Exam 3A, Spring 2018
Exam 3A, Spring 20181. Complete and balance the following in aqueous solution.
a. cyanate ion plus perchloric acid
b. NH3(aq) + HClO(aq)
c. Cu(NO3)2(aq) + Li2S(aq)
d. CdC2O4(s) + HNO3(aq)
e. Ni2+(aq) + H2O(l)
f. Cu2+(aq) + NH3(aq)
a. OCN–(aq) + HClO4(aq) → HOCN(aq) + ClO4–(aq)
b. NH3(aq) + HClO(aq) → ← NH4+(aq) + ClO–(aq) (Kc = 2.9×10–8/5.6×10–10 = 52 < 100)
c. Cu(NO3)2(aq) + Li2S(aq) → CuS(s) + 2 Li+(aq) + 2 NO3–(aq)
d. CdC2O4(s) + 2 HNO3(aq) → Cd2+(aq) + 2 NO3–(aq) + H2C2O4(aq)
e. Ni2+(aq) + 2 H2O(l) → ← NiOH+(aq) + H3O+(aq)
f. Cu2+(aq) + 4 NH3(aq) → ← [Cu(NH3)4]2+(aq)
2. Which is the stronger acid, SiH4 or PH3? Provide a brief explanation.
PH3 is the stronger acid: P is more electronegative than Si.
3. Is a solution of Na3PO4 acidic, basic, or neutral? Show reactions that support your conclusion.
Na3PO4(aq) → 3 Na+(aq) + PO43–(aq)
Na+(aq) + H2O(l) → NR
PO43–(aq) + H2O(l) → ← HPO42–(aq) + OH–(aq)
The solution is basic since hydroxide ion is generated.
4. A solution is prepared with [NH3] = 0.010 M and [NH4+] = 0.050 M. Find the pH.
NH4+(aq)+ H2O(l) → ← H3O+(aq)+ NH3(aq)
Ka = [H3O+]e[NH3]e [NH4+]e = 5.6×10–10
Initial0.05000.010
Change–x+x+x
Equilibrium0.050 – xx0.010 + x
Approximate? 0.010/5.6×10–10 = 1.8×107 > 100 so Yes
5.6×10–10 = x(0.010)/0.050
x = [H3O+]e = 2.8×10–9 M
pH = –log[H3O+] = –log(2.8×10–9) = 8.55
5. Consider ZnCO3:
a. Find the molar solubility of ZnCO3 in water at 25 °C. Ksp = 1.4×10–11
b. Find the molar solubility of ZnCO3 in a 0.10 M solution of NH3 at 25 °C. Assume that the ammonia concentration does not change significantly. Is the assumption valid? Why or why not?
a. solubility in water
ZnCO3(s) → ← Zn2+(aq)+ CO32–(aq)
Ksp = [Zn2+]e[CO32–]e = 1.4×10–11
Initial00
Change+x+x
Equilibriumxx
1.4×10–11 = x2
x = 3.7×10–6 M = molar solubility of zinc carbonate in water.
b. solubility in 1.0 M ammonia
Solubility: ZnCO3(s) → ← Zn2+(aq) + CO32–(aq)
Complex Ion: Zn2+(aq) + 4 NH3(aq) → ← [Zn(NH3)4]2+(aq)
Net: ZnCO3(s) + 4 NH3(aq) → ← [Zn(NH3)4]2+(aq) + CO32–(aq)
Kc = [[Zn(NH3)4]2+]e[CO32–]e [NH3]e4
Kc = Kf×Ksp = (4.1×108)×(1.4×10–11) = 5.7×10–3
Initial0.1000
Change– 4x+x+x
Equilibrium0.10 – 4xxx
Assume that 0.10 – 4x ~ 0.10
5.7×10–3 = (x)(x)/(0.10)4
Taking the square root of both sides and solving for x gives:
x = 7.5×10–4 M
The molar solubility of ZnCO3 in 0.10 NH3 is found to be 7.5×10–4 M.
The assumption is good: for the zinc carbonate to reach 7.5×10–4 M would require the ammonia concentration to be 4×7.5×10–4 = 0.0030. M, and 0.10 – 0.0030 = 0.10.