Analytical Chemistry
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1 Analytical Chemistry Lecture No. 4 Date :16 /12/ 2012 Dr. Mohammed Hamed Buffers solutions: - buffers are solutions that resist changing ph when small amounts of acid or base are added they resist changing ph by neutralizing added acid or base The components of a buffer solution :- It is a mixture of weak acid and its salt:- CH 3 COOH is weak acid, CH 3 COONa is Salt of weak acid The Mechanisms of buffer solution to remain stable: - If base is added (OH - ): It will be buffered by the following reaction CH 3 COOH + OH CH 3 COO - + H 2 O So the ph will not change. the weak acid present in the buffer mixture can neutralize added base. - If acid (H + ) is added: it will be buffered by the following reaction, CH 3 COO - + H CH 3 COOH The ph will not alter significantly because the CH 3 COOH formed is a weak acid. Two important characteristics of a buffer include: 1. ph Range The ph range is the range of ph values over which a buffer system works effectively. The ph range of an acid or base can be seen as the region of little ph change vs. volume of base/acid added on a titration curve It is best to choose an acid with a pka close to the desired ph. 2. Buffering Capacity The buffering capacity of a buffered solution represents the amount of acid or base that can be absorbed by the solution without a significant change in ph. A buffer with a large capacity contains large concentrations of buffering components. Therefore the capacity of a buffered solution is determined by the magnitudes of [HA] and [A - ].
2 Many organic acids are weak acids. Weak acids do not completely dissociate in water. The dissociation of an organic acid is described by the following reaction: AH H + + A - weak acid conjugate base CH 3 COOH CH3COO - + H + weak acid conjugate base H 2 PO H 3 PO 4 + H + weak base conjugate acid One can analyze the strength of a weak acid. This means that the amount of hydrogen ion released can be determined. To do this, one can use the following expression: Ka=[H + ] [A - ] / [HA] where K a is defined as the acid dissociation constant. The larger the value of K a, the stronger the acid is. Because K a values vary over a wide range, they are usually expressed using a logarithmic scale: pka = - log Ka The hydrogen ion is one of the most important ions in biological systems. The concentration of this ion affects most cellular processes. For example, the structure and function of most biological macromolecules and the rates of most biochemical reactions are strongly affected by [H + ]. The ph scale has been devised as a convenient method of expressing hydrogen ion concentration. ph has been defined as the negative logarithm of the hydrogen ion concentration: ph = - log [H + ] The Henderson-Hasselbalch equation provides a convenient way to think about buffers and ph: ph = pka + log[a-] / [HA] If one were examining the dissociation of acetic acid, the Henderson- Hasselbalch relationship, then, would be: ph = pka + log[ch3coo - ] / [CH3COOH]
3 The Henderson-Hasselbalch equation can be used to determine if an aqueous solution of a conjugate acid/base pair is functioning as a buffer. If the concentration of the weak acid is equal to that of its conjugate base, the ratio of these two components is one. When this is the case, the Henderson-Hasselbalch equation reduces to ph = pka because the log(1) is equal to zero. When the ph of the solution is equal to the pka of the ionizing group, the solution is functioning at maximum buffering capacity (best buffer). An aqueous solution of a conjugate acid/base pair functions as a good buffer when the ratio of the conjugate base to weak acid ranges from 1:9 to 9:1. Substituting these ratios into the Henderson-Hasselbalch equation, one finds that this aqueous solution functions as a good buffer when the ph of the solution is within approximately one ph unit of the ionizing group s pka. ph = pka + 1 because the log (1/9) is and the log of (9/1) is Using the Henderson-Hasselbalch Equation Example :- How would you prepare 10mL of a 0.01M phosphate buffer, ph 7.40, from stock solutions of 0.10M KH 2 PO 4 and 0.25M K 2 HPO 4? pka of KH 2 PO 4 = The following approach may be helpful in solving this type of buffer problem in which both the conjugate acid and base are added separately. Please note that the numbers are not rounded off until the very end and are rounded based on the limits of the pipets, cylinders, etc. that are required to accurately measure the calculated volumes. 1. Use the Henderson Hasselbalch equation to find the ratio of A - to HA. ph = pka + log [A - ] / [HA] 7.40 = log [A - ] / [HA] 0.20 = log [A - ] / [HA] = [A - ] / [HA]* *Since [A - ] / [HA] = , we can say that [A - ] / [HA] = / 1. In this case [A - ] = ; [HA] = Calculate the decimal fraction (part/whole) of each buffer component.
