CHEM 112 Exam 3 Practice Test Solutions

Transcription

1 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 asc; 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 ojectve wll e to determne the ph (and then the poh) whch wll requre the numer 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) moles NaOH Snce there are less moles of ase, t s the lmtng reagent n the reacton wth the acd and wll run out frst. Therefore, gves moles of acdc acd formed. The moles of OH- from NaOH led to the producton of moles of acetc acd s conjugate ase (the acetate on). The total volume n ths soluton s L, whch the numer of moles of acetc acd and ts conjugate ase must each e dvded nto to get ther concentraton values. Now, plug those values nto the Henderson- Hasselalch equaton to determne ph, sutract that value from 14, and you get your poh. 4E A low concentraton of hydronum ndcates a asc soluton. So we are lookng for the most asc soluton whch wll have the hghest ph.

2 5B The chemcal formula for ths strong ase s Sr(OH). In soluton, every molecule wll release two hydroxde ons and so the [OH - ] 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 ka values), we can skp straght to the equlrum formula: k a x M where x s equal to [H3O + ], and M s the molarty of the acd. Solvng for x we get: x so ph k a M [ H 3O - log(0.007 ).6 + ]

3 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 [H3O + ]. We use the ph to do ths: x [H3O + ] 10 -ph Lastly we plug n our values to get: k a -4.5 ( 10 ) E The strongest conjugate ase wll come from the weakest acd. So ascally the queston s askng us whch of these s the weakest ase. Straght away we can elmnate answers (a) and () 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 asc solutons are those whch have anons that are the conjugate ases of weak acds. NO3 - and Cl - are oth the conjugate ases of really strong acds (HNO3 and HCl) and so they are not asc at all. CO3 - and F -, however, are oth the conjugate ases of weak acds (HCO3 - and HF) so they wll form asc solutons and thus we pck choces and v.

4 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 ase and was added to water we can skp the ICE tale and go straght to the equlrum constant equaton for any weak ase: x M Rememer M s equal to the ntal molarty of sodum acetate, and snce we re usng a weak ase we need to use, and rememer that x s equal to [OH - ]. We can use the ph to get [OH - ]: poh 14 ph 5.5 [OH - ] 10 -poh x 10-6 M We can use a of acetc acd (the conjugate acd) to get : -14 w Next we solve for M: x a ( M M Lastly, all we need do s multply M y volume to get the numer of moles of sodum acetate: M x 0.05 L 9 x 10-4 moles )

5 11D Ths s a strong acd / strong ase ttraton. To egn, we must compare the numer of moles of acd wth the numer of moles of ase and see whch one s our lmtng reactant: Moles HClO L x 0.1 M mols Moles OH L x 0.5 M 0.05 mols Snce there are fewer moles of acd they wll all e consumed y the ase and we wll e left wth: 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 comned volumes of oth the acdc and asc solutons) poh -log(0.14) 0.85 ph 14 poh A The prmary requrement of any uffer system s the presence of a weak acd and ts conjugate ase. Snce (a) s the only opton that has a weak acd (HF) and ts conjugate ase (F - ) that s the correct answer. Rememer that RF n soluton ecomes R + and F - snce R s a group 1 metal.

6 13B Ths a standard weak acd / strong ase ttraton. Frst we must calculate the numer of moles of weak acd and the numer of moles of ase: Moles HNO 0.1 L x 0.5 M 0.05 mols Moles OH L x 0.05 M mols Now we should look at the reacton that takes place and set up an ICE tale: HNO(aq) + OH - (aq) NO - (aq) + HO(l) I C E As the tale suggests, snce we re dealng wth a strong ase t reacts completely, hence at equlrum 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 ase, NO -. The Henderson Hasselack 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 ase. 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 Hasselack equaton t s not necessary to convert our equlrum mole values to concentratons)

7 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 oth the numer of moles of weak acd, ammonum (NH4 + ), and that of the weak ase, ammona (NH3), so that they could e used n the Henderson Hasselack equaton. Ths s the tme-consumng way. Alternatvely, you can hopefully recognze that oth the acd and ase solutons have the same volumes and concentratons and so ther concentratons n the uffer are equal to one another: [NH4 + ] [NH3] Ths s referred to as the halfway pont, and that means that ph pa. So all we really need to do s determne pa: -14 w 10 a ph pa -log(5.6 x ) D Ths s a weak acd / strong ase ttraton. As wth smlar ttratons we must frst calculate the numer of moles of acd ase: Moles HNO 0.1 L x 0.5 M 0.05 mols Moles OH L x 0.5 M 0.05 mols Immedately, you should recognze that the ntal numer of moles of weak acd are equal to the numer of moles of hydroxde added, and so we are at the equvalence pont. Ths means that all of the weak acd and hydroxde wll e used up and the only thng we ll have left n soluton that can alter the ph s the conjugate ase, NO -. Furthermore, we know that we have the exact same numer 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 tale elow shows what s gong on:

