Exam 2B, Spring 2005
Exam 2B, Spring 20051. Complete and balance the following reactions:
a. H3PO4(aq) + CsOH(aq)
b. C6H5NH2(aq) + HBr(aq)
c. HClO(aq) + H2O(l)
d. Ba(OH)2(s) + HClO4(aq)
a. H3PO4(aq) + 3 CsOH(aq) → 3 H2O(l) + 3 Cs+(aq) + PO43–(aq)
b. C6H5NH2(aq) + HBr(aq) → C6H5NH3+(aq) + Br–(aq)
c. HClO(aq) + H2O(l) → ← H3O+(aq) + ClO–(aq)
d. Ba(OH)2(s) + 2 HClO4(aq) → 2 H2O(l) + Ba2+(aq) + 2 ClO4–(aq)
2. Complete and balance the following reaction and identify the acid, the base, the conjugate acid, and the conjugate base:
bicarbonate ion + cyanide ion
HCO3–(aq) + CN–(aq) → ← CO32–(aq) + HCN(aq)
Acid: HCO3–
Base: CN–
Conjugate Acid: HCN
Conjugate Base: CO32–
3. Identify the Lewis acid and the Lewis base in the following reaction.
Cd2+(aq) + 4 NH3(aq) → ← [Cd(NH3)4]2+(aq)
Lewis Acid: Cd2+
Lewis Base: NH3
4. Identify the stronger acid in pair given below. Briefly (5 words or less) justify your reasoning.
a. H2Se or H2Te
b. H2SO3 or H2SeO3
a. H2Se or H2Te
H2Te because of the Te size
b. H2SO3 vs. H2SeO3
H2SO3 because of the S electronegativity
5. Calculate the pH and pOH of the following solutions:
a. 2.0×10–4 M hydrobromic acid.
b. 3.0×10–4 M lithium hydroxide.
c. 4.0×10–4 M calcium hydroxide.
a. 2.0×10–4 M hydrobromic acid.
HBr(aq) + H2O(l) → H3O+(aq) + Br–(aq)
[H3O+] = [HBr] = 2.0×10–4 M
pH = –log([H3O+]) = –log(2.0×10–4) = 3.30
pOH = 14.00 – pH = 14.00 – 3.30 = 10.70
b. 3.0×10–4 M lithium hydroxide.
LiOH(aq) → Li+(aq) + OH–(aq)
[OH–] = [LiOH] = 3.0×10–4 M
pOH = –log([OH–]) = –log(3.0×10–4) = 3.52
pH = 14.00 – pOH = 14.00 – 3.52 = 10.48
c. 4.0×10–4 M calcium hydroxide.
Ca(OH)2(aq) → Ca2+(aq) + 2 OH–(aq)
[OH–] = 2×[Ca(OH)2] = 2(4.0×10–4) = 8.0×10–4 M
pOH = –log([OH–]) = –log(8.0×10–4) = 3.10
pH = 14.00 – pOH = 14.00 – 3.10 = 10.90
6. Consider the reaction shown below, which has ΔH = 131 kJ/mol.
C(s) + H2O(g) → ← CO(g) + H2(g)
a. Which way will the reaction shift if graphite (solid carbon) is added to the reaction? Explain.
b. Which way will the reaction shift if the temperature is raised? Explain.
C(s) + H2O(g) + heat → ← CO(g) + H2(g) (For an endothermic reaction, ΔH > 0, heat is a reactant)
a. Which way will the reaction shift if graphite (solid carbon) is added to the reaction? Explain.
The reaction will not shift. The solid reactant concentration does not change so the equilibrium is not changed.
b. Which way will the reaction shift if the temperature is raised?
Raising the temperature adds heat so the reaction will shift towards products to remove the added heat.
7. Estimate the pH of a 1.0×10–2 M solution of phenol, C6H5OH. Ka = 1.0×10–10
C6H5OH(aq) + H2O(l) → ← H3O+(aq) + C6H5O–(aq)
Ka = [H3O+]e[C6H5O–]e [C6H5OH]e = 1.0×10–10
Initial 1.0×10–2 0 0
Change –x +x +x
Equilibrium 1.0×10–2 – x x x
Approximate? 1.0×10–2/1.0×10–10 = 1.0×108: yes
1.0×10–10 = x2/(1.0×10–2)
x2 = 1.0×10–12
[H3O+]e = 1.0×10–6 M
pH = –log[H3O+] = –log[1.0×10–6] = 6.00