4 A - = / ( ) = / = HA = / = Find the molarity (M) of each component in the buffer by simply multiplying the molarity of the buffer by the decimal fraction of each component. M A - = 0.01M x = M M HA = 0.01M x = M 4. Calculate the moles of each component in the buffer. Moles = Molarity x Liters of buffer moles A - = M x 0.01L = x 10-5 moles moles HA = M x 0.01L = x 10-5 moles 5. Calculate the volume of each stock solution required to make the buffer Liters of stock = moles of the buffer component / Molarity of the stock L A - = x 10-5 moles / 0.25 M = x 10-4 L = 245μL L HA = x 10-5 moles / 0.10 M = x 10-4 L = 387μL 6. To prepare this buffer, one would use appropriately-sized pipets to measure and transfer each component to a 10mL volumetric flask and bring the solution to volume with dh 2 O. Hydrolysis of Salt Salt : an ionic compound, the product of an acid base neutralization reaction: acid + base water + a salt HCl + NaOH H 2 O + NaCl One way to predict the behavior of a salt in water. A salt may be thought of as the product of a reaction between an acid and a base. 1. When the parent acid and base are strong, a water solution of the salt is neutral. 2. When a salt is formed from the reaction between a weak acid and a strong base, a water solution of that salt will be basic. 3. When a salt is formed from a strong acid and a weak base, a water solution of the salt will be acidic.
5 4. When a salt is formed from a weak acid and a weak base, a water solution of the salt could be acidic, basic or neutral depending on the nature of the ions. 1-Hydrolysis of the Salt of Strong Acid and Strong Base The strength of the acid is determined by how far the equilibrium lies to the right. Qualitatively, this may be judged by the K a of the acid. A large K a indicates a strong acid; a small K a indicates a weak acid. Strong acids, such as HCl, have K a values in the vicinity of infinity. This implies that the dissociation of HCl is virtually complete, and the equilibrium lies completely to the right, therefore, the concentration of the acid equals the concentration of hydronium ions produced. strong bases, such as NaOH, will dissociate completely. The concentration of OH - in solution will be equal to the concentration of the strong base. A typical strong acid problem might be: What is the ph of a M HCl solution? Since HCl is a strong acid, the hydronium ion concentration will be equal to the HCl concentration: [H 3 O + ] = M The ph can be found by taking the negative log of the hydronium ion concentration: ph = -log[h 3 O + ] = -log(0.010) = 2.00 A typical strong base problem might be: What is the ph of a M NaOH solution? Since NaOH is a strong base, the hydroxide ion concentration will be equal to the NaOH concentration: [OH - ] = M The ph can be found by first finding the poh by taking the negative log of the hydroxide ion concentration, and then converting the ph to poh. To find the poh: poh = -log[oh - ] = -log(0.010) = 2.00 The ph can then be calculated from the equation ph + poh = 14: ph = = 12.00
6 2- Hydrolysis of the Salt of Weak Acid and Strong Base This salt will hydrolyze to form a basic solution. The salt sodium acetate will ionize into sodium ions and acetate ions. The sodium ion does not undergo hydrolysis but the acetate ion will react with free hydrogen ions forming acetic acid. Na + C 2 H 3 O HOH <====> HC 2 H 3 O 2 + Na + OH This reaction causes a shift in the [H ]/[OH ] by removing [H ]. This salt is classified as a basic salt because the acetate ion, by definition, is a weak Bronsted-Lowry base (proton acceptor). The solution of such a salt is basic in nature. The anion of the salt is reactive. It reacts with water to form a weak acid and OH - ions. A - + H 2 O; HA + OH - Weak acid for example, the salt CH 3 COONa. It ionises in water completely to give CH 3 COO - and Na + ions. CH 3 COO - ions react with water to form a weak acid, CH 3 COOH and OH - ions. CH 3 COO - + H 2 O CH 3 COOH + OH - C(1-x) Cx Cx Thus, OH - ion concentration increases, the solution becomes alkaline. Applying law of mass action, K h = [CH 3 COOH][OH - ]/[CH 3 CO - ] = (Cx Cx)/C(1-x) = (Cx 2 )/(1-x) )... (i) Other equations present in the solution are: CH 3 COOH CH 3 COO - + H +, K a = [CH 3 COO - ][H + ]/[CH 3 COOH]... (ii) H 2 O H + + OH -, K w = [H + ][OH - ]... (iii) From eqs. (ii) and (iii), log [OH - ] = log K w - log K a + log[salt]/[acid] -poh = -pk w + pk a + log[salt]/[acid]
