H 2 N C COOH H R The general formula for an amino acid 40 Abnormal shape of red

H 2 n c cooh h r the general formula for an amino

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H 2 N C COOH H R The general formula for an amino acid 40
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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
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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
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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
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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
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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
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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
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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
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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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