CHEM 201 TEST 1 KEY
Average ± std deviation = 58 ±23
Rough corresponding grade: A 88 and above; B = 75 + , C = 43 +
(1) a) Solution: this is case II (comparison of replicate measurements)
x1 = 2.3103 s1 = .0011 and x2 = 2.3018 s2 = .0017 spool = { (.0011)2(4-1) +(.0017)2(5-1)
/ (4+5-2)}1/2 = {(3.63x10-6+1.16x10-5)/7}1/2= 1.5x10-3 : so tcalc = {(2.31032.3018)/.0015 } {4(5)/(4+5)}=
tcalc = 5.672.22 = 8.45 ; we can be 99.9% confident that they are different since
8.45>5.408
b) Solution: x1 ± s1 = 2.310 ±.001 and x2 ± s2 = 2.302 ± .002
2) Solution: let x = grams CaC2O4 and y = grams CaCO3 ; also the FW’s of the relevant
substances are:
CaC2O4 = 40.1 + 2(12.0) + 4(16.0) = 128.1; CaCO3 = 100.1; CaO = 56.1
So: (i) x + y = 0.7400 and (ii) x(56.1/128.1) + y(56.1/100.1) = 0.385 =>
0.438x+0.560y=.385
From (i): y = 0.7400-x ; into (ii): 0.438 x + 0.560 (0.7400 – x) = 0.385
=> 0.438 x + 0.415 -0.560x-0.385 = 0 ; x=0.0300/0.122 = 0.246g = g CaC2O4
So, % CaC2O4 = .246g(100%)/.7400 = 33.2 %
b)
m1 = (.246g CaC2O4)(128.1-100.1)/(128.1))=0.054 g
3) a) Solution: MCl = MAgVAg / VCl = (0.0805M)(11.56)/(22.10) = 0.0421M ;
%(w/w) = (0.0421 mol Cl/L) (35.45g/mol) (1L/1000mL)(1mL/1.020g)(100%) = 0.146
Cl%
b) Solution: ppm NaCl = (.0421 mol
NaCl/L)(58.45g/mol)(1000mg/g)(.500/(19.500+0.500) =61.5 ppm NaCl
c) Solution: Mohr titration method; Ag2CrO4
E . (actually Ksp of Ag2CrO4 is larger than AgCl but solubility is less for Ag2CrO4) why?
Because when the ionic compound is of the form MX2 it’s solves differently than MX…
4) Solution: first, get [AgNO3] :
M1V1 = M2V2 => M2 = M1V1 / V2 = (.0530M)(18.10)/(25.00) = 0.0384 M
a) [Ag+] = [Ag+]o(fraction free)(dilution factor) =
=
(0.384M)(18.10 5.00)mL 25.0mL
=0.0232M
18.10mL
25.0 + 5.0mL
pAg = -log(0.0232)=1.635
b) similar: [Ag]= =
(0.384M)(18.10 15.00)mL
25.0mL
= 0.00411
18.10mL
25.0 + 15.0mL
=> pAg = 2.386
c) use Ksp equil AgCl <= = => Ag+ + Cl- : x=[Ag]=[Cl] => x2 =1.8x10-10 ; x =1.34x10-5
pAg= 4.872
d) use Ksp equi:AgCl <= = => Ag+ + Cl- :
[Cl-]= (.0530M)(25.00-18.10)/(25.00+25.00)=0.007314M) so (x)(x+.007314)=1.8x1010
x(.007314)
x 1.8x10-10 / (.007314) = 2.5x10-8 => pAg =7.609
5) Solution: 12.1 M means that in 1 Liter: 12.1 moles HCl are present, so:
12.1molHCl 36.45gHCl 1L
0.441gHCl
=
L solution molHCl 1000mL 1mL sol'n
From 12.1 M we get:
From 16.1 m we get:
And so %wt =
16.1 mol HCl 36.45 g 1 kg
0.587g HCl
=
kg H 2O mol HCl 1000 g
g H 2O
0.587 g HCl
g HCl
x100% = 37.0%
x100% =
0.587g HCl + 1 g H 2O
g solution
b) What is its density? (in g/mL)
Solution:
=
0.587 g HCl
441 g HCl
=
; solve for density () :
1.587 g sol'n 1000 mL sol'n x mass of sol'n
441g HCl 1.586 g sol'n
=
= 1.19
vol of sol'n
1000 mL sol'n 0.586 g HCl
g/mL
37.0 g HCl
1000 mL 1 mol HCl
()
= 12.1 mol/L
100 g sol'n
36.45 g HCl
L
12.1 mol HCl
L
36.45g HCl 100.g sol'n
and so =
= 1.19 g/mL
L sol'n
1000 mL mol HCl 37.0g HCl
Or, you could also say
6) 1) D; the net ionic rxn: Th4+ + 4 IO3- --> Th(IO3)4 (s); @ ep : #eq Th(NO)3 = #eq
Mg(IO3)2;
4 M1V1 = M2V2 => where M1 =[Th4+], etc, M2 =[IO3-] = 2xMMg(IO3)2 =
2(0.0700M)=0.140M, etc
M1 = (1/4)(M2V2)/(V1) = (1/4)(0.140M)(22.0mL)/(10.0mL) = 0.0770 M
Or, can use balanced equation : Th(NO3)4 (aq) + 2 Mg(IO3)2 (aq) --> Th(IO3)4 (s) + 2
Mg(NO3)2(aq). So [Th] = MMg(IO)3)2VMg(NO3)2 (1mol Th/2mol Mg(IO3)2 )/VTh =
(22.0)(0.0700)/2(10.0) = 0.0770M
(2) solution: C: eM = M ( (.2/1.6)2+(.5/10.5)2)1/2 = (0.152)(0.134) = 0.02
(3) solution: A: 0.00100 mol/L(10mL/100mL) (2(52.0g Cr )/mol K2Cr2O7)(1000mg/g) =
10.4 ppm
(4) solution: A: there are 3 equil:
Fe3+ + L2- - - > FeL+ K1
FeL+ + L2- - - > FeL2- K2
FeL2- + L2- - - > FeL33- K3
Since 2 =K1K2 and 1 = K1
then 2/1 = K2 = (2.6x10-15)/(2.9x10-8)=9.0x10-8
(5) solution: C Ag+ starts precipitating at Q = Ksp =1.5x10-5 =[Ag+]2[SO42-] => [SO42-] =
1.5x10-5 /(.01)2 =0.15M; at that concentration of SO42- what’s the [Ca2+] =? Ksp for
CaSO4 = 2.4x10-5 = [Ca2+][0.15] => [Ca2+]=1.6x10-4M. Determine what % of the original
concentration that is:1.6x10-4x100%/0.0100 =1.6% remaining so the % precipitated must
be: 100%-1.6% = 98.4%