Honors Chemistry

Answers to Chapter 4 Study Questions

 

1. a) strong acid: HNO3(aq) H+(aq) + NO3-(aq)

b) weak acid: HClO(aq) H+(aq) + ClO-(aq)

c) weak base: NH3(aq) + H2O NH4+(aq) + OH-(aq)

d) neutral: NaNO3(s) Na+(aq) + NO3-(aq)

e) strong base: Ba(OH)2(s) Ba2+(aq) + 2 OH-(aq)

 

2. a)

b) Volumesolution A x Molaritysolution A = Volumesolution B x Molaritysolution B

125 ml 0.350 M NaCl = x ml 2.00 M NaCl

3. = 90.0 g C6H12O6

 

4. a) Mg(s) + 2 HCl(aq) MgCl2(aq) + H2(g),

[or Mg(s) + 2 H+(aq) Mg2+(aq) + H2(g)]

b) = 0.125 moles H2

5. a) Precipate forms. Fe3+(aq) + 3 OH-(aq) Fe(OH)3(s)

b) No Reaction ((NH4)2CO3 and LiCl are both soluble)

c) Precipate forms. Ni2+(aq) + S2-(aq) NiS(s)

 

6. Lots of possibilities: Pick a soluble Sr2+ and a soluble SO42-, such as: Sr(NO3)2 and Na2SO4.

 

7. any strong acid + strong base: H+(aq) + OH-(aq) H2O

 

8. At neutralization, molesacid = molesbase

Volumeacid Molarityacid = Volumebase Molaritybase

Volumebase 2.00 M = 12.5 ml 0.0800 M

= 0.500 ml

9. a) 3 Pb(NO3)2(aq) + 2 AlCl3(aq) 2 Al(NO3)3(aq) + 3 PbCl2(s).

 

b) 5.58 mL

c) = 0.310 g

10. 2 Fe(NO3)3(aq) + 3 Na2CO3(aq) 6 NaNO3(aq) + Fe2(CO3)3(s)

 

This is a limiting reactant problem, so first determine which reactant is limiting.

 

= 5.20 g Fe2(CO3)3

 

= 8.71 g Fe2(CO3)3

 

Therefore, 5.20 g Fe2(CO3)3 is formed.

 

11. = 66.1 ml solution