Acid & Bases Lecture PDF

SIJ1002 Fundamentals of Chemistry AP Dr Zazali Alias Acids & Bases for teaching and training ONLY Equilibria  Acids produce positively charged hydrogen ions H+ (protons)  > more acidic a compound is, >>>> concentration of protons it produc...

SIJ1002 Fundamentals of Chemistry AP Dr Zazali Alias Acids & Bases for teaching and training ONLY Equilibria  Acids produce positively charged hydrogen ions H+ (protons)  > more acidic a compound is, >>>> concentration of protons it produces.  In water, protons become strongly attached to a water molecule to give the stable hydronium ion H3O+.  In the Brønsted–Lowry definition:  an acid is a substance that will donate a proton  a base is a substance that will accept a proton  in water, the acid HCl ionizes to produce H3O+ and Cl− ions 𝐇𝐀: 𝐇 ̶ 𝐂𝐥 ⇔ 𝐇𝟑𝐎+ : 𝐂𝐥− base acid conjugate acid of H2O conjugate base of HCl (proton acceptor) (proton donor)  H3O+ is termed the conjugate acid (of the base H2O) and Cl− is termed the conjugate base (of the acid HCl).  cleavage of the H–A bond in an acid HA is brought about by a base, generating the conjugate acid of the base + with the conjugate base of the acid 𝐁: 𝐇 ̶ 𝐀 ⇔ 𝐇 ̶ 𝐁+ : 𝐀− base acid conjugate acid conjugate base  The Lewis definition of acids and bases is rather more general than the Brønsted – Lowry  an acid is an electron-pair acceptor  a base is an electron-pair donor  boron trifluoride, which is able to react with trimethylamine to form a salt. − 𝐌𝟑𝐍: 𝐁𝐅𝟑 ⇔ 𝐌𝟑𝐍 + ̶ 𝑩𝑭𝟑 lewis base lewis acid  Brønsted concept it donates its electrons to a proton electrophile, whereas as a Lewis base it donates its electrons to a Lewis acid electrophile. + 𝐑𝟑𝐍: 𝐇 ⇔ 𝐑𝟑𝐍 + ̶ 𝐇 + 𝐑𝟑𝐍: 𝐄 ⇔ 𝐑𝟑𝐍 + ̶ 𝐄 Acidity and pKa values  For the ionization of the acid HA in water  the equilibrium constant K is given by the formula [A−] [H3O+] 𝑲= [HA][H2O] where [HA] signifies the concentration of HA  because the concentration of water is essentially constant in aqueous solution, a new equilibrium constant Ka is defined as [A−] [H3O+] 𝑲𝒂 = [HA][H2O]  Ka is termed the acidity constant  classify  acids as strong acids (a large value for Ka  a high H3O+ concentration)  weak acids (a small value for Ka and, thus, a low H3O+ concentration)  For example, the strong acid HCl has Ka = 107. However, for weak acids, the amount of ionization is much less and, consequently, the value of Ka is rather small.  Thus, acetic acid CH3CO2H has Ka = 1.76 × 10−5. To avoid using such small numbers as these, Ka is usually expressed in the logarithmic form pKa where pKa =− log10 Ka  Accordingly, the pKa for acetic acid is 4.75: pKa = − log(1.76 × 10−5) = −(−4.75) = 4.75  The pKa for hydrochloric acid can similarly be calculated to be −7 pKa = − log(107 ) = −7  This means there is an inverse relationship between the strength of an acid and pKa:  a strong acid has a large Ka and, thus, a small pKa, i.e. A− is favoured over HA;  a weak acid has a small Ka and, thus, a large pKa, i.e. HA is favoured over A−. OR  the smaller the value of pKa, the stronger is the acid.  the larger the value of pKa, the weaker is the acid. Basicity  a base as a substance that will accept a proton by donating a pair of electrons  we need a system to measure the strength of a base  to measure the strength of a base. Accordingly, a basicity scale based on pKb was developed in a similar way to pKa.  