1. A 2.000 grams of a salt are dissolved in enough water to make a solution with
ID: 509015 • Letter: 1
Question
1. A 2.000 grams of a salt are dissolved in enough water to make a solution with a volume of 250.0 mL The solution is titrated with a 0.1111 M sodium hydroxide solution and reached the equivalence point at 30.00 mL. The pH of the analyte solution was determined to be 9.1989 after the addition of 10.00 ml of the sodium hydroxide and 9.8010 after the addition of 20.00 mL of the sodium hydroxide. a) what is the molecular weight of the salt assuming it is monoprotic b) is the salt acidic or basic? c) what is the Ka of the acid form of the salt d) what was the pH before addition of base? (6 pts)
2. Balance the following under both acidic AND basic conditions. Write a reaction quotient for each of the 4 final balanced reactions. A) PbO2(s) + I- (aq) Pb+2(aq) + I2(s) B) Al(s) + MnO4-(aq) MnO2(s) + Al(OH)4- (aq) (8 pts)
3. Use the tables on the next page and your initials to generate a ‘cell diagram’ where the first reaction (first initial) is the anode and the second reaction (second initial) is the cathode and determine whether the reaction is spontaneous. For gasses, assume an inert Platinum electrode. For example Dan Stasko D S Rxn D --> anode (off table, as a reduction) Co+2(aq) + 2e- Co(s) Rxn S --> cathode (off table) PbO2(s) + 4H+(aq) + 2e- Pb+2(aq) + 2H2O(l) a) balanced chemical reaction b) cell diagram (in line notation) c) standard cell potential d) Is the reaction spontaneous or not?
4. A smallish aluminum foundry can run at 9.5 x 106 Ampere pretty much non-stop. Assuming non stop production for 1.0 year at a time, what mass of Aluminum could be made from Aluminum hydroxide via electrolysis in this time frame at this current.
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Half Reaction Ag (ag) 0.80 (ag) 2 e Hg 2 Hg(l) 0.80 Fest ag) e Fe (aq) 0.77 0.76 O2 (g) 2 H (ag) 2 e H2O2 (ag) 0.70 MnO4 (ag) e Mno4 (ag) 0.56 2(s) 2 e 21 (ag) 0.54 0.52 O2 (g) 2 H2O(l) 4 e 4 OH (ag) 0.40 Cu (aq) 2 e Cu(s) 0.34 Biot aq) 2 H (ag) 3 e Bi H2O (1) 0.32 0.27 AgCl(s) e Ag(s) Cl (ag) 0.22 SO42 (ag) 4 H (ag) 2 e H2SO3(ag) H20 0.20 0.16 0.15 S(s) 2H+(ag) 2 e 0.14 AgBr(s) e Ag(s) Br (ag) 0.071 2 H (ag) 2 e H2(g) 0.00 Fe3+(aq) 3 e -0.036 -0.13 Sn (ag) 2 e Sn(s) -0.14 Half-Reaction Agl (s) e Ag(s) I (ag) N2(g) 5 H (ag) 4 e N2H5 ag) 2+ 2 e (ag) Ni(s) 2+ 2 e lag) Co(s) PbSO4 (s) 2 e Pb(s) SO4 (ag) Cd2 (ag) 2 e Cd(s) Fe2 (aq) 2 e 2 2(g) 2 H (ag) 2 e CO2 H2C2O4 (ag) Crs (ag) 3 e (a 2 H2O(l) 2 e H2(g) 2 OH (ag) n (a Al Haq) 3 e Al(s) H2(g) 2 e 2 H (ag) Mg 2+ (aq) 2 e Mg(s) 3+ La (ag) 3 e La(s) Na (ag) e Na(s) 2+ 2 e Ca (ag) Ca(s) (ag) 2 e Bar Ba(s) K (ag) K(s) e -0.15 -0.23 -0.23 -0.28 -0.36 -0.40 -0.45 -0.49 -0.50 -0.73 -0.76 -0.83 1.18 1.66 2.23 2.37 2.38 2.71 2.76 2.90 2.92 3.04Explanation / Answer
At equivalence point moles of base added = moles of acid present
Thus calculate the moles of base = molarity * volume in L
= 0.1111 * 0.030
= 3.333*10^-3 Moles NaOH
moles of acid = 3.333*10^-3 Moles
moles = g/molar mass
molar mass of unknown acid = 2.00 g/3.333*10^-3 Moles = 600 g/mol
Molarity of acid = number of moles / volume in L
= 3.333*10^-3 Moles /0.250 L
= 0.0133 M
at 1/2 equivalence point, the pH = pKa
Assume 250 mL of 0.01333 M acid
moles of acid = 3.333*10^-3 moles
the solution of NaOH is 0.1111 M
moles NaOH required to reach the half-equivalence point
= 3.333*10^-3 / 2
=1.6665*10^-3
volume NaOH = 1.6665*10^-3 / 0.1111 M
=0.015 L
total volume = 0.250 + 0.015 L = 0.265 L
moles of acid in excess = 3.333*10^-3 - 1.6665*10^-3 = 1.6665*10^-3
[H+]= 1.6665*10^-3 / 0.265 =6.288*10^-3 M
pH = - log H+= 2.20
pH will not be equal to pKa= 2.20
Ka= 10^-pKa = 10^-2.20= 6.30*10^-3
The salt id acidic because it is neutralize by base.
HA==> H+ + A-
i: 0.0133,,,,,,,,0 ..........0
c: -x....... .+x.........+x
e: 0.0133-x ........x............x
Ka = [H+][A-]/HA]
6.30*10^-3 = X*X/0.0133-X
8.39*10^-5 -6.30*10^-3X =X^2
X^2 + 6.30*10^-3 X +8.39*10^-5=0
x = 0.00653620M = [H+]
pH = -log [H+]
pH = -log 0.00653620= 2.18
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