.[1) 8,b Using units of ~As N'': <

(a) Convert from mg/L to mole/L: i [No-1- 70 rngNOs X l,molo~ X lgNO~ -11 X 10-a moleNO& O,S' rb thh

• 3 - L 62gN a 1000mg . Os - · . L 1-} ii [NH+] -100 mgl'ffl4 x lmoJ11Nll, X 1,NH.& -5 6 X 10-s moleNHf J.r ' .4 - -i;- l811Nll4 lOO0mgNH~ - ' L

iii [NH l - 0 5 mgrHs X 1roolo§ X lgNH~ - 2 9 X 10-6 molLNH3 • 3 - · 17gNHa 1000Jng 7·11 - •

iv. Total moles/L = 1.1 x 10-3 + 5.6 x 10-3 + 2.9 x 10-6 = 6.7 x 10-3 ~

(b) Convert from mole/L to m':L°N :Note that one mole of ND;, Nllt, and NHa each contain <me mole of NJ

i. [NO-] = 70 ~ X lipolllNOa· X 1 m~ X . ...l!l!L = 16 !Y1i 3 L 02g•N0s lbtO 3 linoleN -i;-

ii. [NHt] = 100 ~ X 1~go~ ~ 1~~~~~4 X 1~~:N =_78~

iii [NB)= 0 5 ~ X lrno Nils X 11110 ~ X ~ = 0 41 !!!Ill: • . 8 • L 17~NH.1 imol.111'1 finoloN' • -i;-

iv. (N03] + (NHt] + fNHs] = 16 "'f N + 78 ~ + 0.41 "'f N = 94 !Df N

(c) Convert from mole/L to~:

, i. 1.1 X 10-s moJo~O'i := 1.1 X 10-a rnolqN ii. 5.6 >< 10-3 moteLNnt = 5.6 x 10-8 mn];N

iii. 2.9·x 10-5 moltHs = 2.9 X 10-6 mo~N

iv. ·fNOsJ + (NHt] + (NHs] = 1.1 x 10-3 + 5.6 x 10-s ~ 2.9 x 10-11 =·6.7 x 10-s morN

(d) Perts (a). and (c) are equivalent b auae there is 1 mole of N per mole of N03, 1 mole + . of N per mole of NH.,1 and 1 mole f N per mole of NH4 • ~ I tJ

6

I I I )

)

@ B f''p

L = fbC, o-+...,· L f-:- n- r:i aJl.(..

JL11 -= [5>2-Ja [o;Jre--i)

_____ .- ----·

[oJ c~f) s lcP-"'iVL j

/ I.

3. {B,h) [VC] - 1!!!g -0 05 g (aq) - 60mL - ·

L

MWvc = 62.5 ~

[VCJ(aq) = 0.05 f X 1r~~ = 8 X 10-4 mte I rt

Ktt = 26.8 Lsatm/mole

(a)

Write the equilibrium relationship of aqueous and gaseous VC-make sure you write the equation in such a way that the units of Ktt work out.

= 26.8 Leatm/mole

Calculate the concentration of VC (moles/L) in the air from Henry's constant:

VC(g) = Ktt x VC(aq) = 26.8x8.0x 10-4 = 0.0214atm

p 0.021 atm [VC]g = VC 106 = -- X 106 = 21, 400 ppm,

X 1 atm

Ptot

21,400 ppm, > > 10 ppm, /

BAD NEWS!

(b) The students should have put the open ( or unsealed bottle) in the fume hood as quickly as possible and told everyone to leave the lab immediately. (;Xf,a.,A- ~ fr..i

4. CS (0) H2S04 is a strong acid:

(a) To calculate moles of [H+], calculate number of moles of H2S04 and multiply by 2:

lo mg H2S04 x 1 g 1 moles 2 mole H+ 4 mole + --- X --- X ---- = 2 X 10- -- H

L 1000 mg 98 g mole H2S04 L

pH= -log[H+] = -log(2 x 10-4) = 3.7 lf r

(b) Normality of H2S04 solution equal the moles per Liter of Ht, which equals 2 x 10-4 ~ ;l.,J:-1 L

6

~ I ti Acld-Baee Equillbrlwn for HOC! in drinking wat«-, ~ 0 r 'Calculate &action of BOCl \hat Js -in the dlsaoclateod form:

plra = 7.5 K.- 10-T~ Sbloe it'• drbiJdng water, ~ume pH - 7. g+ = 10-7 .