For the ionization of the base B in water K 𝐁 + 𝐇𝟐𝐎 ⇔ 𝐁𝐇 + 𝐇𝐎−  the equilibrium constant K is given by the formula [HO−] [BH+] 𝑲= [B][H2O]  since the concentration of water will be essentially constant, the equilibrium constant Kb and the logarithmic pKb may be defined as [HO−] [BH+] 𝑲𝒃 = with pKb =− log10 Kb [B][H2O]  To measure the strength of a base, we use the pKa of its conjugate acid, i.e. we consider the equilibrium K + 𝐁𝐇 + 𝐇𝟐𝐎 ⇔ 𝐇𝟑𝐎+ 𝑩 For which [B] [H3O+] 𝑲𝒂 = [BH+]  It follows that  a strong base has a small Ka and thus a large pKa, i.e. BH+ is favoured over B;  a weak base has a large Ka and thus a small pKa, i.e. B is favoured over BH+ OR  the larger the value of pKa, the stronger is the base  the smaller the value of pKa, the weaker is the base.  The relationship between pKa and pKb can be deduced as follows: [B] [H3O+] 𝑲𝒃 = [BH+] [A−] [H3O+] 𝑲𝒂 = [HA][H2O] [B] [H3O+] [A−] [H3O+] 𝑲𝒂 X 𝑲𝒂 = X = [H3O+] [OH−] [BH+] [HA][H2O]  Thus, Ka × Kb reduces to the ionization constant for water Kw. K 𝐇𝟐𝐎 + 𝐇𝟐𝐎 ⇔ 𝐇𝟑𝐎+ 𝐇𝐎− one molecule of water is acting as a base and accepts a proton from a second water molecule. This second water molecule, therefore, is acting as an acid and donates a proton. The equilibrium constant K for this reaction is given by the formula [H3O+] [OH−] 𝑲= [H2O][H2O] As concentration of water is essentially constant in aqueous solution, the new equilibrium constant Kw is defined as 𝑲𝒘 = [H3O+] [OH−]  For every hydronium ion produced, a hydroxide anion must also be formed, so that the concentrations of hydronium and hydroxide ions must be equal.  In pure water at 25 ◦ C, this value is found to be 10−7 M 𝑲𝒘 = [H3O+] [OH−] = 10−7 × 10−7 = 10−14 pH  The acidity of an aqueous solution is normally measured in terms of pH. pH is defined as pH = − log10[H3O+]  The lower the pH, the more acidic the solution; the higher the pH, the more basic the solution only applies to aqueous solutions, and is only a measure of the acidity of the solution.  It does not indicate how strong the acid is (that is a function of pKa) and the pH of an acid will change as we alter its concentration.  For instance, dilution will decrease the H3O+ concentration, and thus the pH will increase.  In water, the hydronium ion concentration arises by the self-dissociation equilibrium 𝐇𝟐𝐎 + 𝐇𝟐𝐎 ⇔ 𝐇𝟑𝐎+ + 𝐇𝐎−  one molecule of water is acting as a base, accepting a proton from a second water molecule. The second molecule is acting as an acid and donating a proton  For every hydronium ion produced, a hydroxide anion must also be formed, so that the concentrations of hydronium and hydroxide ions must be equal.  In pure water at 25 ◦ C, this value is found to be 10−7 M.  The equilibrium constant K is given by the formula [H3O+] [OH−] 𝑲= [H2O][H2O]  because the concentration of water is essentially constant in aqueous solution, the new equilibrium constant Kw (the ionization constant for water) is defined as 𝑲𝒘 = [H3O+] [OH−] = = 10−7 × 10−7 = 10−14  This means that the pH of pure water at 25 ◦ C is therefore pH = − log 10−7 = 7  Weak acids are not completely ionized in aqueous solutions, and the amount of ionization, and thus hydronium ion concentration, is governed by the equilibrium 𝐇𝐀 + 𝐇𝟐𝐎 ⇔ 𝐇𝟑𝐎+ + 𝐀−  and the equilibrium constant Ka we defined : [A−] [H3O+] 𝑲𝒂 = [HA]  However, since [H3O+] must be the same as [A−], we can write [H3O+]2 𝑲𝒂 = therefore [H3O+] = 𝑲𝒂[HA] [HA]  If we take negative logarithms of both sides, we get − log [H3O+] = − 1/2log 𝑲𝒂 − 1/2log [HA]  which becomes pH = 1/2p𝑲𝒂 − 1/2log [HA]  The calculation of the pH of a weak base may be approached in the same way. The equilibrium we need to consider is 𝐁 + 𝐇𝟐𝐎 ⇔ 𝐁𝐇+ + 𝐇𝐎−  and the equilibrium constant Kb will be defined as [OH−] [BH+] 𝑲𝒃 = [B]  However, since [H3O+] must be the same as [A−], we can write [HO−]2 𝑲𝒃 = therefore [HO−] = 𝑲𝒃[B] [B]  need to remember that 𝑲𝒘 = [H3O+] [OH−]  so that we can replace [HO−] with Kw/[H3O+]; this leads to 𝑲𝒘 𝑲𝒘 = 𝑲𝒃[B] Hence, [H3O+] = [H3O+] 𝑲𝒃[B]  we now take negative logarithms of both sides, we get − log [H3O+] = − 1/2log 𝑲𝒂 − 1/2log [HA] which becomes pH = pKw − 1/2pKb− 1/2log [B] The Henderson–Hasselbalch equation  Ka for the ionization of an acid HA has been defined as [A−] [H3O+] + 𝑲𝒂[HA] 𝑲𝒂 = can be rearranged to give [H3O ] = [HA] [A−]  Taking negative logarithms of each side, this becomes [A−]] − log [H3O+] = − log 𝑲𝒂 + log [HA]  OR [A−] pH= p𝑲𝒂 + log [HA]  >>>>>>> the Henderson–Hasselbalch equation, written as [base] pH= p𝑲𝒂 + log [acid]  Using this relationship, it is possible to determine the degree of ionization of an acid at a given pH.  An immediate outcome from this expression is that the pKa of an acid is the pH at which it is exactly half dissociated. This follows from [A−] pH= p𝑲𝒂 + log [HA]  But when the concentrations of acid HA and conjugate base A− are equal, then [A−] log = log 1 = 0 [HA]  so that pH= p𝑲𝒂 Buffers  a solution that helps to maintain a reasonably constant pH environment  >>>>>>>> countering the effects of added acids or bases.  The simplest type of buffer is composed of a weak acid–strong base combination or a weak base–strong acid combination.  This may be prepared by combining the weak acid (or base) together with its salt.  maximum efficiency >>>>> pH needs to be within about 1 pH unit either side of the pKa of the weak acid / base  acetic acid is only weakly dissociated, the [acetic acid] will be almost the same as the amount put in the mixture.  On the other hand, the sodium acetate component will be almost completely dissociated, so the acetate ion concentration can be considered the same as that of the sodium acetate used for the solution  Addition of an acid such as HCl to the buffer solution provides H+, which combines with the acetate ion to give acetic acid  This has a two fold effect: it reduces the amount of acetate ion present + also increases the amount of undissociated acetic acid  Provided the amount of acid added is small relative to the original concentration of base in the buffer, the alteration in base: acid ratio in the Henderson–Hasselbalch equation is relatively small and has little effect on the pH value. 𝐇𝐎𝐀𝐜 + 𝐇𝟐𝐎 ⇔ 𝐇𝟑𝐎+ + 𝐀𝐜𝐎− ionization of weak acid 𝐀𝐜𝐎− + 𝐇𝐂𝐥 ⇔ 𝐇𝐎𝐀𝐜 + 𝐂𝐥− addition of acid Base reduced non ionized acid increased  Similar considerations apply if a base such as NaOH is added to the buffer solution.  This will decrease the amount of undissociated acid, and increase the amount of acetate ion present. 𝐇𝐀 + 𝐇𝐎− ⇔ 𝐇𝟐𝐎 + 𝐀𝐜𝐎− addition of base non ionized acid decreased  The sodium acetate–acetic acid combination is one of the most widely used buffers, and is usually referred to simply as acetate buffer.  Other buffer combinations commonly employed in chemistry and biochemistry include  carbonate–bicarbonate (sodium carbonate–sodium hydrogen carbonate) citrate (citric acid–trisodium citrate)  phosphate (sodium dihydrogen phosphate–disodium hydrogen phosphate)  tris [tris(hydroxymethyl)aminomethane–HCl]