Summary of Acid, Base, Salt, Buffer and Titration Calculations

Transcription

1 Chem 1721/1821 Summary of Acid, Base, Salt, Buffer and Titration Calculations These descriptions are to serve as a guideline to set up the various problems we have discussed. These guidelines apply only to the fundamental and most straight-forward problems; obviously there are countless possible variations to these calculations and there is no way I could describe every possible problem. The goal here is to get you started and help you organize the primary solution reactions, equilibria, equilibrium constants, and where relationships exist between them. strong acids, HCl, HBr, HI, HNO 3, HClO 4 and H 2 SO 4 completely dissociated therefore [H 3 O + ] = [acid] 0 no equilibrium calculation or K a value is required weak monoprotic acids, HA any monoprotic acid that is NOT HCl, HBr, HI, HNO 3 or HClO 4 the reaction in solution that produces H 3 O + and therefore controls the ph is the acid ionization equlibrium: HA + H 2 O A + H 3 O + K a = [H 3 O + ][A ]/[HA] pk a = logk a typcially, in an equilibrium calculation, x will equal [H 3 O + ] and [A ] and can be used to determine [HA] and ph A is the conjugate base of HA; the weaker the acid the stronger its conjugate base HA/A is a conjugate acid/base pair polyprotic acids, H 2 A (diprotic) or H 3 A (triprotic) more than one acidic H + 2 or 3 dissociation steps; 1 H + donated per step each step is an acid dissociation equilibrium, produces H 3 O +, and has a K a value: H 2 A + H 2 O HA + H 3 O + K a1 = [H 3 O + ][HA ]/[H 2 A] pk a1 = logk a1 HA + H 2 O A 2 + H 3 O + K a2 = [H 3 O + ][A 2 ]/[HA ] pk a2 = logk a2 note: the initial concentrations of H 3 O + and HA in the 2 nd dissociation step are determined in the 1 st dissociation step typically, in an equilibrium calculation based on the 1 st dissociation step, x will equal [H 3 O + ] and [HA ] (carried into the 2 nd dissociation step) and can be used to determine [H 2 A] and ph typically, in an equilibrium calculation based on the 2 nd dissociation step, x will equal [A 2 ] for a polyprotic acid, the acid becomes progressively weaker with each dissociation step; HA is a weaker acid than H 2 A; K a2 < K a1 strong bases, discussed in 2 groups: LiOH, NaOH, KOH, RbOH, CsOH completely dissociated in solution therefore [OH ] = [MOH] 0 Mg(OH) 2, Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 completely dissociated in solution therefore [OH ] = 2 [M(OH) 2 ] 0 no equilibrium calculation or K b value is required weak bases, B typically organic compounds that contain N the reaction in solution that produces OH and therefore controls the ph is the base ionization equilibrium: B + H 2 O BH + + OH K b = [BH + ][OH ]/[B] pk b = logk b

