CHEM 11 Exam 3 Practce Test Solutons 1A No matter what temperature the reacton takes place, the product of [OH-] x [H+] wll always equal the value of w. Therefore, f you take the square root of the gven w, you can get the concentraton of H+. Take the log to get the ph. C Most metal oxdes are basc; they form metal hydroxdes n water. The other statements all follow from the perodc trends outlned n the packet and n notes. 3B The man objectve wll be to determne the ph (and then the poh) whch wll requre the number of moles of acetc acd and NaOH present. That calculaton s as follows: (0.05 L)(0. M) 0.01 moles acetc acd (0.035 L)(0.1 M) 0.0035 moles NaOH Snce there are less moles of base, t s the lmtng reagent n the reacton wth the acd and wll run out frst. Therefore, 0.01 0.0035 gves 0.0065 moles of acdc acd formed. The 0.0035 moles of OH- from NaOH led to the producton of 0.0035 moles of acetc acd s conjugate base (the acetate on). The total volume n ths soluton s 0.085 L, whch the number of moles of acetc acd and ts conjugate base must each be dvded nto to get ther concentraton values. Now, plug those values nto the Henderson- Hasselbalch equaton to determne ph, subtract that value from 14, and you get your poh. 4E A low concentraton of hydronum ndcates a basc soluton. So we are lookng for the most basc soluton whch wll have the hghest ph.
5B The chemcal formula for ths strong base s Sr(OH). In soluton, every molecule wll release two hydroxde ons and so the [OH - ] 0.050 M x 0.10 M. Lastly, to get poh: poh -log[oh - ] -log(0.1) 1 6A Snce we are dealng wth a weak acd (strong acds don t have k a values), we can skp straght to the equlbrum formula: k a x M where x s equal to [H 3 O + ], and M s the molarty of the acd. Solvng for x we get: x so ph k a M 0.00 7 [ H 3O log(0.007 ).6 + ] 7B Just lke the prevous queston we are dealng wth a weak acd and so we can wrte: k a x M Snce we ve already got M 0.4 M all we need s x, whch s equal to [H 3 O + ]. We use the ph to do ths: x [H 3 O + ] 10 -ph 10-4.5 Lastly we plug n our values to get: k a 4.5 ( 10 ) 9 0.4.5 10
8E The strongest conjugate base wll come from the weakest acd. So bascally the queston s askng us whch of these s the weakest base. Straght away we can elmnate answers (a) and (b) snce they are very strong acds. The trend for these weaker oxo-acds s that acdty decreases as you go down the group. Snce odne s lowest n the group, HIO s the weakest acd. 9C The only compounds that could form basc solutons are those whch have anons that are the conjugate bases of weak acds. NO 3 - and Cl - are both the conjugate bases of really strong acds (HNO 3 and HCl) and so they are not basc at all. CO 3 - and F -, however, are both the conjugate bases of weak acds (HCO 3 - and HF) so they wll form basc solutons and thus we pck choces and v.
10C To fgure out how many moles were added we ll need to get the ntal concentraton of sodum acetate. Snce sodum acetate s a weak base and was added to water we can skp the ICE table and go straght to the equlbrum constant equaton for any weak base: b x M Remember M s equal to the ntal molarty of sodum acetate, and snce we re usng a weak base we need to use b, and remember that x s equal to [OH - ]. We can use the ph to get [OH - ]: poh 14 ph 5.5 [OH - ] 10 -poh 10-5.5 3. x 10-6 M We can use a of acetc acd (the conjugate acd) to get b : 14 w 10 b 5.6 10 5 1.8 10 Next we solve for M : x a (3. 10 6 M 10 b 5.6 10 ) 10 0.018M Lastly, all we need do s multply M by volume to get the number of moles of sodum acetate: 0.018 M x 0.05 L 9 x 10-4 moles
11D Ths s a strong acd / strong base ttraton. To begn, we must compare the number of moles of acd wth the number of moles of base and see whch one s our lmtng reactant: Moles HClO 4 0.15 L x 0.1 M 0.015 mols Moles OH - 0.1 L x 0.5 M 0.05 mols Snce there are fewer moles of acd they wll all be consumed by the base and we wll be left wth: 0.05 0.015 0.035 mols OH -. All we need do now s calculate [OH - ] and then convert to ph: [OH - ] (0.035 mol) / (0.5 L) 0.14 M (note that the volume s the combned volumes of both the acdc and basc solutons) poh -log(0.14) 0.85 ph 14 poh 14 0.85 13.15 1A The prmary requrement of any buffer system s the presence of a weak acd and ts conjugate base. Snce (a) s the only opton that has a weak acd (HF) and ts conjugate base (F - ) that s the correct answer. Remember that RbF n soluton becomes Rb + and F - snce Rb s a group 1 metal.
