Pharmaceutical Analytical Chemistry I (PA 101) Lecture Notes PDF

# Pharmaceutical Analytical Chemistry I (PA 101) ## Zagazig National University Faculty of pharmacy Pharm D clinical program ## Dr/ Yasmine Sharaf Ass. Prof of Analytical Chemistry Analytical Chemistry Department Faculty of Pharmacy Zagazig University Tel: 01066816943 Email: [email protected]...

# Pharmaceutical Analytical Chemistry I (PA 101) ## Zagazig National University Faculty of pharmacy Pharm D clinical program ## Dr/ Yasmine Sharaf Ass. Prof of Analytical Chemistry Analytical Chemistry Department Faculty of Pharmacy Zagazig University Tel: 01066816943 Email: [email protected] ## Arrhenius theory of ionization or Dissociation - Non-electrolytes are compounds that do not ionize at all in solution. - Examples of Non-electrolytes: Alcohols and glucose (sugar) readily dissolves in water, but because it does not dissociate into ions in solution and do not conduct electricity. - Electrolytes when dissolved in water split into two types of electrically charged particles called ions. - Examples of electrolytes (acids, bases and salts) dissociate into oppositely charged ions on dissolution in water. | Chemical Equation | Dissociation | |---|---| | NaCl → Na+ + Cl- | | | HCI → H+ + Cl | | | NaOH → Na+ + OH- | | ## Arrhenius theory of Dissociation or ionization - Ionization: is the process of splitting of the electrolyte molecules into ions. - The degree of dissociation: is the fraction of the total number of molecules present in solution as ions - Not all electrolytes dissociate in the same extent. - Arrhenius introduced the degree of dissociation as "α", as follows - α = n (number of molecules dissociated into ions) / N (Total number of molecules) x 100% - In case of strong electrolyte (Strong acids and bases) α will be close to 1 - In case of weak electrolyte (weak acids and bases) α will be less than 1 ## Arrhenius theory of ionization or Dissociation - **Dissociation α = 20%** - **Dissociation α = 60%** - **Dissociation α = 40%** - **Dissociation α = 80%** ## Law of mass action - A + B <=> C + D - V₁ = [A].[B]. K Vb = [C]. [D]. Kb - at equilibrium that is when V₁ = Vb K₁[A]. [B] = Kb [C]. [D] Kf [C][D] / Kb = [A][B] K = [C][D] / [A][B] "The rate of a chemical reaction is proportional to the active masses of the reacting substances." ➤Active mass may be expressed by the concentrations of the reacting species Where K denotes the "equilibrium constant" of the reaction ## The Dissociation of Water - Water is itself a very weak electrolyte. - H₂O = H+ + OH - According to the law of mass action: - K = [H+][OH-] / [H₂O] - Because pure water hardly ionises, [H₂O] is considered equal 1 - [H+]. [OH]- = Kw "The ionic product of water" = = 10-14 - Since in pure water the concentrations of H⁺ and OH- ions are evidently equal, we find that each is 10-7 - [H+]2 = Kw = 1 x 10-14 - [H+] = √10-14 = 10-7 - Thus one litre of water contains only 10-7 mole of H+ ions (and the same quantity of OH-); - -Log K = -Log [H+] + -Log [OH+] = -Log 10-14 - pKw = pH + POH = 14 ## The dissociation of water Kw "The ionic product of water” [****H****+****] =√10-14**** = 10-7**** - if the [H+] = [OH-] = 10-7 the solution is described as "neutral", - if [H+] is more than 10-7, (10-6, 10-5..., etc.) the solution is "acidic" - if [H+] is less than 10-7, (10-8, 10-9 ..etc.) the