K. [H+l(QOJ-J - 10-'l',II 3,,.t,- ,.., (HOOi) - I

Rearrange equ/ltion, end substitute wluea for K. and a+:

(001-1 = 10-o.a >< (HOOl] = 0'.316(HO01)

~b I

_ (BOOl] =- [HOOi 1 Q/h aactionDiss~~ted -:- fHOOlJ + [OCl-J . 001 + 0.316(HOCI • l + O;:fl~ == ~

1

(~}' ~ ,ts E,qm,,,9 50 mg/L 9f HCO5 aa: • eq/L:

..: mg lmole l eq lg -4 · [HCOa] = 60-L x -61 x -1 -1 x 1000 == 8.2 x 10 eq/L • ·g Ill O mg

• mole/L Since 1 eq of HOOa1 = 1 mole of HCOi1,

• mg/L as OaCOa: (HC01] = 8.2 x 10""' mole/L kcf rf:S

[HOOjj] = 8.2 x 10-'eq/L x !iOg oq Oa x lO 10mg = '1.mg/LuO..00.

Calculate alkalinity in mg/L 88 CaC08:

[HCO-J • 111 ~ X l mole lg X l (!(II = 1.8 X 10-a eqL of alkalinity I L Gig X lOOOtn mo . .

~ mg lmole lg 2e J -4 eq (CVs ] = 11 y x 60 g x 1000 mg x ~ - 15, 1 x 10 1 of alkallnity

Approximate Alkalinity (eq/L) = 1.8 x 10-a + 5.7 >< 10-• = 2.4 x 10-8 ! ~k Approxlme.t e Alkalinity (eq/L) = 2., >< 10-8 f X 100,n!CO;, x lleq = 119 IJ:I u CaOOa, ~

First Order Reaction:

(a) 90% of A destroyed, SQ 10% of initial concentration remains.

0.1.Ao = Ao X e-0.1'

ln(Y:.!) .... 1 t = =230d- - G.o)d ay -1 , ~-

(b) Same•as part (a) except that 99% destroyed, BO 1% temainmg:

-~) 3 J.. t = o.o• = 400 days ,~

(c) Same as pa.rt (a) except tha~ 00.9% destroyed, BO 0.1% remaining:

_ t.la.ooO ~Ah t= o,o\ =690days -=,-~

(Q) - . ,~,~ "'6il Spill nineteen yea.rs ago:

Co = 400 '¥ C alter 19y = 400 i;' i. 'D:y a zero order rate equation:

J dC -=-k dt

After integration:

C=Co-kt where Co = A00 ~ 0=20 !!!ft g

- t = 19 year

Solve for k = 20 t-1, and substitute k into the integrated zero rate eque.ti'on above to obtain

C = Co - 20y8&1'-l x 20year = 0

Answer: Yes the engineer ie correct if the degradation rate is zero order. i,-li ii. To find the "worst-case ecenario," calcula.te the concentration of the pollutant after

twenty yea.re using a first order and second order rate equation.

First Order:

Solve fork:

· ln.52. · ·,ln~ k= -~ = - = 0.l6y-1 t · 19y

. Use k and solve for the time it will take to for C = 1 1e:, assuming first.order kinetics:

1n 0. }n...L t = -~ = - 400 = 38y k 0.16yea.r-\ ·

Second Order:

Co C=l+C

Rearrange and solve fork:

k = l:_i = (t - i&i) ~ = 0.0025 kg t 19y mg xy

Uae k and solve for the time it "7ill t• for q = 1 f, aasum~ second-order kinetics:

I - I (-!--..L)k t = L.]i!_ = = 1 400 I< IJIII = 39!)y

k 0.003~

In conclusion the "worst-case scena.rio" is second order, in which case, it would take ~99 y for the pollutant to degrade. However, first order is more likely,, ,3 f +s

(D -~C ~ ~ - -

I I I de -}Ldb - J{_ rrf-, - - C 2.

- c.. t r -·2. -l•f' /; ) C JL - Co 0

-J:: ~- - -~t' tJ - f -J C,.;,

-[t-t] - -}l-t- 1 tl-J - I - .J_ -t- ~t - - G G,

_L.._ - + {01'-t (_

Co

\ C -=- Co

)+ (o }..t ! ~ rh

0