A weak acid Propionic Acid CH CH COOH ₃ ₂ abbreviated as HPr is added to water

# A weak acid propionic acid ch ch cooh ₃ ₂

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A weak acid (Propionic Acid ( CH CH COOH ), abbreviated as HPr ) is added to water to produce a HPr final concentration of 10 -3 M. Calculate the pH of the solution from mass balance and charge equations. Assume ideal solution. Cubic equation in 1 unknown {H + }. Can be solved by trial error. The correct solution is the one that yields the positive value ( {H + }=10 -3.963 and pH = 3.963 ) Using the spreadsheet approach and the mass balance and charge balance equations, determine the pH of a solution obtained when a weak acid, acetic acid (H 3 C-COOH or HAc), is added to water to a final concentration of 10 -2 M – Assume ideal solution. Just replace HPr by Hac, and Pr by AC

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Non-ideal solutions 1.0 Example Practice Problem-2 : Compare the pH and the concentrations of HF and F- in a solution prepared by adding 10 -3 M HF to pure water, with that of a solution prepared the same way, except using water containing 10 - 1 M NaCl. Assume that the ionic strength is low enough in the NaCl-free solution that the solution behaves ideally. To estimate the activity coefficients in the solution containing NaCl, use Davies’ equation . ) 3 . 0 1 ( log 2 I I I Az i i
NaCl-water : non-ideal solution Type-a : H, OH - Type-b : HF, F - Type-c : Na+, Cl- Type-d : N/A Type-e : H+ OH- F- and HF HF ↔ H + + F - H 2 O ↔ H + + OH - Ionic strength, I=0.5 S m i *z i 2 & H+ = OH- = F- = 0.78 (using Davies Eq.) Equilibrium constant Equations Mass balance Eq : TOTF = 10 -3 M = [HF] + [F - ] TOTNa = 10 -1 M = [Na + ] TOTCl = 10 -1 M = [Cl - ] Charge balance Eq : [H + ] + [Na + ] = [OH - ] + [F - ] + [Cl - ] Pure Water : ideal solution =1 HF ↔ H + + F - K a = 10 -3.17 H 2 O ↔ H + + OH - K w = 10 -14.0 Type-a: H, OH- Type-b: HF, F- Mass balance: TOTF = 10 -3 M =[HF] + [F - ] Charge balance: [H + ] = [OH - ] + [F - ] Species Conc. Activity H + 5.49*10 -4 5.49*10 -4 OH - 1.82*10 -11 1.82*10 -11 HF 4.51*10 -4 4.51*10 -4 F - 5.49*10 -4 5.49*10 -4 pH =3.26 Species Conc. Activity H + 6.39*10 -4 4.98*10 -4 OH - 2.62*10 -11 2.04*10 -11 HF 3.61*10 -4 3.61*10 -4 F - 6.39*10 -4 4.98*10 -4 pH =3.30

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Example diprotic acid : H 2 S Diprotic Ionization Fractions = 1.0 a a  a c c c 2 a1 a1 a2 0 1 2 2 c c c a1 a1 a2 [H ] K K K [H ] D D D D [H ] [H ] K K K
Consider the mono-protic ( n =1 proton ) acid HA HA H + + A - And if H+ = A- = 1.0 ; MASS BALANCE (MB) equation of species A in the aqueous solution of HA is: TOT A = [ HA ] + [ A - ] The above equations can be manipulated to establish the relationships between a fractions and both H + and K a . Note: Mass balances relate concentrations Note: For solution with ideal behavior, K a can relate concentrations as {a i }=[a i ] Note: K a relates activities and not concentrations   a HA H A H A K HA K a [H + ][A ] [HA]

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STEP-2 : Adding an expression with a value of 1.0 to each side of the above equation: STEP-3 : Inverting both sides of the above equation yields expressions that relate a to {H + } and K a and { ?
• Fall '08
• pH, kA, propionic acid

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