8 HNO(aq) + OH - (aq) NO - (aq) + HO(l) I C E 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 prolem as we would any other weak ase equlrum prolem y usng the equaton: x M Rememer that snce we re dealng wth a weak ase we ll e needng (not a) and also that x [OH - ] (not [H3O + ]). We get usng the same old equaton: -14 w a Next we need the ntal molarty, M, whch s the concentraton of our weak ase: [NO - ] (0.05 mol) / (0.15 L) 0.17 M (note that the volume s the comned volumes of oth the acdc and asc solutons) Now we can solve for x: -11 x x M, M -11 (. 10 ) ( 0.17) [ OH - ] poh - log[ OH ph 14 - poh ] -log ( 10 ) 5.7

9 16B CuCl would e least solule n the soluton that already contans the most Cu + and/or Cl -. Opton (a) has 0.5 M Cu + Opton () 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 solulty at all. Opton (e) has 0.4 M Cl - So clearly answer choce () 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 ase of the weak acd, and so snce t s a ase we ca expect the soluton to e asc too. Thus ph > 7. Note that f ths were a STRONG acd / strong ase ttraton, then the ph would equal 7 at the equvalence pont. 18B For optmum uffer capacty we frst need to fnd weak acd / conjugate ase pars. Answer choces (a) and (c) oth use HCl (a strong acd) and cannot e canddates for the est uffer. Snce () and (d) oth use the same legtmate weak acd / conjugate ase 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 () t s 0.5M, so the most relale of the two uffers s () as t has the hghest mnmum concentraton.

10 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 neglgle and to not nclude t. Once we get the hydronum concentraton we can smply convert to ph y takng ts negatve log. 0E The one requrement to e a lews ase s to have a lone par of electrons. Snce opton E has asolutely no lone pars of electrons when you draw ts lews structure, t s least lkely to act as a ase. 1A Ths s a strong acd / strong ase ttraton. We fst need to determne the numer of moles of hydronum and of hydroxde: H3O + : 0.05M x 0.13L moles OH - : (0.05M x 0.055L) x mol OH / 1 mol Ba(OH) moles If we set up an ICE tale for ths reacton we ll see that all the hydroxde s used up and we are left wth moles of H3O +. Thus the ph can now e determned y calculatng the fnal concentraton of hydronum and then convertng to PH: [H3O + ] ( moles / L) M, and ph -log[h3o + ].7 A Understand that alone n soluton, HClO does not reak up, and so t wll e present n the net reacton. On the other hand, NaOH does reak up even efore 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.

11 3C By recognzng that ths soluton s essentally a weak acd, HF, and ts conjugate ase, whch s the F - on, we can always calculate the ph usng the Henderson Hasselach equaton: ph -log(6.8 x 10-4 ) + log([0.4]/[0.5]) A A common on exsts etween 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( x)]/(0.70 x) Plug n the value for a gven and dsregard the + x and x n the equaton ecause the value of x s so small that part can e gnored to make for an easer calculaton. The value for x equals [CN-]. 5E Smply use the Henderson-Hasselalch equaton ecause t s a uffer system of a weak acd and ts conjugate ase. The concentraton of the acd s 0.15 M and 0.3 M for the ase. ph pa + log [ase]/[acd] 6A Ths queston can e answered easly usng the Henderson-Hasselalch equaton ecause t nvolves a weak acd/ase conjugate par. The was gven, so smply fnd the a frst y takng (1.0 x /) to get a.5 x Now plug n the known values nto: ph pa + log [ase]/[acd] Solve for [ase]. THEN, multply that numer y.0 ecause the queston asks for numer of MOLES n a.0l soluton, not the concentraton. Trcky, trcky, trcky

12 7D The queston s really askng us to fnd out the molar solulty 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 e taken nto account when calculatng solulty at equlrum. Snce ph 11.5, we know poh , and so [OH - ] M To help calculate solulty we ll use the followng ICE tale: Mn(OH)(s) Mn + (aq) + OH - (aq) I C - +x +x E - x x We then set up the equaton for the solulty product constant, sp: sp [Mn + ][OH - ] [x][ x], where x s the solulty Snce sp s so small we can assume that x s neglgle and the equaton ecomes: sp [x][0.003] x (1.6 x ) / (0.003) 1.6 x 10-8 M By examnng the ICE tale we can see that [Mn + ] x, and so [Mn + ] 1.6 x 10-8 M