Exam 3A, Spring 2006
Exam 3A, Spring 20061. Complete and balance the following reactions:
a. H2SO4(aq) + KOH(aq)
b. thiocyanate ion plus water
c. Fe2+(aq) + EDTA4–(aq)
d. CoCl2(aq) + Na2CO3(aq)
e. HOCN(aq) + NH3(aq)
f. Mn2+(aq) + H2O(l)
g. Ba(OH)2(s) + H3PO4(aq)
a. H2SO4(aq) + 2 KOH(aq) → 2 K+(aq) + SO42–(aq) + 2 H2O(l)
b. SCN–(aq) + H2O(l) → ← HSCN(aq) + OH–(aq)
c. Fe2+(aq) + EDTA4–(aq) → ← [Fe(EDTA)]2–(aq)
d. CoCl2(aq) + Na2CO3(aq) → CoCO3(s) + 2 Na+(aq) + 2 Cl–(aq)
e. HOCN(aq) + NH3(aq) → OCN–(aq) + NH4+(aq) (Kc = 3.5×10–4/5.6×10–10 = 6.3×105 > 100)
f. Mn2+(aq) + 2 H2O(l) → ← MnOH+(aq) + H3O+(aq)
g. 3 Ba(OH)2(s) + 2 H3PO4(aq) → Ba3(PO4)2(s) + 6 H2O(l)
2. Identify if the following ionic salts are acidic, basic, or neutral in aqueous solution. Show the reaction that defines the acidity or basicity, if the salt is not neutral.
a. Cu(ClO4)2
b. KClO3
c. NaNO3
a. Cu(ClO4)2(aq) → Cu2+(aq) + 2 ClO4–(aq)
Cu2+(aq) + 2 H2O(l) → ← CuOH+(aq) + H3O+(aq)
ClO4– + H2O(l) → NR
Acidic
b. KClO3(aq) → K+(aq) + ClO3–(aq)
K+(aq) + 2 H2O(l) → NR
ClO3–(aq) + H2O(l) → ← HClO3(aq) + OH–(aq)
Basic
c. NaNO3(aq) → Na+(aq) + NO3–(aq)
Na+(aq) + 2 H2O(l) → NR
NO3–(aq) + H2O(l) → NR
Neutral
3. A buffer is prepared to be 0.10 M in HClO2 and 0.10 M in NaClO2.
a. Calculate the pH of the solution, showing all of your work.
b. 0.1 g of HCl(g) is added to 1 L of the buffer. Calculate the new pH.
a. Calculate the pH of the solution, showing all of your work.
NaClO2(aq) → Na+(aq)+ ClO2–(aq)
HClO2(aq) +H2O(l) → ← H3O+(aq) +ClO2–(aq)
Ka = [H3O+]e[ClO2–]e /[HClO2]e = 1.1×10–2
Initial0.1000.10
Change–x+x+x
Equilibrium0.10 – xx 0.10 + x
Approximate? 0.10/1.1×10–2 = 9 < 100 No
1.1×10–2 = x(0.10 + x)/(0.10 – x)
1.1×10–3 – 1.1×10–2x = 0.10x + x2
x2 + 0.111x – 1.1×10–3 = 0
x = [H3O+]e = 9.2×10–3 (x = –0.12 is chemically unreasonable)
pH = –log([H3O+]) = –log(9.2×10–3) = 2.04
b. 0.1 g of HCl(g) is added to 1 L of the buffer. Calculate the new pH.
[HCl] = 0.1 g/(1.0 g/mol + 35.5 g/mol)/1.0 L = 0.003 M
NaClO2(aq) → Na+(aq)+ ClO2–(aq)
HClO2(aq) +H2O(l) → ← H3O+(aq) +ClO2–(aq)
A/BHClO2(aq) +Cl–(aq) ← HCl(aq) +ClO2–(aq)
Ka = [H3O+]e[ClO2–]e /[HClO2]e = 1.1×10–2
Initial0.1000.10
A/B+ 0.0030 – 0.003
Change–x+x+x
Equilibrium0.10 – xx 0.10 + x
Approximate? 0.10/1.1×10–2 = 9 < 100 No
1.1×10–2 = x(0.10 + x)/(0.10 – x)
1.1×10–3 – 1.1×10–2x = 0.10x + x2
x2 + 0.111x – 1.1×10–3 = 0
x = [H3O+]e = 9.2.×10–3 (x = –0.12 is chemically unreasonable)
pH = –log([H3O+]) = –log(9.2×10–3) = 2.04
4. A mixture of Cd(OH)2 in water is prepared. Will the solubility in increase, decrease, or stay the same if the mixture is added to the following solutions. Show a chemical reaction to justify your answer if the solubility increases or decreases.
a. 1 M HNO3
b. 1 M NaNO3
c. 1 M NH3
a. 1 M HNO3
Cd(OH)2(s) + 2 HNO3(aq) → Cd2+(aq) + 2 NO3–(aq) + 2 H2O(l)
Solubility increases because of the acid/base reaction.
b. 1 M NaNO3
Cd(OH)2(s) + NaNO3(aq) → NR
There is no change in the solubility because no reactions occur.
c. 1 M NH3
Cd(OH)2(s) + 4 NH3(aq) → ← [Cd(NH3)4]2+(aq) + 2 OH–(aq)
Solubility increases because of the complex ion formation.
5. Determine the solubility of Li3PO4 in a 0.10 M solution of LiNO3.
LiNO3(aq) → Li+(aq) +NO3–(aq)
Li3PO4(s) → ← 3 Li+(aq) +PO43–(aq)
Ksp = [Li+)]e3[PO43–]e = 3.2×10–9
Initial0.10 0
Change+3x+x
Equilibrium0.10 + 3x x
0.10 + 3x ~ 0.10 since x will be small
3.2×10–9 = (0.10)3(x)
x = molar solubility = 3.2×10–6 M