Chapter four. According to this definition, HCl is an acid NaOH is a base as shown below :. HCl H + + Cl - NaOH Na + + OH -

Transcription

1 Chapter four A cids, Bases and buffers Acids base Theories 1-Arrhenius Theory(1884) Acids: is any substance that ionized in water ( partially or completely ) to give hydrogen ions Base: is any substance that ionized in water to give hydroxyl ions. According to this definition, HCl is an acid NaOH is a base as shown below :. HCl H + + Cl - NaOH Na + + OH - Note: acid chemically reacts with a base as follow :. HCl + NaOH Na + + H 2 O + Cl - 2- Bronsted lowry Theory (1923) Acids: is any substance that can donate a proton. Base: is any substance that can accept a proton. 3- Lewis Theory(1923) Acids: is a substance which can accept an electron pair by combining with a second substance with an unshared pair of electrons. A base:. Is a substance which shares it unshared electron pair during chemical reaction, for example BCl 3 + :NH 3 = Cl 3 B:NH 3 acid base accept has pair electron of electron 1

2 Conjugate pairs:. Acids and base pairs in the ionization reaction are called "conjugate pairs". the conjugate base of the acid and acid is the conjugate of the base Acid = H + + base For. e.g H 2 O = H + + OH -, + NH 4 = H + + NH 3 H 2 O : conjugated acid for OH - NH 3 : is conjugated base for (NH + 4 ) + or NH 4 = is conjugated acid for (NH 3 ) base Acid- Base strength: When acid or base is dissolved in water it will dissociate or ionize. the degree of ionization depends on the strength of the acid. A strong: electrolyte is completely dissociated Aweak: electrolyte is partially dissociated weak acid has a relatively small dissociation constant (Ka) where's. strong acid has a large dissociation constant. Note: The strength of an acid depends on its type and is not related to the concentration. e.g : HCl : strong acid regardless of whether its concentration. 1M or 10-4 M Strong base : is a base with relatively large Kb Weak base : is a base with relatively small Kb Strong electrolytes weak electrolytes HCl CH 3 COOH HClO 4 C 6 H 5 OH phenol H 2 SO 4 HCOOH formic acid HNO 3 C 6 H 5 NH 2 aniline NaOH NH 3 CH 3 COONa 2

3 Acid- Base equilibria in water: Kw = [H + ] [OH - ] = ( ionization contant of water) [H + ] [OH - ] = [H + ] = [OH - ] = Conc. of [H + ] in a solution is often expressed as the PH of the solution which is the negative of the logarithim of [H + ] ph = -log [H + ] poh = -log [OH - ] pkw = ph + poh = 14 note: Examples: 1-4 chap.2 : examples of strong acids and strong bases : which are completely ionizedthus H + Or OH - is directly determined Inorder to determine the ph of weak. acids or bases which are partially ionized, different calculations must be followed as shown in the following examples:- The ph (or acidity) scale The concentration of H + and OH - in aqueous solution in vary over a wide ranges from 1M or greater to M or less. The ph of a solution is defined as :- ph = - log [ H + ] [ H + ] = is hydrogen ion concentration in eq/l poh = - log [OH - ] it will be noted that ph and poh = 14 at 25Cᵒ A ph of 7 means neutral solution Acidity is increased with decreasing the ph below 7 increasing the ph above 7 is an indication for increasing alkalinity. 3

4 Examples(1) Calculate the ph of M of hydrochloric acid solution? HCl: strong electrolyte and it is completely ionized [H + ] = M ph = - log ( ) = 3 log 2 = 2.7 Examples(2) Calculate the poh and the ph of a M solution of NaOH? [OH - ] = M poh = - log ( ) = 2- log 5 = 1.30 ph + poh = 14 ph = = 12.7 Examples (3) Calculate the hydrogen ion concentration of a solution with ph 9.67? ph = - log [H + ] = 9.67 [H + ] = = M Examples (4) Calculate the ph of a solution prepared by mixing 2ml of a strong acid solution of ph 3 and 3ml of a strong base of ph 10? [H + ] of acid solution = M Number of moles of [H + ] = M vol = = mmol 4

5 Similarly poh of the base solu. = = 4 [OH] = m M Number of moles of [OH - ] = mmol Excess quality of acid = = mmol [H + ] of resulting solution = = M ph = - log ( ) = 4 log 3.4 = 3.47 Examples(5) Calculate the ph and poh of a solution prepared by dissolving 6 gm of acetic acid in distilled water to make 250 ml of solution? CH 3 COOH = H + + CH 3 COO - Or HOAC = H + + OAC - OAC - = CH 3 COO - Ka = = Initial conc of acetic acid = = 0.4 M Let the equilibrium conc. of [H + ] = [ CH 3 COO - ] = X Let the equilibrium conc. of [CH 3 COOH] = 0.4-X Since Ka is smaller than 1/100 of the initial concentration (i.e less than (0.004), hence 0.4-x =

