Please show steps for problems! A sample weighing 640 mg is comprised of 12.0 %
ID: 993309 • Letter: P
Question
Please show steps for problems! A sample weighing 640 mg is comprised of 12.0 % KC1 (74 56 g/mol). 8.71 %CaCl_2 (1110 g/mol) and inert material What weight of AgCI (143.3 g/mol) can be precipitated from this sample? Calculate the molarity of nitrate in a 5 .00 ppm Ca(NO_3)_2 solution {Ca(NO_3)_2 = 164g/mol; NO_3^-= 62 g/mol). Use activities to calculate the molar solubility of Zn(OH)_2 (Ksp = 3 times 10^-16) in 0.01 M KCl. Calculate the pH of the solution when 20.00 mL of 0.1 M CH_3COOH (K_a = 1.76 times 10^5) is titrated with 10 and 30 mL of 0.1 M NaOH. Using 1 times 10^-6 M as the criterion for quantitative removal, is it feasible to use IO_3^- to separate In^3+ and Tl^+ in a solution that is initially 0.11 M in In^3+ and 0 06 M in TI^+? (Ksp: In(IO_3)_3 = 3.3 times 10^-11; TIIO_3= 3.1 times 10^-6). 50.0 mL of 0.01 M Sr^2+ solution was buffered at pH = 11 and titrated with 0.02 M EDTA. Calculate pSr^2+ after the addition of 10 and 30 mL of EDTA. (K_sry = 4.3 times 10^8. a_4 = 0.85). Calculate the theoretical potential and the equilibrium constant for the spontaneous reaction of the following cell. Cu|Cu^2+ (0 12 M) || Pd^2+ (0.03M>|Pd pd^2+ + 2e^- rightarrow pd E^0 = 0.987 v Cu^2+ + 2e^- rightarrow Cu E^0 = 0.342 V A solution containing 3.75 mg/100 mL of compound A (220 g/mol) has a transmittance of 39.6% in a 1.5 cm cell. Calculate the molar absorptivity of A.Explanation / Answer
2. Molarity of Ca(NO3)2 solution = 5 ppm = 5 mg/L = 0.005 g/164 g/mol = 3.05 x 10^-5 M
molarity of NO3- in solution = 2 x 3.05 x 10^-5 = 6.1 x 10^-5 M
3. Ionic strength (I) of KCl solution = 1/2(Cizi^2)
= 1/2(1 x 0.01 + 1 x 0.01)
= 0.01 M
activity coefficient for,
log[Y] = -0.51 x zi^2 x sq.rt.(I)/(1 + 3.3 x alpha(r) x sq.rt.(I))
alpha(r) = effective diameter for ion
log[Y(K+)] = -0.51 x 1^2 x sq.rt.(0.01)/(1 + 3.3 x 0.3 x sq.rt.(0.01))
Y(K+) = 0.898
Y[Cl-] = 0.898
Zn(OH)2 + 2KCl ---> ZnCl2 + 2KOH
let x amount of salt is in solution,
Ksp = [Cu2+][OH-]^2
3 x 10^-16 = (x)(0.01 x 0.898 x x)^2 = 8.06 x 10^-5x^3
x^3 = 3.72 x 10^-12
molar solubility = x = 1.55 x 10^-4 M
4. pH
(a) 10 ml 0.1 M NaOH added
This is half equivalence point
[CH3COOH] left in solution = [CH3COO-] formed in solution
pH = pKa = 4.75
(b) 30 ml of 0.1 M NaOH added
This is past equivalence point
excess [OH-] = 0.1 M x 10 ml/50 ml = 0.02 M
pOH = -log[OH-] = 1.70
pH = 14 - pOH = 12.3
5. Starting with the most soluble salt,
TlIO3
Ksp = [Tl+][IO3-]
3.1 x 10^-6 = [0.06][IO3-]
[IO3-] required for precipitation of more soluble salt = 5.16 x 10^-5 M
Amount of IO3- required for precipitation of les soluble salt,
In(IO3)3
Ksp = [In3+][IO3-]^3
3.3 x 10^-11 = [0.11][IO3-]^3
[IO3-] = 6.7 x 10^-4 M
So the 1 x 10^-6 M concentration is below the concentration of the two values and thus it is not possible to separate the two salts quantitaively from solution.
6. EDTA titration
Kf' = Kf x alpha[Y4-] = 4.3 x 10^8 x 0.85 = 3.655 x 10^8
(a) 10 ml of 0.02 M EDTA added
moles of Sr2+ = 0.01 M x 50 ml = 0.5 mmol
moles of EDTA = 0.02 M x 10 ml = 0.2 mmol
[Sr2+] excess = 0.3 mmol/60 ml = 0.005 M
pSr2+ = -log[Sr2+] = -log(0.005) = 2.30
(b) 30 ml of 0.02 M EDTA added
moles of Sr2+ = 0.01 M x 50 ml = 0.5 mmol
moles of EDTA = 0.02 M x 30 ml = 0.6 mmol
[SrY2-] formed = 0.5 mmol/80 ml = 6.25 x 10^-3 M
[EDTA] excess = 0.1 mmol/80 ml = 1.25 x 10^-3 M
Kf' = [SrY2-]/[Sr2+][EDTA]
3.655 x 10^8 = (6.25 x 10^-3)/[Sr2+](1.25 x 10^-3)
[Sr2+] = 1.37 x 10^-8 M
pSr2+ = -log[Sr2+] = 7.86
7. Eo = 0.987 - 0.342 = 0.645 V
E = Eo - 0.0592/n logK
= 0.645 - 0.0592/2 log(0.12/0.03)
= 0.627 V
nFEo = -RTlnK
2 x 96485 x 0.645 = 8.314 x 298 lnK
K = 6.57 x 10^21
8. Absorbance = 2 - log(%T) = 2 - log(39.6) = 0.402
A = ebc
c = 3.75 mg/1000 x 0.1 L x 220 = 1.70 x 10^-4 M
molar absorptivity = e = 2358.4 M-1.cm-1