Acid Base Equilibria

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Acid Base Equilibria N Goalby Chemrevise.org ACIDBASE : Arrhenius The most basic of the acidbase concepts is the Arrhenius theory Acids are substances that dissociate in water to produce hydronium ions, H 3 + and bases are substances that dissociate in water to produce hydroxide ions, H 1

Acid Base: BronstedLowry A BronstedLowry acid is defined as a substance that can donate a proton. A BronstedLowry base is defined as a substance that can accept a proton. HCl (g) + H 2 (l) H 3 + (aq) + Cl (aq) acid 1 base 2 acid 2 base 1 Each acid is linked to a conjugate base on the other side of the equation. Calculating ph Indicates the acidity of the solution ph = log [H + ] Where [H + ] is the concentration of hydrogen ions in the solution From the French pouvoir hydrogene ( hydrogen power or power of hydrogen) 2

Calculating ph ph = log[h + (aq)] If [H + (aq)] = 0.01M then ph = log(0.01) =2.00 If [H + (aq)] = 0.000001M then ph = log(0.000001) =6.00 Always give ph to 2d.p. Calculating ph of strong acids Strong acids completely dissociate. For HCl and HN 3 the [H + (aq)] will be the same as the original concentration of the acid. For 0.1M HCl the ph will be log[0.1] =1.00 For strong acids with a concentration >1M the ph will be negative. What is the ph of a solution of a) 0.02M HCl b) 0.05M HN 3 c) 0.001M HBr 3

ph Range 1M H + 0.01M H + 0.000001M H + 1M H 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 Acidic Basic [H [H + ]>[H Neutral [H ] [H + ] = [H [H ]>[H + ] ] = [H ] [H Finding [H + ] from ph [H 3 + ] = 1 x 10 ph n most calculators this is done by pressing Inv (or 2 nd function) log number(ph) What is the concentration of the following acids HCl with a ph of 1.50 HN 3 with a ph of 2.10 4

Ionic Product for water In all aqueous solutions and pure water the following equilibrium occurs 2 H 2 (l) H 3 + (aq) + H (aq) This can be simplified to H 2 (l) H + (aq) + H (aq) This equilibrium has the following equilibrium expression Kc= [H + (aq)][h (aq)] [H 2 (l)] Rearrange to Kc x [H 2 (l)] = [H + (aq) ][H (aq) ] Kw = [H + (aq) ][H (aq) ] Because [H 2 (l)] is much bigger than the concentrations of the ions, we assume its value is constant Kw = [H + (aq) ][H (aq) ] At 25 o C the value of Kw for all aqueous solutions is 1x10 14 mol 2 dm 6 This expression can be used to calculate [H + (aq) ] ions if we know the [H (aq) ] ions In pure water [H + (aq) ] = [H (aq) ] = 1x10 7 moldm 3 In a solution of 0.1M NaH what is the concentration of the H + and H ions? In a strong base there is complete dissociation which means that [H (aq) ] = 0.1M 5

Calculating ph of strong base For bases we generally know the concentration of hydroxide ion. To work out the ph we need to work out [H + (aq)] using the kw expression. Example: What is the ph of the strong base 0.1M NaH Assume complete dissociation. Kw = [H + (aq) ][H (aq) ] = 1x10 14 [H + (aq)] = kw/ [H (aq) ] = 1x10 14 / 0.1 = 1x10 13 M ph = log[1x10 13 ] =13.00 Weak acids Weak acids slightly dissociate when dissolved in water, giving an equilibrium mixture. HA + H 2 (l) H 3 + (aq) + A (aq) We can simplify this to HA (aq) H + (aq) + A (aq) Ka = [H + (aq)][a (aq)] [HA (aq) ] For weak acids such as ethanoic acid the ka is 1.7 x 10 5 mol dm 3. The larger ka the stronger the acid 6