solution is "alkaline". ## The Dissociation of Acids and Bases ### 1- Weak acids: - HA(aq) + HOH(aq) <=> H3O+(aq) + A-(aq) - HC2H3O2 + HOH(aq) <=> H3O+ + C2H3O2- K = [H+][A-] / [HA] K₂ is the equilibrium constant for the dissociation of an acid ### 2- Weak base: - B + HOH(aq) <=> BH+(aq) + OH-(aq) - NH3 + HOH <=> NH4+ + OH- Kp = [BH+][OH-] / [B] K is the equilibrium constant for the dissociation of a base Acid and base equilibrium (ionization) constants are the measure of the strengths of acids and bases. pKa = -logka pKp = -logKb larger K₂ => smaller pK₂ => stronger acid --- larger K₁ => smaller pK₁ => stronger base ex: -log 0.1 = 0.9 and -log 0.01= 1.9 ## Importance of Dissociation constant of acids and bases Acid and base equilibrium (ionization) constants are the measure of the strengths of acids and bases. pKa = -logka pKp = -logK larger Ka => smaller pKa => stronger acid larger K => smaller pK => stronger base **strong acid** Ka increases pKa decreases **weak acid** Ka decreases pKa increases ## Difference between pKa and pH | | pKa | pH | |---|---|---| | | pka is the negative value of the logarithmic of ka Indicates if the acid is strong or weak Gives details about the acid dissociation in water Higher pka for weak acid Lower pka for strong acid Depend on concentration of acid, conjugate base and H+ | pH is the negative value of the logarithmic of H+ concentration Indicates if the soln is acidic or alkaline Gives details about H+ concentration in soln Higher pH for alkaline soln LowerpH for acidic soln Depend on concentration of H+ | ## pH calculations **pH** Calculator Important abbreviations ✓pH= - log [H+] ✓pOH= Kw – pH ✓Ka = acid ionization constant and pKa=-log Ka ✓K₁ = base ionization constant and pK₁= -log Kb ✓Ca = conc. Of acid and pCa= -log Ca ✓C₁ = conc. Of base and pC₁= -log Съ ✓C₃ = Conc. Of salt and pCs = -log Сь ## pH calculations | | pH calculation | |---|---| | Strong acid | pH= pC₂= -log[H+] | | Strong base | pH = pKw-pCb | | Weak acids e.g. CH3COOH | pH = ½ (pKa+ pCa) | | Weak bases e.g. NH4OH | pH = pK-1½ (pK + pC₁) | Example 1. Calculate the pH value of a solution of a completely ionised 1.0 N solution of acid; or base. ? - [H+] = 1M - pH = -log 1 = 0 (zero) similarly, 1.0 N solution of base - [OH-] = 1 M - POH = -log 1 = 0 (zero) **Remember** Log 0.1= -1 Log 0.01= -2 Log 1=0 Log 10= 1 Log 100= 2 ## Example 2 Calculate the [H+] and pH of 0.009 N hydrochloric acid? - [H+] = 0.009 N - pH = -log (9.0 X 10-3) = 2.05 ## Example 3 Calculate the pH values of a solution of sodium hydroxide whose [OH-] is 1.05 x 10-3? - pOH = - (log 1.05 x 10-3) = 2.98 - pH =14-2.98=11.02 ## Example 4 Calculate the hydrogen ion concentration of a solution of pH 5.3? - pH = -log[H+] 5.3 = -log[H+] - [H+] = 5.01 x 10–6 M ## Example 5 Calculate the hydroxyl ion concentration of a solution of pH 10.75? - pOH = 14 – 10.75 = 3.25 - [OH-] = the antilog of -3.25 - [OH-] = 5.62 x 10-4 M ## 2. Solution of weak acids and bases ### A) Calculation of pH of solution of weak acids pH = 1/2 (pKg + pCa) ## Example 6. Calculate the pH and [H+] of 0.10 N acetic acid (pK = 4.76)? - pCa -log (0.1) = 1 - pH = 1½ (pK + pCa) - = 1/2 (4.76 + 1) = 2.88 ## Example 7 Calculate the pH and [H+] of a 0.0045 M solution of phenobarbital (pK = 7.41)? - PC = -log (4.5 x 10-3) = 2.35 - pH = 1/2 (7.41 + 2.35) = 4.88 - [H+] = antilog - 4.88 = 