H 2 N C COOH H R The general formula for an amino acid 40
Abnormal shape of red blood cells in sickle cell anemia. In sickle cell anemia, the hemoglobin has two amino acids with neutral R groups instead of charged groups. The abnormal hemoglobin causes the red blood cells to have a sickle shape, as seen here. Several amino acids have charged R groups in addition to the NH 2 and COOH group. These are essential to the normal structure of many proteins. 41
Equilibria of Slightly Soluble Ionic Compounds Any “insoluble” ionic compound is actually slightly soluble in aqueous solution. We assume that the very small amount of such a compound that dissolves will dissociate completely. For a slightly soluble ionic compound in water, equilibrium exists between solid solute and aqueous ions . PbF 2 ( s ) Pb 2+ ( aq ) + 2F - ( aq ) Q c = [Pb 2+ ][F - ] 2 [PbF 2 ] Q sp = Q c [PbF 2 ] = [Pb 2+ ][F - ] 2 42
Q sp and K sp Q sp is called the ion-product expression for a slightly soluble ionic compound. For any slightly soluble compound M p X q , which consists of ions M n + and X z - , Q sp = [M n + ] p [X z - ] q When the solution is saturated, the system is at equilibrium, and Q sp = K sp , the solubility product constant . The K sp value of a salt indicates how far the dissolution proceeds at equilibrium (saturation). 43
Metal Sulfides Metal sulfides behave differently from most other slightly soluble ionic compounds, since the S 2- ion is strongly basic. We can think of the dissolution of a metal sulfide as a two-step process: MnS( s ) Mn 2+ ( aq ) + S 2- ( aq ) S 2- ( aq ) + H 2 O( l ) → HS - ( aq ) + OH - ( aq ) MnS( s ) + H 2 O( l ) Mn 2+ ( aq ) + HS - ( aq ) + OH - ( aq ) K sp = [Mn 2+ ][HS - ][OH - ] 44
Example Writing Ion-Product Expressions SOLUTION: PROBLEM: Write the ion-product expression at equilibrium for each compound: (a) magnesium carbonate (b) iron(II) hydroxide (c) calcium phosphate (d) silver sulfide (a) MgCO 3 ( s ) Mg 2+ ( aq ) + CO 3 2- ( aq ) K sp = [Mg 2+ ][CO 3 2- ] (b) Fe(OH) 2 ( s ) Fe 2+ ( aq ) + 2OH - ( aq ) K sp = [Fe 2+ ][OH - ] 2 (c) Ca 3 (PO 4 ) 2 ( s ) 3Ca 2+ ( aq ) + 2PO 4 3- ( aq ) K sp = [Ca 2+ ] 3 [PO 4 3- ] 2 (d) Ag 2 S( s ) 2Ag + ( aq ) + S 2- ( aq ) S 2- ( aq ) + H 2 O( l ) → HS - ( aq ) + OH - ( aq ) K sp = [Ag + ] 2 [HS - ][OH - ] Ag 2 S( s ) + H 2 O( l ) 2Ag + ( aq ) + HS - ( aq ) + OH - ( aq ) 45
Solubility-Product Constants ( K sp ) of Selected Ionic Compounds at 25°C Name, Formula K sp Aluminum hydroxide, Al(OH) 3 Cobalt(II) carbonate, CoCO 3 Iron(II) hydroxide, Fe(OH) 2 Lead(II) fluoride, PbF 2 Lead(II) sulfate, PbSO 4 Silver sulfide, Ag 2 S Zinc iodate, Zn(IO 3 ) 2 3x10 -34 1.0x10 -10 4.1x10 -15 3.6x10 -8 1.6x10 -8 4.7x10 -29 8x10 -48 Mercury(I) iodide, Hg 2 I 2 3.9x10 -6 46
Example Determining K sp from Solubility PROBLEM: (a) Lead(II) sulfate (PbSO 4 ) is a key component in lead- acid car batteries. Its solubility in water at 25°C is 4.25x10 -3 g/100 mL solution. What is the K sp of PbSO 4 ? (b) When lead(II) fluoride (PbF 2 ) is shaken with pure water at 25°C, the solubility is found to be 0.64 g/L. Calculate the K sp of PbF 2 . K sp = [Pb 2+ ][SO 4 2- ] K sp = [Pb 2+ ][SO 4 2- ] = (1.40x10 -4 ) 2 SOLUTION: = 1.96x10 -8 (a) PbSO 4 ( s ) Pb 2+ ( aq ) + SO 4 2- ( aq ) Converting from g/mL to mol/L: 4.25x10 -3 g PbSO 4 100 mL soln x 1000 mL 1 L x 1 mol PbSO 4 303.3 g PbSO 4 = 1.40x10 -4 M PbSO 4 Each mol of PbSO 4 produces 1 mol of Pb 2+ and 1 mol of SO 4 2- , so [Pb 2+ ] = [SO 4 2- ] = 1.40x10 -4 M 47
K sp = [Pb 2+ ][F - ] 2 K sp = [Pb 2+ ][F - ] 2 = (2.6x10 -3 )(5.2x10 -3 ) 2 = 7.0x10
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