13B Ths a standard weak acd / strong base ttraton. Frst we must calculate the number of moles of weak acd and the number of moles of base: Moles HNO 0.1 L x 0.5 M 0.05 mols Moles OH - 0.1 L x 0.05 M 0.005 mols Now we should look at the reacton that takes place and set up an ICE table: HNO (aq) + OH - (aq) NO - (aq) + H O(l) I 0.05 0.005 0 - C -0.005-0.005 +0.005 - E 0.045 0 0.005 - As the table suggests, snce we re dealng wth a strong base t reacts completely, hence at equlbrum there s no hydroxde left and all we ve got n soluton s some of the weak acd we started wth, HNO, and some of ts conjugate base, NO -. The Henderson Hasselback Equaton s perfect for stuatons lke ths snce t s a drect lnk to ph when all you ve got n soluton s weak acd / conjugate base. So we plug n our values: ph -log(4.5 x 10-4 ) + log(0.005/0.045).4 (note that when usng the Henderson Hasselback equaton t s not necessary to convert our equlbrum mole values to concentratons)
14B There are a couple of ways to solve ths. One s tme consumng, the other s not. Under most crcumstances you would need to calculate both the number of moles of weak acd, ammonum (NH 4 + ), and that of the weak base, ammona (NH 3 ), so that they could be used n the Henderson Hasselback equaton. Ths s the tme-consumng way. Alternatvely, you can hopefully recognze that both the acd and base solutons have the same volumes and concentratons and so ther concentratons n the buffer are equal to one another: [NH 4 + ] [NH 3 ] Ths s referred to as the halfway pont, and that means that ph p a. So all we really need to do s determne p a : 14 w 10 a 5.6 10 5 1.77 10 b ph p a -log(5.6 x 10-10 ) 9.5 10 15D Ths s a weak acd / strong base ttraton. As wth smlar ttratons we must frst calculate the number of moles of acd base: Moles HNO 0.1 L x 0.5 M 0.05 mols Moles OH - 0.05 L x 0.5 M 0.05 mols Immedately, you should recognze that the ntal number of moles of weak acd are equal to the number of moles of hydroxde added, and so we are at the equvalence pont. Ths means that all of the weak acd and hydroxde wll be used up and the only thng we ll have left n soluton that can alter the ph s the conjugate base, NO -. Furthermore, we know that we have the exact same number of moles of NO - (0.05 mols) as we had of the hydroxde ons that were all used up durng the ttraton. Ths s all useful nformaton f you have any nterest n savng tme durng your test. Stll, for those who want to see a more detaled analyss, the ICE table below shows what s gong on:
HNO (aq) + OH - (aq) NO - (aq) + H O(l) I 0.05 0.05 0 - C -0.05-0.05 +0.05 - E 0 0 0.05 - As you can see, all we have left n soluton s 0.05 moles of NO -, and so to calculate ph we just have to treat ths problem as we would any other weak base equlbrum problem by usng the equaton: b x M Remember that snce we re dealng wth a weak base we ll be needng b (not a ) and also that x [OH - ] (not [H 3 O + ]). We get b usng the same old equaton: 14 w 10 b. 10 4 4.5 10 a Next we need the ntal molarty, M, whch s the concentraton of our weak base: [NO - ] (0.05 mol) / (0.15 L) 0.17 M (note that the volume s the combned volumes of both the acdc and basc solutons) Now we can solve for x: 11 b x x M b, M 11 (. 10 ) ( 0.17) 1.9 10 6 [ OH ] poh log[ OH ph 14 poh 8.3 ] log 1.9 6 ( 10 ) 5.7
16B CuCl would be least soluble n the soluton that already contans the most Cu + and/or Cl -. Opton (a) has 0.5 M Cu + Opton (b) has 0.6 M Cl - (snce the rato of chlorde to calcum chlorde s :1) Optons (c) and (d) contan no common ons and wll not reduce solublty at all. Opton (e) has 0.4 M Cl - So clearly answer choce (b) has the hghest concentraton of the common on and s thus the rght answer. 17D At the equvalence of a weak acd ttraton statement s correct. Statement s true at the half-way pont. Also, at the equvalence pont, the only thng left n soluton that could affect ph s the conjugate base of the weak acd, and so snce t s a base we cab expect the soluton to be basc too. Thus ph > 7. Note that f ths were a STRONG acd / strong base ttraton, then the ph would equal 7 at the equvalence pont. 18B For optmum buffer capacty we frst need to fnd weak acd / conjugate base pars. Answer choces (a) and (c) both use HCl (a strong acd) and cannot be canddates for the best buffer. Snce (b) and (d) both use the same legtmate weak acd / conjugate base pars we must then look to see whch pars have the hghest mnmum concentratons. The mnmum concentraton for answer (d) s 0.15M, whereas for answer (b) t s 0.5M, so the most relable of the two buffers s (b) as t has the hghest mnmum concentraton.
