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CH 221โZiegler Chapter 4 WORKSHEET Precipitation, Acids and Bases, and Molarity
- Oxidation numbers Give the oxidation state of each element in the following compounds
Zn Zn= 0
Pb(NO 3 ) 2 Pb= +2; N= +5; O= -
Zn(NO 3 ) 2 Zn= +2; N= +5; O= -
HI H= +1; I= -
HIO 4 H= +1; O= -2; I= +
H 2 SO 4 H= +1; S= +6; O= -
CO C= +2; O= -
CO 2 C= +4; O= -
CO 3 -2^ C= +4; O= -
NH 3 N= -3; H= +
H 2 O O= -2; H= +
H 2 O 2 H= +1; O= -
NO 3 -^ N= +5; O= -
O 2 O= 0
N 2 N= 0
CH 4 C= -4; H= +
- In the following reactions, identifiy: the element which is oxidized, the element reduced, the oxidizing agent, and the reducing agent.
Cu + 2AgNO 3 ร 2Ag + Cu(NO 3 ) 2
Cu loses electrons to form Cu+2^ ; Ag+^ gains electrons to form Ag Cu is oxidized, and is the reducing agent Ag+^ is reduced, and is the oxidizing agent.
Mg + Ca(OH) 2 ร Ca + Mg(OH) 2
Mg loses electrons; Ca+2^ gains electrons Mg is oxidized, and is the reducing agent. Ca+2^ is reduced, and is the oxidizing agent.
2AgCl + Co ร CoCl 2 + 2Ag
Ag+^ gains electrons; Co loses electrons to form Co+ Co is oxidized, and is the reducing agent. Ag+^ is reduced, and is the oxidizing agent.
Cu + Zn+2^ ร Zn + Cu+
Cu loses electrons to form Cu +2; Zn+2^ gains electrons to form Zn Cu is oxidized, and is the reducing agent. Zn+2^ is reduced, and is the oxidizing agent.
- Write net ionic equations for any precipitation reaction reactions that occur on mixing aqueous solutions of the following substances.
a. KCl and AgNO 3
Ionic: K+^ + Cl-^ + Ag+^ + NO 3 -^ โ K+^ + AgCl (^) (s) + NO 3 -
Net ionic: Cl-^ + Ag+^ โ AgCl(s)
b. NaOH and MgCl 2
Ionic: 2Na+^ + 2OH-^ + Mg+2^ + 2Cl-^ โ Mg(OH) 2 + 2Na+^ + 2Cl-
Net ionic: 2OH-^ + Mg+2^ โ Mg(OH)2(s)
c. Fe 2 (SO 4 ) 3 and BaCl 2
Ionic: 2Fe+3^ + 3SO4-2^ + 3Ba+2^ + 6Cl-^ โ 2Fe+3^ + 6Cl-^ + 3BaSO4(s)
Net ionic: Ba+2^ + SO4-2^ โ BaSO4(s)
d. KCl and Mg(NO 3 ) 2
Ionic: 2K+^ +2Cl-^ + Mg +2^ + 2NO 3 -^ โ 2 K+^ + 2NO 3 -^ + Mg+2^ + 2Cl-
Net ionic: NO REACTION
- Write net ionic equations for each of the following acid/base reactions. a. HF + Ca(OH) 2 ร 2H 2 O + CaF2(aq)
Ionic: 2HF + Ca+^ + 2OH-^ โ 2H 2 O + Ca+2^ + 2F- Net ionic: HF + OH-^ โ H 2 O + F-
b. H 2 SO 4 + Mg(OH) 2 ร 2H 2 O + MgSO4(aq)
Ionic: 2H+^ + SO 4 -2^ + Mg(OH) 2 โ 2H 2 O + Mg+2^ + SO 4 -
Net ionic: 2H+^ + Mg(OH) 2 โ 2H 2 O + Mg+
b. How many mL of this solution is required to provide 0.0367 mol of MgCl 2?
2 2
1.0L 1000mL ? mL 0.0367mole MgCl 146.6mL 147 mL of solution 0.250 mole MgCl 1.0L
x x
โ โ ^ โ^ โ
- Calculate the mass of MgCl 2 required to prepare 250. mL of a 0.100 M MgCl 2 solution.
2 2 2 2
2 2
0.100 mole MgCl 1.0L 95.1g MgCl
?g MgCl = 250mL
1.0L 1000mL 1mole MgCl
2.375g MgCl 2.38g MgCl
x x x
โ โ โ โ โ^ โ
- What is the molarity of Cl-^ ions in a 0.100 M MgCl 2 solution?
2 -
? moles Cl 0.100moles MgCl 2moles Cl 0.200moles Cl
L 1.0L 1mole MgCl 1L
x
โ โ โ^ โ
- A solution is prepared by taking 25.00 mL of a 0.300 M Ca(OH) 2 solution, and adding enough water to a total volume of 0.500 L of solution.
a. What is the molarity of Ca(OH) 2 in this solution? This is a dilution problem and can be worked either of 2 ways:
1 1 2 2
1 1 2 2 2
2 2
M V = M V
M V (0.300 moles/L)(25mL) 1.0L 0.015moles Ca(OH)
M =
V 0.500L 1000mL L
Or you can do it with dimensional Analysis alone.
? moles Ca(OH) 0.300moles Ca(OH)^ 1L
25.00 mL
L 1.0L 100
x
x x
โ โ =^ โ โ โ โ=
โ โ โ^ โ ^ โ^ โ
2
0mL 0.500L
0.0150 moles Ca(OH)
L
x
b. Of Ca+2^ ion?
+2 + (^2) +
2
? moles Ca 0.0150 moles Ca(OH) 1mole Ca
0.0150moles Ca /L
L 1L 1mole Ca(OH)
x
โ โ โ^ โ
c. Of OH-^ ion?
2
2
? moles OH 0.0150moles Ca(OH) 2moles OH 0.0300moles OH
L 1L 1mole Ca(OH) 1L
x
โ โ โ^ โ^ โ^ โ
- Give the net ionic equation describing what happens when 25.00 ml of the solution in 0.300 M Ca(OH) 2 is mixed with 25.00 ml of the solution in 0.100 M MgCl 2. 2OH-^ + Mg+2^ โ Mg(OH) 2
a. What is the ionic equation for the reaction between these two solutions?
Ca+2^ + 2OH-^ + Mg+2^ + 2Cl-^ โ Mg(OH) 2 + Ca+2^ + 2Cl-
b. How many moles of the precipitate is formed?
2 2
0.300moles Ca(OH) 2moles OH 1mole Mg(OH) If Ca(OH) limits : 25.00mL 1000mL 1mole Ca(OH) 2moles OH 0.00750mole Mg(OH)
0.100 mole MgCl 1moleMg(OH) If MgCl2 limits: 25.00mL 1000mL 1moleMg
x x x
x x
โ โ โ^ โ^ โ^ โ
2 2 2 2
Cl
58g Mg(OH) 0.00250mole Mg(OH) 0.437g Mg(OH) 1mole Mg(OH)
x
c. What is the molarity of each spectator ion in the resulting solution?
0.100mole MgCl 25.00mL 2mole Cl
?mole Cl =
1000mL 1 1mole MgCl
0.00500mole Cl 0.100moles Cl /L
0.050L
x
โ
โ โ โ โ โ^ โ
โ โ โ โโ^ โ=
