Advanced Chemistry Practice Problems

Transcription

1 Aqueous Equilibria: olar Solubility and the Common Ion Effect 1. Question: Which of the following compounds will decrease the solubility of lead(ii) bromide in water? a. Lead(II) nitrate b. Sodium chloride c. Sodium bromide. Answer: To decrease the solubility of PbBr, the following reaction must be shifted to the left. PbBr (s) Pb + (aq) + Br (aq) a. The addition of lead(ii) nitrate will add Pb + ions to the solution and shift the reaction to the left (decreasing the solubility). b. Addition sodium chloride will add Na + and Cl - ions to the solution. Neither are in the above equilibrium and therefore the NaCl will have no effect on the solubility of lead(ii) bromide. c. The addition of sodium bromide will add Na+ and Br - ions to the solution. The Br - ions will shift the equilibrium to the left (decreasing the solubility). 3. Question: Calculate the solubility of lead(ii) bromide, PbBr, in a. 10 Pb(NO 3) b. 10 NaBr c. 10 gbr K sp (PbBr ) = Answer: For each part of this question, the following equilibrium is used: PbBr (s) Pb + (aq) + Br - (aq)

2 a. 10 Pb(NO 3) will dissociate in water to give [Pb + ] = 10, prior to any PbBr dissolving. Use an ICE table to determine solubility. olarity PbBr (s) Pb + (aq) + Br - (aq) I Some 10 0 C -s +s +s E Still some 10+s s The equilibrium expression is: K sp Pb Br ( 10 s)( s) 10( 4s ) 40 s s s olar solubility of PbBr in 10 NaBr is

3 b. 10 NaBr will dissociate in water to give [Br - ] = 10, prior to any PbBr dissolving. Use an ICE table to determine solubility. olarity PbBr (s) Pb + (aq) + Br - (aq) I Some 0 10 C -s +s +s E Still some s 10+s The equilibrium expression is: K sp Pb Br s( 10 s) s( 010) 010 s 4 s olar solubility of PbBr in 10 Pb(NO 3) is

4 c. 10 gbr will dissociate in water to give [Br - ] = 0, prior to any PbBr dissolving. Use an ICE table to determine solubility. olarity PbBr (s) Pb + (aq) + Br (aq) I Some 0 0 C -s +s +s E Still some s 0+s The equilibrium expression is: K sp Pb Br s( 0 s) s( 0) 040 s s olar solubility of PbBr in 10 gbr is

5 Aqueous Equilibria: The Effect of ph on Solubility 1. Question: Which of the following salts will be more soluble in an acid than in water? a. PbCl b. CuS c. BaSO 4 d. AgI Answer: For a salt to be more soluble in an acid than in water, the anion must be a conjugate base of a weak acid. a. The anion of PbCl is Cl -. This is the conjugate base of the strong acid HCl. Therefore it is not more soluble in an acidic solution than in water. b. The anion of PbS is S -. This is the conjugate base of HS -, a weak acid. It is more soluble in acidic solution. c. The anion of BaSO4 is SO4 -. This is the conjugate base of HSO4 -, a weak acid. (HSO4 is strong, however HSO4 - is weak.) This salt is more soluble in acidic solution. d. The anion of AgI is I -. This is the conjugate base of the strong acid HI. Therefore it is not more soluble in an acid solution than in water.

6 Aqueous Equilibria: Precipitation Reaction and Selective Precipitation 1. Question: A solution containing 15.0 ml of 0010 Pb(NO 3) is added to a flask containing 1.50 ml of 0010 NaCl. Determine if a precipitate will form. K sp (PbCl ) = Answer: To determine if a precipitate will form, calculate Q and compare the value to K sp. Q = [Pb + ] o[cl - o A common mistake is to use the 0010 for [Pb + ] o and 0010 for [Cl - ] o. However, the two solutions, when combined, will each dilute the other. Use 1V 1 = V to find the concentration of each ion after the solutions are mixed.. For [Pb + ]: For [Cl - ]: 1V V 1V V ( 15. 0mL) mL (1.50 ml) ml Q = [Pb + ] o[cl - ] o = ( )( ) = K sp = The value is Q < K sp and therefore no precipitation will occur.

7 . A test tube contains a solution that is 10 chloride ions and 10 bromide ions. A solution of silver nitrate is added drop wise until a precipitate is observed. K sp (AgCl) = ; K sp (AgBr) = a. What solid will precipitate first? b. What is the concentration of Ag + required to cause the solid (in part a) to precipitate. Answer: a. The K sp for AgBr is smaller than the value for AgCl. Therefore, AgBr will precipitate first. b. K sp=[ag + ][Br - ] = [Ag + ](10 ) = [Ag + ]