19D Snce we are dealng wth a weak acd alone n soluton, we can skp straght to the equaton: a (x ) / (M x). To get ph we ll need to solve for x, the hydronum concentraton, and so we are allowed to assume the x n the denomnator s neglgble and to not nclude t. Once we get the hydronum concentraton we can smply convert to ph by takng ts negatve log. 0E The one requrement to be a lews base s to have a lone par of electrons. Snce opton E has absolutely no lone pars of electrons when you draw ts lews structure, t s least lkely to act as a base. 1A Ths s a strong acd / strong base ttraton. We fst need to determne the number of moles of hydronum and of hydroxde: H 3 O + : 0.05M x 0.13L 0.0065 moles OH - : (0.05M x 0.055L) x mol OH / 1 mol Ba(OH) 0.0055 moles If we set up an ICE table for ths reacton we ll see that all the hydroxde s used up and we are left wth 0.0010 moles of H 3 O +. Thus the ph can now be determned by calculatng the fnal concentraton of hydronum and then convertng to PH: [H 3 O + ] (0.0010 moles / 0.185 L) 0.0054 M, and ph -log[h 3 O + ].7 A Understand that alone n soluton, HClO does not break up, and so t wll be present n the net reacton. On the other hand, NaOH does break up even before t reacts wth anythng else, and so we really have Na + and OH - floatng around separately. Snce only the OH - s needed for the reacton wth HClO, the Na + s consdered a spectator on and wll not appear n the net reacton. Thus opton A s the approprate answer choce.
3C By recognzng that ths soluton s essentally a weak acd, HF, and ts conjugate base, whch s the F - on, we can always calculate the ph usng the Henderson Hasselbach equaton: ph -log(6.8 x 10-4 ) + log([0.4]/[0.5]) 3.07 4A A common on exsts between HCN, a weak acd, and HI, a strong acd that wll completely dssocate. Therefore, to calculate the concentraton of CN- from the a equaton of HCN, an ntal concentraton of H+ exsts from HI (0.10 M). Therefore, you get an equaton that looks lke: a [x(0.10 + x)]/(0.70 x) Plug n the value for a gven and dsregard the + x and x n the equaton because the value of x s so small that part can be gnored to make for an easer calculaton. The value for x equals [CN-]. 5E Smply use the Henderson-Hasselbalch equaton because t s a buffer system of a weak acd and ts conjugate base. The concentraton of the acd s 0.15 M and 0.3 M for the base. ph pa + log [base]/[acd] 6A Ths queston can be answered easly usng the Henderson-Hasselbalch equaton because t nvolves a weak acd/base conjugate par. The b was gven, so smply fnd the a frst by takng (1.0 x 10-14 /b) to get a.5 x 10-9. Now plug n the known values nto: ph pa + log [base]/[acd] Solve for [base]. THEN, multply that number by.0 because the queston asks for number of MOLES n a.0l soluton, not the concentraton. Trcky, trcky, trcky
7B An atom n ts elemental form always has an oxdaton state of zero; therefore, Al(s) has an oxdaton state of zero. Al(s) loses electrons (s oxdzed) when t becomes Al(OH) 4 -, because t then has an oxdaton state of +3. A substance that s oxdzed s a reducng agent. 8D To balance a redox reacton, frst separate t nto the reducton and oxdaton half reactons. Oxdaton: Mn+ MnO4- Reducton: NaBO3 B3+ Balance each of these separately. Frst balance all atoms other than O and H (for example add Na+ to the rght sde of the oxdaton to balance the Na atoms). Then, balance the O atoms, by addng the same number of HO molecules to the other sde. Then, balance the H atoms (ncludng those from the HO s just added) by addng H+ ons to the other sde. Fnally, balance the charge on each sde by addng electrons to ether the reactant sde (reducton equaton), or the product sde (oxdaton equaton). The fnal step s to add both equatons together, makng sure that the number of electrons on the reactant sde and product sde are equal. For ths to be true n ths equaton, the oxdaton equaton had to be multpled by and the reducton equaton had to be multpled by 5. Therefore, the fnal coeffcent n front of B3+ s fve. 9E When balancng a redox reacton n basc soluton you use the same approach as n acdc soluton untl the end when everythng else has already been balanced. Then, you add OHons to each sde for every H+ present. In ths case, there should be 14H+ on the reactant sde, so we add 14 OH- to both sdes. After the H+ and OH- on the reactant sde neutralze to form HO, you wll then have 14 HO on the reactant sde and 7HO on the product sde, whch cancel out to leave 7 HO on the reactant sde.
