = pk + log a [ anion] A = pk a + log [ acid] [ HA] where A = C 2 H 3 O 2 and HA = HC 2 H 3 O 2. We note further that the log term involves a ratio of concentrations, but that the volume remains constant in a process such as that to be analyzed here. Thus the log term may be replaced by a ratio of mole amounts, since volumes cancel: ph = pk + log a ( moles C2H3O2 ) ( moles HC H O ) 2 3 2

10 Thus we need only determine the number of moles of acid and conjugate base that remain in the buffer after the addition of a certain amount of H + or OH, in order to determine the ph of the buffer mixture after that addition. The buffer is changed in the following way by the addition of OH : HC 2 H 3 O 2 + OH H 2 O + C 2 H 3 O 2 In other words, if moles of OH are added to the buffer, the amount of HC 2 H 3 O 2 goes down by moles, whereas the amount of C 2 H 3 O 2 goes up by moles. In addition, the ph of the solution will increase upon the addition of base. The buffer is changed in the following way by the addition of H + : C 2 H 3 O 2 + H + HC 2 H 3 O 2 If mol of H + are added, then the amount of C 2 H 3 O 2 goes down by moles and the amount of HC 2 H 3 O 2 goes up by moles. As before, the addition of acid will decrease the ph of the buffer system. For the general buffer mixture that contains x mol of C 2 H 3 O 2 and y mol of HC 2 H 3 O 2, we can apply the Henderson Hasselbach equation, noting the maximum amount by which we want the ph to change (namely 0.10 units): For the case of added base: = log ( x + ) ( x ) moles C2H3O moles HC2H3O2 Simplifying and taking the antilog of both sides of the above equation gives: [x ]/[y ] = 3.0, which is now designated eq. 1. For the case of added acid: = log ( x ) ( x + ) moles C2H3O moles HC2H3O2 Simplifying and taking the antilog of both sides of the equation gives: [x ]/[y ] = 1.9, which is now designated eq. 2. The equations 1 and 2 as designated above are solved simultaneously, since they are two equations containing two unknowns: x = initial mol C 2 H 3 O 2 = 0.15 mol y = initial mol HC 2 H 3 O 2 = mol

11 These values are converted to grams as follows: mol HC 2 H 3 O g/mol = 3.8 g HC 2 H 3 O mol NaC 2 H 3 O g/mol = 18 g NaC 2 H 3 O 2 2H 2 O These are the minimum amounts of acid and conjugate base that would be required in order to prepare a buffer that would change ph by only 0.10 units on addition of either mol of OH or mol of H +. (b) mol/0.250 L = 0.25 M HC 2 H 3 O mol C 2 H 3 O 2 /0.250 L = 0.60 M NaC 2 H 3 O 2 (c) mol OH /0.255 L = M OH poh = log[oh ] = log( ) = ph = poh = = (d) mol H + /0.255 L = M H + ph = log[h + ] = log(3.92 x 10 2 ) = Using the equilibrium for the conjugate base of NH 4 + (K b = ) and the conjugate acids of C 2 H 3 O 2 (K a = ). Since the two equilibrium constants are equivalent, and the concentrations of the anion and cation are equivalent, therefore the concentrations of H + and OH are equivalent and the ph is For the given mixture, ph does not equal pk a because the concentration of the acid does not equal the concentration of its salt. For each OH ion added an additional F will form and an HF molecule will dissociate: HF + H 2 O H 3 O + +F As OH is added it reacts with the H 3 O + causing additional HF to dissociate. For each H + ion added an F ion will react with it forming additional HF. Using the equation above, the added H + reacts with F producing HF. Initially, there are 5HF and 4F ions. After addition of 3OH ions we will have 2HF and 6F. If 2H + are added we will have 7HF and 2F.

12 The moles of HCl is determined from the ideal gas law: n = ( 734 torr)( 1 atm )( L) PV = = mol HCl RT K 760 torr 4 L atm mol K The molarity of the solution is: M of solution = mol HCl L solution = M HCl ph = log[h + ] = log( M) = 2.813

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