Calculating ph of a weak acid What is the ph of a solution of 0.01M ethanoic acid (ka is 1.7 x 10 5 mol dm 3) Write the expression for ka CH 3 C 2 H (aq) CH 3 C 2 (aq) + H + (aq) [H + (aq)][ch 3 C 2 (aq) ] Ka = [CH 3 C 2 H (aq) ] To make the calculation easier two assumptions are made: 1) [H + (aq)] eqm = [CH 3 C 2 (aq) ] eqm because they have dissociated according to a 1:1 ratio 2) As the amount of dissociation is small we assume that the initial concentration of the ethanoic acid has remained constant. So [CH 3 C 2 H (aq) ] eqm = [CH 3 C 2 H (aq) ] initial Rewrite expression using assumptions Ka = [H + (aq)][h + (aq)] [CH 3 C 2 H (aq) ] = [H + (aq)] 2 [CH 3 C 2 H (aq) ] intial Replace [CH 3 C 2 (aq) ] with [H + (aq)] using assumption 1 7

What is the ph of a solution of 0.01M ethanoic acid (ka is 1.7 x 10 5 mol dm 3) Ka = = [H + (aq)] 2 [CH 3 C 2 H (aq) ] intial 1.7 x 10 5 = [H + (aq)] 2 0.01 [H + (aq)] 2 = 1.7 x 10 5 x 0.01 [H + (aq)] = (1.7 x 10 7 ) = 4.12 x10 4 ph = log [H + ] = log (4.12 x10 4 ) =3.38 pka Sometimes Ka is quoted as pka pka = log (Ka) What is the pka of ethanoic acid (ka 1.7 x10 5 mol dm 3 ) pka = log (1.7 x10 5 ) = 4.75 The larger the value of Ka the smaller the value of pka The stronger the acid the smaller the value of pka 8

Name Equation with conjugate base Ka mol dm 3. Pka Hydrofluoric HF (aq) F (aq) + H + (aq) 5.60 x 10 4 3.25 Ethanoic acid CH 3 C 2 H (aq) CH 3 C 2 1.7 x 10 5 4.75 (aq) + H + (aq) Hydrocyanic acid HCN (aq) CN (aq) + H + (aq) 4.9 x 10 10 9.31 The larger the value of Ka the smaller the value of pka The stronger the acid the smaller the value of pka Why do some acids react with sodium carbonate to give C 2? Ethanoic acid will react with sodium carbonate CH 3 C 2 H (aq) CH 3 C 2 (aq) + H + (aq) The H + ions produced in this reaction force the following equilibria involving carbonate ions to the left (by le Chatelier), producing C 2 H 2 + C 2 H + + HC 3 4.5 x 10 7 mol dm 3. HC 3 H + + C 3 2 4.8 x 10 11 mol dm 3 However if the acid is a weaker acid than carbonic i.e. has a ka < 4.5 x 10 7 mol dm 3 then it will not be able to donate protons to the carbonic acid and no C 2 will be produced. Eg Boric acid has a ka of 5.8 x 10 10 mol dm 3 and does not react with sodium carbonate. H 3 B 3 H + + HB 3 9

Strength of Acids Trichloroethanoic acid 2.3 x10 1 Dichloroethanoic acid 5.0 x 10 2 Chloroethanoic acid 1.3 x 10 3 Benzoic acid 6.3 x10 5 Ethanoic acid 1.7 x 10 5 Carbonic acid 4.5 x 10 7 Boric acid 5.8 x 10 10 Phenol 1.28 x 10 10 Decreasing acid strength Acids below carbonic acid will not give off C 2 with sodium carbonate Strength of Carboxylic acids H 3 C CH 2 C delocalised H Alkyl groups H 3 C CH2 C electron releasing Increasing chain length pushes electron density on to the C ion, making it more negative and less stable. This make the acid less strong. H Cl C H C H Chlorine electron withdrawing delocalised Propanoic acid less acidic than ethanoic acid C H 3 C Electronegative chlorine atoms withdraw electron density from the C ion, making it less negative and more stable. This make the acid more strong. chloroethanoic acid more acidic than ethanoic acid 10