7 pk w - poh = pk a + log[salt]/[acid] ph = pk a + log[salt]/[acid] Considering eq. (i) again, K h = cx 2 /(1-x) or K h = Ch 2 /(1-h) When h is very small, (1-h) 1 or h 2 = K h /C or h = K h /C [OH - ] = h C = (CK h ) = (C*K w /K a ) [H + ] = K w /[OH - ] = K w / (C*K w /K a ) = (K a *K w )/K c -log [H + ] = -1/2log K w - 1/2log K a + 1/2log C ph = 1/2pK w + 1/2pK a + 1/2log C = 7 + 1/2pK a + 1/2log C. 3- Hydrolysis of the Salt of a Strong Acid and a Weak Base The solution of such a salt is acidic in nature. The cation of the salt which has come from weak base is reactive. It reacts with water to form a weak base and H + ions. B + + H 2 O BOH + H + Weak base Consider, for example, NH 4 Cl. It ionises in water completely into NH 4 and Cl ions. ions react with water to form a weak base (NH 4 OH) and H + ions. NH H 2 O NH 4 OH + H + C(1-x) Cx Cx Thus, hydrogen ion concentration increases and the solution becomes acidic. Applying law of mass action,
8 K h = [H x ][NH 4 OH]/[NH 4 + ]=(Cx.Cx)/C(1-x) = (x 2 C)/((1-x))... (i) where C is the concentration of salt and x the degree of hydrolysis. Other equilibria which exist in solution are NH 4 OH NH OH -, K b = [NH + 4][OH - ]/[NH 4 Oh]... (ii) H 2 O H + + OH -, K b = [H + ][H - ]... (iii) From eqs. (II) and (iii) K w /K b =[H + ][NH 4 OH]/[NH 4 + ] =K h... (iv) [H + ] = [H + ][NH 4 + ]/[NH 4 OH] = K w /K b [NH 4 + ]/[NH 4 OH] log [H + ] = log K w - log K b + log[salt]/[base] -ph = -pk w + pk b + log[salt]/[base] pk w - ph = pk b + log[salt]/[base] poh = pk b + log[salt]/[base] Relation between Hydrolysis constant and Degree of hydrolysis The extent to which hydrolysis proceeds is expressed as the degree of hydrolysis and is defined as the fraction of one mole of the salt that is hydrolysed when the equilibrium has been attained. It is generally expressed as h or x. h = (Amount of salt hydrolysed)/(total salt taken) Considering again eq. (i), K h = x 2 C/(1-x) or K h = h 2 C/(1-h) When h is very small (1-h) 1, H 2 = K h 1/c or h = (K h /C) = (K w /K b * C) [H + ] = h C = (C*K h )/K b
9 log [H + ] = 1/2 log K w + 1 1/2log C - 1/2log K b ph = 1/2pK w - 1/2 log C - 1/2 pk b = 7-1/2 pk b - 1/2log C 4-Hydrolysis of the Salt of Weak Acid and a Weak Base The ions of this type of salt will hydrolyze to form a solution that may be acidic, basic, or neutral, depending on the strength (ionization constant) of the weak acid and weak base that formed the salt. If the ionization constant of the weak acid is greater Maximum hydrolysis occurs in the case of such a salt as both the cation and anion are reactive and react with water to produce H + and OFT ions. The solution is generally neutral but it can be either slightly acidic or slightly alkaline if both the reactions take place with slightly different rates. Consider, for example, the salt CH 3 COONH 4. It gives CH 3 COO - and ions in solution. Both react with water. Other equilibria which exist in solution are: CH 3 COOH CH 3 COO - + H +, K a = [CH 3 COO - ][H + ]/[CH 3 COOH]... (i) NH 4 OH NH OH -, Kb = [NH + 4] [OH - ]/[NH 4 OH]... (ii) H 2 O H + + OH -, K w = [H + ][OH - ]... (iii) From Eqs. (i), (ii) and (iii), K h = Kw/K a.k b = [CH 3 COOH][NH 4 OH]/[CH 3 COO - ][NH + 4]... (iv) Let C be the concentration and h be the degree of hydrolysis
10 K h = h 2 /(1-h) 2 When h is small, (1-h) 1. K h = h 2 h = Kh = K w /K a* K b [H + ] K a h = K a K w /K a* K b = K w * K a /K b -log [H + ] = -1/2log K a - 1/2log K w + 1/2log K b ph = 1/2pK a + 1/2pK w - 1/2pK b = 7 + 1/2pK a - 1/2pK b When pk a = pk b, ph = 7, i.e., solution will be neutral in nature. When pk a > pk b. The solution will be alkaline as the acid will be slightly weaker than base and ph value will be more than 7. In case pk a < pk b, the solution will be acidic as the acid is relatively stronger than base and ph will be less than 7.
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