Question

g You observed the formation of several precipitates in the Reactions in Solution lab exercise. Identify the precipitate in each of the following reactions: a. The yellow precipitate formed in the reaction between KI and Pb(NO3)2 is . b. The white precipitate formed in the reaction between BaCl2 and H2SO4 is . c. The brown precipitate formed in the reaction between NaOH and FeCl3 is . d. The blue precipitate formed in the reaction between CuSO4 and NaOH is .

Answer 1

Answer:

For a: Lead iodide is a yellow precipitate.

For b: Barium sulfate is a white precipitate.

For c: Ferric hydroxide is a brown precipitate.

For d: Copper (II) hydroxide is a blue precipitate.

Explanation:

Precipitation reaction is defined as the reaction where a solid precipitate (solid substance) is formed at the end of the reaction. It is insoluble in water.

For the given options:

• For (a):

The chemical reaction between KI and lead (II) nitrate follows:

$2KI(aq)+Pb(NO_3)_2(aq)\rightarrow PbI_2(s)+2KNO_3(aq)$

The iodide of lead is generally insoluble in water. Thus, lead iodide is a yellow precipitate.

• For b:

The chemical reaction between barium chloride and sulfuric acid follows:

$BaCl_2(aq)+H_2SO_4(aq)\rightarrow BaSO_4(s)+2HCl(aq)$

The sulfate of barium is insoluble in water. Thus, barium sulfate is a white precipitate.

• For c:

The chemical reaction between NaOH and ferric chloride follows:

$3NaOH(aq)+FeCl_3(aq)\rightarrow Fe(OH)_3(s)+3NaCl(aq)$

The hydroxide of iron is insoluble in water. Thus, ferric hydroxide is a brown precipitate.

• For d:

The chemical reaction between NaOH and copper sulfate follows:

$CuSO_4+2NaOH\rightarrow Cu(OH)_2+Na_2SO_4$

The hydroxide of copper is insoluble in water. Thus, copper (II) hydroxide is a blue precipitate.

Answer 2

(a) The yellow precipitate formed in the reaction between KI and Pb(NO3)2 would be PbI2 according to the equation:

$Pb(NO_3)_2(aq) + 2KI(aq) ---> PbI2(s) + 2KNO_3(aq)$

(b) The white precipitate formed in the reaction between BaCl2 and H2SO4 would be  BaSO4 according to the equation:

   $BaCl_2 (aq) + H_2SO_4 (aq) ---> BaSO_4 (s) + 2 HCl (aq)$

(c) The brown precipitate formed in the reaction between NaOH and FeCl3 would be Fe(OH)3 according to the equation:

$FeCl_3 (aq) + NaOH (aq) ---> Fe(OH)_3 (s) + NaCl (aq)$

(d) The blue precipitate formed in the reaction between CuSO4 and NaOH would be Cu(OH)2 according to the equation:

$CuSO_4(aq) + 2 NaOH (aq) ---> Cu(OH)_2 (s) + Na_2SO_4 (aq)$

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