6 x 2 = x = M = [H + ] = [CH 3 COO - ] ph = - log [H + ] = -log ( ) = = 2.58 poh = 14- ph = = Example (6) The dissociation constant for ammonia (Kb= ) at 25Cᵒ Calculate the ph and poh for a M of ammonia solution? NH 3 + H 2 O = NH OH - Kb = = Since Kb is less than , i.e less than , hence similar simplification of solution can be used i.e = = X 2 = x = [OH - ] poh = -log [OH - ] = = 2.73 ph = = Salt of weak acids and bases:. Sodium acetate (NaOAC), salt of weak acid, astrong electrolyte like almost all salts, and completely ionized. The union of the salt of a weak acid is aprnsted base, which will accept proton HOAC + OH - OAC - + H 2 O Ionization is known as hydrolysis of the salt and the equilibrium constant is called hydrolysis constant (Kh) of the salt which is the same as the basicity constant (Kb) Kh= Kb = 6

7 The value of Kb can be calculated from Ka of acetic acid and Kw by multiplying the numerator and denominator [H + ] Kb = Kb = remember that Ka = and Kw = [H + ][OH - ] Hence Kb = Kw / Ka or Ka Kb = Kw i.e The product of Ka of any weak acid and Kb of its conjugated base is always equal to Kw The ph and poh of such salt (aprnsted base) is calculated in the same manner as for any other weak base as shown in the following example: Example(7) Calculate the ph of a 0.10M sodium acetate solution? NaOAC Na + + OAC - OAC - + H 2 O = HOAC + OH - Kb = = = = If the initial conc. of OAC - = ci hence Kb = = ci-x = ci if ci > 100 Kb which will be the general case for weakly ionized bases, hence OH - = x = Ci = Kb Ci [OH - ] = x = = Kb Ci Note: The quadratic formula must be used to solve the problem if the condition of Ci >100 Kb hence, In this example Ci = 0.1 > 100Kb, hence 7

8 [OH - ] = x= KbCi = = poh= -log [OH - ] = = 5.12 ph= = 8.88 Similar equations can be derived for the cations of salts of weak bases (the salts are completely dissociated. These are Brnsted acids which ionized (hydrolysis) in water BH + + H 2 O = B + H 3 O + Kh = Ka = Multiply the numerator and denominator by [OH - ] Ka = = Similarly, if ci > 100ka, than [H + ] = Ci = Ka Ci Example(8) Calculate the ph of a 0.25 M ammonium chloride solution? + NH 4 Cl NH 4 + Cl - + NH 4 + H 2 O NH 4 OH + H + or NH H 2 O NH 3 + H 3 O + Ka = = = = Since Ci > Ka, Than [H + ]= x = KaCi = = ph= -log [H + ] = = 4.92 Mixtures of weak acid and its salt or a weak base and its salt For a mixture (weak acid + its salt) or (weak base + its salt), salt is completely ionized, hence the concentration of the anion increases and chemical equilibrium will displace : Accordingly the effect of salt must be considered in computing the PH. A solution of acetic acid and sodium acetate. The [H + ] [OAC - ] 8

9 HOAC H + + OAC - Ka = [H + ] = Ka -log [H + ] = -log Ka = - log ph = pka + log For the general case HA H + +A - Henderson equation useful for calculation ph of weak acid + its salt ph = pka + l og ph = pka + log ph = pka + log For weak base + its salt :B = base BH + = B + H + Ka = = BH= conjugate acid [H + ] = Ka =. Log [H + ] = -log Ka log ph = pka + log = -log = -log[ ] = (pkw- pkb) + log ph = pka + log poh = pkw ph = (pkw pkb) + log 9

10 poh = pka + log Example (9) = pkb + log Calculate the ph of a solution prepared by mixing 10 ml of 0.1M acetic acid and 20ml of 0.1M sodium acetate? mol of HOAC = 0.1mol/L 10ml = 0.001mol mol of OAC - - = = mol Tatol volume of solution = = 30 ml [HOAC] = [OAC - ] = = mmol/ml = 0.033mmol/ml Note:. The amount of acid dissociated is very small in the presence of the added salt, and can be neglected.hence, the initial conc. can be assumes equal to the equilibrium conc. Use hendrerson equ. ph = pka + log = -log log = log 2= = 5.06 Example (10) It is required to prepare 100ml of a solution with ph= 10 by mixing concentration ammonia (14.8M) and ammonium chloride. The salt concentration in the final solution 0.2M. Calculate the volume of concentrated ammonia and the mass of NH 4 Cl required to prepare the solution? ph = (pkw- pkb) + log 11

