General Chemistry Lec 4+5 PDF

College of Health and Medical Technologies Anesthesia Department 2024-2025 (Methods of Expressing analytical concentrations) General Chemistry-The first stage Lec (4+5) Dr. Ahmed...

College of Health and Medical Technologies Anesthesia Department 2024-2025 (Methods of Expressing analytical concentrations) General Chemistry-The first stage Lec (4+5) Dr. Ahmed Jaber solution is defined as a homogeneous mixture of two or more substances, of which the concentrations can vary within limits. * Concentrated Solution has a large amount of solute. * Dilute Solution has a small amount of solute * When a substance called solute dissolved in another substance called the , solvent a solution is formed. Dr. ahmed jaber Solute: The substance which dissolves in a solution. Example :- Sugar, Salt, NaOH, Na2CO3 Solvent: The substance which dissolves another to form a solution. Example:- water, Acetone, Ether, Ethanol. Solution: A mixture of two or more pure substances. In a solution one pure substance is dissolved in another pure substance homogenously. Example:- in a sugar and water solution, the solution has the same concentration throughout. Dr. ahmed jaber Mole: A fundamental unit of mass, used by chemists. This term refers to a large number of elementary particles (atoms, molecules, ions, electrons, etc) of any substance. 𝒘𝒆𝒊𝒈𝒉𝒕 (𝒈𝒓𝒂𝒎) 𝐌𝐨𝐥𝐞 = 𝐌𝐨𝐥𝐞𝐜𝐮𝐥𝐚𝐫 𝐖𝐞𝐢𝐠𝐡𝐭 (𝐠𝐫𝐚𝐦/𝐦𝐨𝐥) Dr. ahmed jaber The Molecular Weight: The Molecular Weight (M.wt) of substance is the sum of the atomic weight s of all the atoms in a molecule of the substance, Measure in (gram/mol). example Dr. ahmed jaber Molarity (M): is the concentration of a solution expressed as the number of moles of solute per liter of solution: 𝐌𝐨𝐥𝐞 𝐌= 𝐕𝐨𝐥𝐮𝐦𝐞(𝐋) 𝐰𝐞𝐢𝐠𝐡𝐭 × 𝟏𝟎𝟎𝟎 𝐌= 𝐌. 𝐖𝐭 × 𝐯𝐨𝐥𝐮𝐦𝐞(𝐦𝐥) Dr. ahmed jaber Normality (N): is the concentration of a solution expressed as the number of gram equivalents of solute per liter of solution: 𝐍𝐨. 𝐨𝐟 𝐞𝐪𝐮𝐢𝐯𝐚𝐥𝐞𝐧𝐭𝐬 (𝐞𝐪) 𝐍𝐨𝐫𝐦𝐚𝐥𝐢𝐭𝐲 (𝐍) = 𝐕𝐨𝐥𝐮𝐦𝐞 (𝐋) 𝐰𝐞𝐢𝐠𝐡𝐭 × 𝟏𝟎𝟎𝟎 𝐍𝐨𝐫𝐦𝐚𝐥𝐢𝐭𝐲(𝐍) = 𝐞𝐪. 𝐰𝐭 × 𝐯𝐨𝐥𝐮𝐦𝐞(𝐦𝐥) 𝐰𝐞𝐢𝐠𝐡𝐭 𝐠𝐫𝐚𝐦 𝑵𝒐. 𝒐𝒇 𝒆𝒒𝒊𝒗𝒂𝒍𝒆𝒏𝒕𝒔 = 𝐄𝐪𝐮𝐢𝐯𝐚𝐥𝐞𝐧𝐭 𝐰𝐞𝐢𝐠𝐡𝐭 𝐠/𝐞𝐪 Dr. ahmed jaber Equivalent weights: (1)Equivalent weight of Acid : Ex: equivalent weight of H2SO4 =M.Wt H2SO4/2 Ex: equivalent weight of H3PO4 =M.Wt H3PO4/3 Ex: eqivalent weight HCl=M.Wt HCl/1 Dr. ahmed jaber (2) Equivalent weight of Base: Ex: equivalent weight of NaOH =M.Wt NaOH/1 Ex: equivalent weight of Ca(OH)2=M.Wt Ca(OH)2/2 Dr. ahmed jaber (3) Equivalent weight for salts: Ex: equivalent weight of NaCl =M.Wt NaCl/1 Ex: equivalent weight of CaCl2=M.Wt CaCl2/2 Dr. ahmed jaber Part per million(ppm):Milligrams of solute per liter of solution. 