Transition Metals

Figure 23.5 Aqueous oxoanions of transition elements. Mn 2+ MnO 4 2− MnO 4 − VO 4 3− Cr 2 O 7 2− MnO 4 − One of the most characteristic chemical properties of these elements is the occurrence of multiple oxidation states.

Figure 23.6 Colors of representative compounds of the Period 4 transition metals. Titanium(IV) oxide sodium chromate potassium ferricyanide nickel( II ) nitrate hexahydrate zinc sulfate heptahydrate scandium oxide vanadyl sulfate dihydrate manganese( II ) chloride tetrahydrate cobalt( II ) chloride hexahydrate copper( II ) sulfate pentahydrate

Sample Problem 19.13Calculating the Concentration of a Complex Ion SOLUTION: PROBLEM:An industrial chemist converts Zn(H 2 O) 4 2+ to the more stable Zn(NH 3 ) 4 2+ by mixing 50.0 L of M Zn(H 2 O) 4 2+ and 25.0 L of 0.15 M NH 3. What is the final [Zn(H 2 O) 4 2+ ]? K f of Zn(NH 3 ) 4 2+ is 7.8x10 8. PLAN:Write the reaction equation and K f expression. Use a reaction table to list various concentrations. Remember that components will be diluted when mixed as you calculate final concentrations. It is obvious that there is a huge excess of NH 3 and therefore it will drive the reaction to completion. Zn(H 2 O) 4 2+ ( aq ) + 4NH 3 ( aq ) Zn(NH 3 ) 4 2+ ( aq ) + 4H 2 O( l ) K f = [Zn(NH 3 ) 4 2+ ] [Zn(H 2 O) 4 2+ ][NH 3 ] 4 [Zn(H 2 O) 4 2+ ] initial = (50.0 L)( M) 75.0 L = 1.3x10 -3 M [NH 3 ] initial =(25.0 L)(0.15 M) 75.0 L = 5.0x10 -2 M

Sample Problem 19.13Calculating the Concentration of a Complex Ion Zn(H 2 O) 4 2+ ( aq ) + 4NH 3 ( aq ) Zn(NH 3 ) 4 2+ ( aq ) + 4H 2 O( l )Concentration (M) Initial Change Equilibrium 1.3x x ~(-1.3x10 -3 )~(-5.2x10 -3 ) Since we assume that all of the Zn(H 2 O) 4 2+ has reacted, it would use 4 times its amount in NH 3. [NH 3 ] used = 4(1.3x10 -3 M) = 5.2x10 -3 M ~(+1.3x10 -3 )- x4.5x x10 -3 [Zn(H 2 O) 4 2+ ] remaining = x(a very small amount) - K f = [Zn(NH 3 ) 4 2+ ] [Zn(H 2 O) 4 2+ ][NH 3 ] 4 = 7.8x10 8 = (1.3x10 -3 ) x(4.5x10 -2 ) 4 x = 4.1x10 -7 M

Sample Problem 19.14Calculating the Effect of Complex-Ion Formation on Solubility SOLUTION: PROBLEM:In black-and-white film developing, excess AgBr is removed from the film negative by “hypo”, an aqueous solution of sodium thiosulfate (Na 2 S 2 O 3 ), through formation of the complex ion Ag(S 2 O 3 ) Calculate the solubility of AgBr in (a) H 2 O; (b) 1.0 M hypo. K f of Ag(S 2 O 3 ) 2 3- is 4.7x10 13 and K sp AgBr is 5.0x PLAN:Write equations for the reactions involved. Use K sp to find S, the molar solubility. Consider the shifts in equilibria upon the addition of the complexing agent. AgBr( s ) Ag + ( aq ) + Br - ( aq ) K sp = [Ag + ][Br - ] = 5.0x S = [AgBr] dissolved = [Ag + ] = [Br - ]K sp = S 2 = 5.0x ; S = 7.1x10 -7 M(a) (b)AgBr( s ) Ag + ( aq ) + Br - ( aq ) Ag + ( aq ) + 2S 2 O 3 2- ( aq ) Ag(S 2 O 3 ) 2 3- ( aq ) AgBr( s ) + 2S 2 O 3 2- ( aq ) Br - ( aq ) + Ag(S 2 O 3 ) 2 3- ( aq )

Sample Problem 19.14Calculating the Effect of Complex-Ion Formation on Solubility K overall = K sp x K f = [Br - ][Ag(S 2 O 3 ) 2 3- ] [S 2 O 3 2- ] 2 = (5.0x )(4.7x10 13 )= 24 AgBr( s ) + 2S 2 O 3 2- ( aq ) Br - ( aq ) + Ag(S 2 O 3 ) 2 3- ( aq )Concentration (M) Initial Change Equilibrium S S 00 +S+S+S+S SS K overall = S2S2 ( S) 2 = 24; S S = (24) 1/2 S = [Ag(S 2 O 3 ) 2 3- ] = 0.45 M