2 typically, in an equilibrium calculation, x will equal [OH ] and [BH + ] and can be used to determine [B], poh, ph BH + is the conjugate acid of B; the weaker the base the stronger its conjugate acid B/BH + is a conjugate acid/base pair salts, MA or BHX cations can be neutral or acidic neutral cations are the conjugate acids or the strong bases: Li +, Na +, K +, Rb +, Cs +, Mg 2+, Ca 2+, Sr 2+ and Ba 2+ acidic cations are of 2 groups: 1. conjugate acid of a weak base (B), BH + B and BH + are a conjugate acid/base pair the reaction in a solution containing BH + that produces H 3 O + and controls the ph is an acid ionization equilibrium: BH + + H 2 O B + H 3 O + K a = K w /K b of B = [B][H 3 O + ]/[BH + ] pk a = 14 pk b typcially, in an equilibrium calculation, x will equal [H 3 O + ] and [B] and can be used to determine [BH + ] and ph also, typcially, the euqilibrium constant you will be given or look up will be K b (or pk b ) of the base, B you will probably need to calculate K a or pk a of BH + : K a = K w /K b for B; OR pk a = 14 pk b 2. small, highly charged metal ion in aqueous solution, M(H 2 O) z n+ the reaction in solution containing M(H 2 O) n+ z that produces H 3 O + and controls the ph is an acid ionization equilibrium (z is an integer, z may = 2 8): M(H 2 O) z n+ + H 2 O M(OH)(H 2 O) z 1 (n 1)+ + H 3 O + K a = [M(OH)(H 2 O) z 1 (n )+ ][H 3 O + ]/[ M(H 2 O) z n+ ] typcially, in an equilibrium calculation, x will equal [H 3 O + ] and can be used to determine ph hydrated metal ions, M(H 2 O) z n+, have K a values that will be given or tablulated anions can be neutral or basic (*see exception below) neutral anions are the conjugate bases or strong monoprotic acids: Cl, Br, I, NO 3 and ClO 4 a basic anion is the conjugate base of a weak acid (HA), A HA and A are a conjugate acid/base pair the reaction in solution containing A that produces OH and controls the ph is a base ionization equilbrium: A + H 2 O HA + OH K b = K w /K a of HA = [HA][OH ]/[A ] pk a = 14 pk b typcially, in an equilibrium calculation, x will equal [OH ] and [HA] and can be used to determine [A ], poh, ph also, typcially, the euqilibrium constant you will be given or look up will be K a (or pk a ) of the base, HA you will probably need to calculate K b or pk b of A : K b = K w /K a for HA; OR pk b = 14 pk a *the exception here is that an anion that is the conjugate base of a polyprotic acid is acidic HA + H 2 O A 2 + H 3 O + K a2 = [A 2 ][H 3 O + ]/[HA ] common ion solutions: HA/A, OR B/BH + common ion solutions contain both a weak acid and its conjugate base (HA/A ) OR a weak base and its conjugate acid (B/BH + ) the common ion is the conjugate (A or BH + ) added to a solution of the weak acid or base (HA or B) in the form of a salt; i.e. MA or BHX; M n+ or X n are trypically spectator ions and DO NOT CONTRIBUTE to the acidic or basic properties of the solution the reaction in solution that controls the ph is either the acid ionization equilbirum (for HA/A solution), or the base ionization equilibrium (for B/BH + solution)

3 in a solution containing both HA and A, the acid ionization equilibrium of HA produces H 3 O + and controls the ph: HA + H 2 O A + H 3 O + ; use K a of HA typically initial [HA] and [A ] will NOT be zero; initial [H 3 O + ] will be 0 M typically, in an equilibrium calculation, x will equal [H 3 O + ] and can be used to determine [HA], [A ] and ph in a solution containing both B and BH +, the base ionization equilibrium of B produces OH and controls the ph: B + H 2 O BH + + OH ; use K b of B typically initial [B] and [BH + ] will NOT be zero; initial [OH ] will be 0 M typically, in an equilibrium calculation, x will equal [OH ] and can be used to determine [B], [BH + ], poh and ph buffer solutions buffer solutions contain an acid and a base; typically this is a conjugate acid/base pair (HA/A OR B/BH + ) the reaction in solution that produces H 3 O + or OH and controls the ph, is EITHER the acid ionization equlibrium (for an HA/A buffer), OR the base ionization equilibrium (for a B/BH + buffer) determine the ph of a buffer solution as described above for a common ion solution: HA + H 2 O A + H 3 O + ; use K a of HA typically initial [HA] and [A ] will NOT be zero; initial [H 3 O + ] will be 0 M typically, in an equilibrium calculation, x will equal [H 3 O + ] and can be used to determine [HA], [A ] and ph B + H 2 O BH + + OH ; use K b of B typically initial [B] and [BH + ] will NOT be zero; initial [OH ] will be 0 M typically, in an equilibrium calculation, x will equal [OH ] and can be used to determine [B], [BH + ], poh and ph the ph of a buffer solution is dependent on the pk a of the acid and the relative concentrations of the acid and base components in the solution; given by the Henderson-Hasselbalch equation: ph = pk a + log ([base]/[acid]) where: in an HA/A buffer solution the base is A and the acid is HA in a B/BH + buffer solution the base is B and the acid is BH + pk a = logk a of the acid component in the solution (either HA or BH + ); remember that for BH +, pk a of BH + = 14 pk b of B a buffer solution has a job... to maintain a relatively constant ph when small amount of strong acid (H + ) of strong base (OH ) are added addition of H + or OH will change the relative concentrations of base and acid to calculate change in ph upon addition of H + or OH 1. write the net ionic equation for the neutralization reaction that occurs 2. calculate # mol of all reactants and product before and after reaction; mol = volume [ ] 3. calculate the concentrations of the acid and base (in the buffer sol n) after the reaction is complete; [ ] = mol total sol n volume 4. plug the [base] and [acid] into the Henderson-Hasselbalch equation; calculate ph for an HA/A buffer solution: neutralization reaction: when H + is added: H + + A HA then Henderson-Hasselbalch equation: when OH is added: OH + HA A + H 2 O ph = pk a of HA + log([a ]/[HA])