30D Frst we need to separate the reacton nto two half reactons, and then apply materal and charge balances to each: MnO 4 - (aq) + 1e - MnO 4 - (aq) I - (aq) + 3H O(l) IO 3 - (aq) + 6H + (aq) + 6e - Once ths s done we must multply each reacton by the smallest whole number to get the electrons transferred to be equal. Ths means that we just multply the frst reacton by 6 and leave the other one alone to get: 6MnO 4 - (aq) + 6e - 6MnO 4 - (aq) I - (aq) + 3H O(l) IO 3 - (aq) + 6H + (aq) + 6e - Remember that when balancng we add an H O for every oxygen needed and an H + for every hydrogen needed. Now we combne the reactons agan by addng together everythng on the left sde and everythng on the rght sde (make sure to cancel out anythng that exsts on both sdes): 6MnO 4 - (aq) + I - (aq) + 3H O(l) 6MnO 4 - (aq) + IO 3 - (aq) + 6H + (aq) Note that ths would represent the balanced reacton n acdc soluton. The next step s to take nto consderaton that we re n basc soluton and so we must add an OH - on to each sde of the reacton for every H + that s present to get: 6MnO 4 - (aq) + I - (aq) + 3H O(l) + 6OH - (aq) 6MnO 4 - (aq) + IO 3 - (aq) + 6H + (aq) + 6OH - (aq) As expected the 6H + (aq) and 6OH - (aq) wll neutralze to form 6H O and the reacton becomes: 6MnO 4 - (aq) + I - (aq) + 3H O(l) + 6OH - (aq) 6MnO 4 - (aq) + IO 3 - (aq) + 6H O(l) Lastly, we cancel out any extra H O that mght be on both sdes to get: 6MnO 4 - (aq) + I - (aq) + 6OH - (aq) 6MnO 4 - (aq) + IO 3 - (aq) + 3H O(l) Ths s our balanced reacton n basc soluton. The sum of the coeffcents s 3.
31A The oxdzng agent s the speces that gets reduced. Ths can only be a reactant so (d) and (e) aren t even optons to begn wth. The quckest way to answer ths problem s to assgn oxdaton numbers to each element on both the product s sde and the reactant s sde and see whch elements show a change n number. If ths s done correctly you wll notce that the Cr n Cr O 7 - has an oxdaton number of +6 on the reactant s sde and +3 on the product s sde. Ths means that t snce ts charge s decreasng t must be ganng electrons and so t s beng reduced. Thus, Cr O 7 - must also be the oxdzng agent. As a sde note, I d lke to pont out that the oxdaton number for S n S O 3 - s + but on the product s sde t s +.5. Ths s very unusual, and yet we stll see an ncrease n oxdaton number, whch means that S O 3 - s stll beng oxdzed and s thus the reducng agent. 3A The oxdaton state of Ag+ s 1+ and t s reduced to Ag(s), whch has an oxdaton state of zero. When the oxdaton state of an on/molecule s lowered, that on/molecule s REDUCED.
33D The queston s really askng us to fnd out the molar solublty of Mn(OH) n a soluton that already has hydroxde ons floatng around. Ths s the common on effect, and means that the ntal concentraton of hydroxde has to be taken nto account when calculatng solublty at equlbrum. Snce ph 11.5, we know poh 14 11.5.5, and so [OH - ] 10 -.5 0.003 M To help calculate solublty we ll use the followng ICE table: Mn(OH) (s) Mn + (aq) + OH - (aq) I - 0 0.003 C - +x +x E - x 0.003 +x We then set up the equaton for the solublty product constant, sp : sp [Mn + ][OH - ] [x][0.003 + x], where x s the solublty Snce sp s so small we can assume that x s neglgble and the equaton becomes: sp [x][0.003] x (1.6 x 10-13 ) / (0.003) 1.6 x 10-8 M By examnng the ICE table we can see that [Mn + ] x, and so [Mn + ] 1.6 x 10-8 M