Page 150 - Theory and Problems of BEGINNING CHEMISTRY

CHAP. 9] NET IONIC EQUATIONS 139

9.13. A student used H 3 PO 4 to prepare a test solution that was supposed to contain phosphate ions. Criticize
this choice.
Ans. Phosphoric acid is weak, and relatively few ions are present. The student should have used an ionic
phosphate—sodium phosphate or ammonium phosphate, for example.

9.14. Write one or more complete equations for each of the following net ionic equations:

+ + − 2−
(a)NH 3 + H −→ NH 4 (c)CO 2 + 2OH −→ CO 3 + H 2 O
(b)Cu 2+ + S 2− −→ CuS

Ans. (a)NH 3 + HNO 3 −→ NH 4 NO 3 or NH 3 + HCl −→ NH 4 Cl
or ammonia plus any other strong acid to yield the ammonium salt.
(b) CuSO 4 + K 2 S −→ CuS + K 2 SO 4 or Cu(NO 3 ) 2 + BaS −→ CuS + Ba(NO 3 ) 2
or CuSO 4 + (NH 4 ) 2 S −→ CuS + (NH 4 ) 2 SO 4
or any soluble copper(II) salt with any soluble sulfide.
(c)CO 2 + 2KOH −→ K 2 CO 3 + H 2 O
or CO 2 plus any other soluble hydroxide to give a soluble carbonate. Note that the following equation
cannot be used, because BaCO 3 is insoluble: CO 2 + Ba(OH) 2 −→ BaCO 3 + H 2 O
9.15. What chemical would you use to prepare a solution to be used for a test requiring the presence of Ca 2+
ions?
Ans. Ca(NO 3 ) 2 , CaCl 2 , or any other soluble calcium salt.

9.16. What chemical would you use to prepare a solution to be used for a test requiring the presence of Br −
ions?
Ans. NaBr, BaBr 2 , HBr(aq), or any other soluble ionic bromide.

2−
9.17. What chemical would you use to prepare a solution to be used for a test requiring the presence of SO 4
ions?
Ans. Na 2 SO 4 , FeSO 4 , or any other soluble sulfate.

Supplementary Problems

9.18. Write a net equation for the following overall equation:
4Au + 16 KCN + 6H 2 O + 3O 2 −→ 4 KAu(CN) 4 (aq) + 12 KOH
−
Ans. 4Au + 16 CN + 6H 2 O + 3O 2 −→ 4Au(CN) 4 + 12 OH −
−
9.19. Write a net ionic equation for each of the following equations:

(a) CdS(s) + I 2 −→ CdI 2 (aq) + S
(b) 4 KOH + 4 KMnO 4 (aq) −→ 4K 2 MnO 4 (aq) + O 2 + 2H 2 O
(c) 2 HI + 2 HNO 2 −→ 2NO + 2H 2 O + I 2
(d) AlCl 3 + 4 NaOH −→ NaAl(OH) 4 (aq) + 3 NaCl
Ans. (a) CdS(s) + I 2 −→ Cd 2+ + 2I + S
−
− − 2−
(b)4 OH + 4 MnO 4 −→ 4 MnO 4 + O 2 + 2H 2 O
(c) 2 H + 2I + 2 HNO 2 −→ 2NO + 2H 2 O + I 2 (HNO 2 is weak)
+
−
(d)Al 3+ + 4OH −→ Al(OH) 4 −
−