𝐰𝐭(𝒎𝐠) 𝐩𝐩𝐦 = 𝐕(𝐋) 𝐰𝐭(𝐠) 𝐩𝐩𝐦 = × 106 𝐕(𝐦𝐋) The relationship between molarity (M) , normality (N) and part per million (ppm):- 𝑝𝑝𝑚 = 𝑀 𝑥 𝑀. 𝑤𝑡 𝑥 1000 𝑝𝑝𝑚 = 𝑁 𝑥 𝐸𝑞. 𝑤𝑡 𝑥 1000 Dr. ahmed jaber The relationship between molarity (M) & normality (N) 𝑵=𝒏𝒙𝑴 n = No. of (H) for acid = No. of (OH) for base Dr. ahmed jaber Density: is the weight per unit volume at the specified temperature, usually (gm/mL) or (gm/cm3) or (gm.cm-3) in 20 C is the ratio of the mass in (gm) and volume (mL). Specific gravity (sp. gr.): defined as the ratio of the mass of a body (e.g. a solution) usually at 20C to the mass of an equal volume of water at 4C (or sometimes 20C) or (is the ratio of the densities of the two substances). Dr. ahmed jaber Dilution: is the process in which more solvent is added to a solution in order to lower its concentration. Dilution always lowers the concentration of a solution. The same amount of solute is present, but it is now distributed in a larger amount of solvent N1×V1=N2×V2 M1×V1=M2×V2 Dr. ahmed jaber Example(1): prepare a standard solution of a Sodium Hydroxyl (NaOH), 0.2 M in 100 ml of water. (atomic weight : Na=23 , O=16 , H=1)? SOLVE:- 1-Calculate M.wt :- M.wt (NaOH) =(1*23)+(1*16)+(1*1) = 40 gm /mol 2- Calculate a Weight (Mass) :- weight (g)×1000 M= 𝑀.𝑊𝑡×𝑣𝑜𝑙𝑢𝑚𝑒(𝑚𝑙) 0.2 M = Wt. * 1000 / 40(gm /mol) * 100(ml) Wt. = 0.2(mol/ml) * 40(gm /mol) * 100(ml)/1000 Wt. = 0.8 (gm) Dr. ahmed jaber Example(2):- prepare a standard solution of a Carbonate Sodium (Na2CO3), 1.6 N in 0.25 L of water. (atomic weight(A.W): Na=23 , C=12 , O=16 ) ? SOLVE:- 1-Calculate Eq.wt :- M.wt (Na2CO3) =(2*23)+(1*12)+(3*16) = 106 gm /mol n=2 (eq/mol) Eq.wt = M.wt/n = 106/2 = 53 (gm/eq) 2-Calculate a Weight (Mass) :- weight Normality(N) = eq. wt × volume(L) 1.6 N = Wt. (gm)/ 53(gm/eq) * 0.25 (L) Wt. = 1.6 (eq/L) * 53(gm /eq) * 0.25 (L) Wt. = 14 (gm) Dr. ahmed jaber Example(3): prepare a standard solution of a Sodium Chloride (NaCL), 58000 PPM in 200 ml of water. And then find molarity (M) and normality(N) for solution ? (atomic weight : Na=23 , CL=35 ) solve :- 𝐰𝐭(𝐠) 1- 𝐩𝐩𝐦 = × 106 𝐕(𝐦𝐋) 𝐰𝐭(𝐠) 𝟓𝟖𝟎𝟎𝟎 = × 106 𝟐𝟎𝟎 𝒎𝒍 wt(g)= 11.6 g 2-Calculate M.wt :- M.wt (NaCL) =(1*23)+(1*35) = 58 gm /mol 𝑝𝑝𝑚 = 𝑀 𝑥 𝑀. 𝑤𝑡 𝑥 1000 M= 𝑝𝑝𝑚 / 𝑀. 