11 10 = (14-PKb) + log Kb = , pkb = -log Kb = = 4.75 [NH + 4] = 0.200M Log = = 0.75 :. [NH 3 ] = 5.6(0.20) = 1.12M :. moles of (NH 3 ) = 1.12 = mol :. Volume of conc. Ammonia = 1000 ml = 7.6ml mass of NH 4 Cl = mol Mwt 0.2 = 1.07 gm BUFFERS:- Abuffer solution is a solution which resist change ph when a small amount of an acid or base is added or when the solution is diluted. it consists amixture of weak acid and its conjugate base or a weak base and its conjugate acid at specific concentrations or ratios. A buffer is prepared as a mixture of weak acid and its salt or weak base and its salt. In general, the change in [H + ] can be from the following relation. [H + ] = ka HA = weak acid HA = H + + A - The range of between BUFFER CAPACITY:- Is defined as the number of moles of astrong base strong acid (containing one acidic or basic functional groups for each mole of acid or base which is required to change the ph of one liter of the buffer solution by one unit 11

12 As the buffer capacity increases, the amount strong base or strong acid required to change The ph by afixed amount increases, hence, the slope of titration curve decreases. Similarly, a mixture of weak base and its salt acts as a buffer in the same manner as a weak acid and its salt The buffering capacity is maximum at a ph = pka (or at poh= pkb) with a useful range of pka+1 Example (11):- Calculate the buffer capacity of a solution which is 0.10M acetic acid and 0.10M sodium acetate ph = pka + log log = 4.75 If NaOH is added until the ph is increased by one unit i.e ph= = log = 10 [OAC- ] = 10[HOAC] The total conc of [OAC - ] and [HOAC] before the addition of NaOH = = 0.2M = Total conc of [OAC] [HOAC] after the addition of NaOH [HOAC] + 10[HOAC] = 0.2 [HOAC] = 0.2/11= M :.[OAC] = 0.182M HOAC + OH - = H 2 O + OAC - HOAC reacted= = M= quantity of NaOH reacted Hence moles of NaOH is required for one liter of solution to change the ph by one unit i.e buffer capacity = mol NaOH/literpH 12

13 H.W 1- Calculate the ph of a solution prepared by adding 25 ml of 0.1M of NaOH solu. to 30 ml of 0.2M of acetic acid? 2- Calculate the change in ph upon adding 1 ml of 0.1M of hydrochloric acid to 10 ml of buffer solution consists of 0.2M acetic acid * 0.2 M sodium acetate PROBLEMS:- 1- Calculate the ph, poh of the following strong acid solutions? a M HCl b- 0.02M HClO 4 C M HNO 3 Ans : a) 3.87, 10.13, b) 1.7, 12.3 C)3.89, Calculate the ph, poh of the following strong base solutions? a) 0.05M NaOH b) 0.14M Ba(OH) 2 C) 2.4M NaOH d) M KOH Ans: a) 12.70, 1.3, b) 13.45,0.55, c) 14.38,-0.38 d) 7.48, Calculate the ph, poh of a solution obtaining by mixing equal volumes of 0.10M H 2 SO 4 and 0.3M NaOH? Ans: 12.70, Calculate the hydrogen ion conc. and ph of a neutral solution at 50 if Kw at 50Cᵒ = Ans: ,, Calculate the ph of 0.01M acetic acid solution? Ans: The ph of an acetic acid solution is 3.26 what is the conc of solution and what is the percent acid ionized? Ans: M, 3.2% 7- Asolution at ph 4.0 was analyzed and found to contain 0.012M formic acid (HCOOH) and 0.02M formate COOH - calculate the pka of formic acid?ans: Calculate the ph of 0.1M of aniline solution (weak base)? Ans: Calculate the ph of a solution contains 0.10M benzoic acid (C 6 H 5 COOH) and 0.1M benzoate ion (C 6 H 5 COO - ) Ka= Ans: Calculate the ph of a M NaOH solution? Ans: 10.57, Ka =

14 11-Calculate the ph of the solution obtained by adding 20ml of 0.10M HOAC to 20ml of 0.10 M NaOH? Ans: Sodium hydroxide was added to 0.10M of acetic acid solution until the ph of the solution was 6.0.Calculate the conc of acetate ion and acetic acid in the final solution; neglecting dilution effect? Ans: [OAC - ] = M [HOAC] = M 13- Abuffer solution is prepared by adding 25 ml of 0.05M sulfuric acid solution to 50ml of 0.10m ammonia solution. what is the ph of the buffer? Ans: Calculate the buffer capacity of a solution which is 0.1 m in formic acid (HCOOH) and format ion (HCOO - )? Ans :0.082 mole NaOH/(liter ph) 15- Calculate the buffer capacity of a solution consists of 0.15 mole benzoic acid (C 6 H 5 COOH) and 0.22 m of sodium benzoate (C 6 H 5 COONa)? Ans: 0.13 mole NaOH /(liter. ph),ka