4 for a B/BH + buffer solution: neutralization reaction: when H + is added: B + H + BH + when OH is added: BH + + OH B + H 2 O then Henderson-Hasselbalch equation: ph = pk a of BH + + log([b]/[bh + ]) strong acid + strong base titrations the concentration of the excess reactant will determine the ph of the resulting solution to determine the ph: 1. write the net ionic equation for the neutralization reaction; H + + OH H 2 O 3. calculate concentration of excess reactant after reaction is complete; [ ] = mol total sol n volume 4. calculate ph or poh based on [H + ] or [OH ] at the stoichiometric point, ph = 7.00 weak acid + strong base titrations calculation of ph will be different at different points in the titration to determine the ph of the solution before the stoichiometric point: 1. write the net ionic equation for the neutralization reaction; HA + OH A + H 2 O 3. calculate concentration of excess HA remaining and A formed after reaction is complete; [ ] = mol total sol n volume 4. after neutralization the solution is an HA/A buffer solution; calculate ph using the Henderson-Hasselbalch equation: ph = pk a of HA + log([a ]/[HA]) 5. before the stoichiometric point, but AT THE HALF-WAY POINT... ph = pk a of HA because [HA] = [A ] to determine the ph of the solution at the stoichiometric point: 1. write the net ionic equation for the neutralization reaction; HA + OH A + H 2 O 2. at the stoichiometric point mol HA = mol OH 3. calculate concentration of A in the solution after reaction is complete; [ ] = mol total sol n volume 4. A is the conjugate base of HA, therefore the ph of the solution will be determined by the OH produced by the base ionization equilibrium of A ; A + H 2 O HA + OH use K b for A ; K b for A = K w /K a for HA, or pk b for A = 14 pk a for HA typically, in an equilibrium calculation, x will equal [OH ] and [HA] and can be used to determine [A ], poh and ph of the solution 5. at the stoichiometric point, ph > 7.00 to determine the ph of the solution beyond the stoichiometric point: 1. write the net ionic equation for the neutralization reaction; HA + OH A + H 2 O 3. calculate concentration of excess reactant (OH ) after reaction is complete; [ ] = mol total sol n volume 4. calculate poh and ph based on [OH ]

5 weak base + strong acid titrations calculation of ph will be different at different points in the titration to determine the ph of the solution before the stoichiometric point: 1. write the net ionic equation for the neutralization reaction; B + H + BH + 3. calculate concentration of excess B remaining and BH + formed after reaction is complete; [ ] = mol total sol n volume 4. after neutralization the solution is a B/BH + buffer solution; calculate ph using the Henderson-Hasselbalch equation: ph = pk a of BH + + log([b]/[bh + ]) 5. before the stoichiometric point, but AT THE HALF-WAY POINT... ph = pk a of BH + because [B] = [BH + ] to determine the ph of the solution at the stoichiometric point: 1. write the net ionic equation for the neutralization reaction; B + H + BH + 2. at the stoichiometric point mol B = mol H + 3. calculate concentration of BH + in the solution after reaction is complete; [ ] = mol total sol n volume 4. BH + is the conjugate acid of B, therefore the ph of the solution will be determined by the H 3 O + produced by the acid ionization equilibrium of BH + ; BH + + H 2 O B + H 3 O + use K a for BH + ; K a for BH + = K w /K b for B, or pk a for BH + = 14 pk b for B typically, in an equilibrium calculation, x will equal [H 3 O + ] and [B] and can be used to determine [BH + ], ph of the solution 5. at the stoichiometric point, ph < 7.00 to determine the ph of the solution beyond the stoichiometric point: 1. write the net ionic equation for the neutralization reaction; B + H + BH + 3. calculate concentration of excess reactant (H + ) after reaction is complete; [ ] = mol total sol n volume 4. calculate ph based on [H + ]