10-4.88 - = 1.32 x 10-5 M ## B) Calculation of pH of solution of weak bases: pH = pKw - 1/2 (рК + рСь) ## Example 8. Calculate the pH, and the [H+] of 0.13 N ammonia solution (pK = 4.76)? - PCb=-log (0.13) = 0.89 - pH = 14-1/2 (4.76 + 0.89) = 11.18 - [H+] = antilog -11.18 = 6.6 x 10-12 M ## Example 9: Calculate the pH and the [H+] of a 0.0037M solution of cocaine base (pK = 5.59)? - PCb = - log (0.0037)= 2.43 - pH = 14.00-1½ (5.59 + 2.43) = 9.99 - [H+] = antilog of -9.99 = 1.03 x 10-10 ## Remember | | | |---|---| | Strong acid | pH= pCa= -log[H+] | | Strong base | pH = pKw-pCb | | Weak acids e.g. CH3COOH | pH = ½ (pKa+ pCa) | | Weak bases e.g. NH4OH | pH = pK-1½ (pK + pC₁) | | Salts of strong acids or bases | pH = 7 | ## Salts - Salts are ionic compounds formed in the reaction between an acid and a base. - The reaction of an ion with water to produce the conjugate acid and hydroxide ion or the conjugate base and hydrogen ion is called hydrolysis. - The pH of a salt depends on the strengths of the original acids and bases: | Acid | Base | pH of the salt | Equation | Problem contain salt (name is two parts) | |---|---|---|---|---| | Strong | Strong | 7 = Neutral | pH = 0.5 pKw | The two parts strong | | Weak | Strong | >7 = Basic | pH = 0.5 (pK + pKa - pCs) | The first part strong | | Strong | Weak | <7 = Acidic | pH = 0.5 (pK - PK + pCs) | The second part strong | | Weak | Weak | pH = 0.5 (pK + pKa - pKb) | | The two parts weak | ## Degree of hydrolysis Is the fraction of total salt which is hydrolyzed at equilibrium. h = y/Cs Y is H+ conc if salt is derived from strong acid Y is OH conc if salt is derived from strong base Cs is the salt conc The percentage of the salt hydrolyzed = 100 x (y/C) ## 1- Neutral Salts NaCl - Na+ is derived from NaOH, a strong base and Cl is derived from HCl, a strong acid - NaCl + H₂O => Na+ + Cl- - Na+ and Clions do not react with water so the solution is neutral. ## 2- Basic Salts KCN - K+ is derived from KOH, a strong base and CN is derived from HCN, a weak acid - KCN + H₂O => K++ OH + HCN - The OH ions are produced, so the solution is basic. - hydrolysis ## 3- Acidic Salts NH₄Cl - NH4+ is derived from NH3, a weak base and Cl- is derived from HCl, a strong acid - NH4Cl (s) + H₂O => NH4 + OH + H+ + Cl- - The H+ ions are produced, so the solution is acidic - hydrolysis ## Example 9. Calculate the pH, the [H+], the [OH-], and the degree of hydrolysis of a 0.1 M solution of sodium acetate (pK = 4.76)? - PC =- log 0.1 = 1 - pH = 1½ (14.00 - 1.00 +4.76) = 8.88 - [H+] = 1.3 x 10-9 M - [OH-] = 10-14/ 1.3x10-9 = 7.5x10-6 M - |h = 7.5x10-6 / 0.1 =7.5 x 10-5 - %h_ = 100x(7.5x10¯*) = 0.007% hydrolysed ## Example 10: Calculate the pH of a 0.165 M solution of sodium sulphathiazole (pK = 7.12)? - pC₁ = -log 0.165 = 0.78 - pH = 1½ (14.00 – 0.78 + 7.12) = 10.17 ## D) Salts of weak bases and strong acids: NH4+ + H₂O = NH4OH + H+ pH = 1/2 (pKw + pCs – pKb) ## Example 11. Calculate the pH, the [H+] and the degree of hydrolysis of a 0.05 N solution of ammonium chloride, pKb=4.76.? - pH = 1/2 (14.00 +1.30-4.76) = 5.27 - [H+] = 5.4 x 10-6 M - Degree of hydrolysis = 5.4×10-6 / 0.05 = 1.1×10-4 - Percentage of hydrolysis = 100 x (1.1 x 10-4) = 0.011 % ## Example 12: Calculate the pH of a 0.025 M solution of ephedrine sulphate (pK = 4.64)? - pC = - log 2.5 x 10-2 = 1.58 - pH = 1/2 (14.00 + 1.58 -4.64) = 5.47