𝑤𝑡 𝑥 1000 = 58000 / 58 𝑥 1000 = 1 M NOTE :- if n=1 , the molecular weight ( M.wt) = Equivalent weight ( Eq.wt) 3-Calculate Eq.wt :- Eq.wt (NaCL) = M.wt/n = 58/1 = 58 (gm/eq) 𝑝𝑝𝑚 = 𝑁 𝑥 𝐸𝑞. 𝑤𝑡 𝑥 1000 N= 𝑃𝑃𝑀 / 𝐸𝑞. 𝑤𝑡 𝑥 1000 = 58000 / 58 𝑥 1000 = 1 N Dr. ahmed jaber Example (4):-How many millilitres(ml) of a 2000 ppm solution of CaCl2 prepare from 5 mg ? Solve:- 𝐰𝐭(𝒎𝐠) 1- by using 𝐩𝐩𝐦 = 𝐕(𝐋) 𝐰𝐭(𝒎𝐠) 𝟓 V(L) = = = 0.0025 L 𝐏𝐏𝐌 𝟐𝟎𝟎𝟎 0.0025 * 1000 = 2.5 ml --------------------------------------------------------------------- 𝐰𝐭(𝐠) 2- or by using 𝐩𝐩𝐦 = 𝐕(𝐦𝐋) × 106 𝐰𝐭(𝐠) 𝟎.𝟎𝟎𝟓 𝐕(𝐦𝐋) = × 106 = × 106 = 2.5 ml 𝐩𝐩𝐦 𝟐𝟎𝟎𝟎 Dr. ahmed jaber Example (5):-Calculate the molarity of 28% NH3, specific gravity 0.898 (atomic weight : N=14 , H=1 ) Solve:- M.wt (NH3) =(1*14)+(3*1) = 17 gm /mol % 𝑥 𝑠𝑝. 𝑔𝑟. 𝑥 1000 𝑀= 𝑀. 𝑤𝑡 28/100 𝑥 0.898𝑥 1000 𝑀= = 14.7 M 17 Dr. ahmed jaber Example (6) :-How many millilitres of concentrated sulphuric acid, 94% , density 1.831 g/cm3, are required to prepare 1 liter of a 0.1 M solution? Note ( M.wt of sulphuric acid = 98.1 g/mol) solve:- 1- Calculate Molarity (M) of sulphuric acid % 𝑥 𝑑𝑒𝑛𝑠𝑖𝑡𝑦 𝑥 1000 94/100 𝑥 1.831 𝑥 1000 𝑀= = = 17.5 M 𝑀.𝑤𝑡 98.1 2- using diluting low 𝑀1 𝑥 𝑉1 𝑐𝑜𝑛𝑐 = 𝑀2 𝑥 𝑉2 𝑑𝑖𝑙𝑢 17.5 𝑥 𝑉1 = 0.1 𝑥 1000 𝑉1 = 𝟓. 𝟕 𝒎𝒍 Dr. ahmed jaber Example (7) :How many milliliters of concentration hydrochloric acid , 38 %, specific gravity 1.19, are required to prepare 0.5L of a 0.2 M solution? Note ( M.wt of hydrochloric acid = 36 g/mol) solve:- 1- Calculate Molarity (M) of hydrochloric acid % 𝑥 𝑠𝑝.𝑔.𝑥 1000 38/100 𝑥 1.19 𝑥 1000 𝑀= = = 12.5 M 𝑀.𝑤𝑡 36 2- using diluting low 𝑀1 𝑥 𝑉1 𝑐𝑜𝑛𝑐 = 𝑀2 𝑥 𝑉2 𝑑𝑖𝑙𝑢 12.5𝑥 𝑉1 = 0.2 𝑥 500 𝑉1 = 𝟖𝒎𝒍 Dr. ahmed jaber Example (8):- prepare 500 mL of 0.1 N (K2Cr2O7) solution from a 0.250 N solution? Solve:- 𝑁1 𝑥 𝑉1 𝑐𝑜𝑛𝑐 = 𝑁2 𝑥 𝑉2 𝑑𝑖𝑙𝑢 0.250𝑥 𝑉1 𝑐𝑜𝑛𝑐 = 0.1𝑥 500 𝑑𝑖𝑙𝑢 0.1 𝑥 500 𝑉1 = = 𝟐𝟎𝟎 𝐦𝐥 0.250 Dr. ahmed jaber Example (9):-What is the molarity and normality of a 13% H2SO4 , specific gravity 1.090 (atomic weight : H=1 , S=32 , O=16 )? Solve:- M.wt (H2SO4) =(2*1)+(1*32)+(4*16) = 98 gm /mol % 𝑥 𝑠𝑝. 𝑔𝑟. 𝑥 1000 𝑀= 𝑀. 𝑤𝑡 13/100 𝑥 1.090𝑥 1000 𝑀= = 1.4 M 98 2- for calculate normality (N) 𝑵=𝒏𝒙𝑴 𝑵 = 𝟐 𝒙 𝟏. 𝟒𝟓 = 𝟐. 𝟖 𝐍 Dr. ahmed jaber Example (10):- Calculate the molarity(M) & normality(N) of each of the following solutions: (a) 10 gm H2SO4 in 250 ml of solution (M.wt (H2SO4) =98 g/mol) (b)6 gm NaOH in 500 ml of solution (M.wt (NaOH) = 40 g/mol) (c)25 gm AgNO3 in 1L of solution (M.wt (AgNO3) = 170 g/mol) solve:- 𝐰𝐞𝐢𝐠𝐡𝐭×𝟏𝟎𝟎𝟎 (a) 𝐌 = 𝐌.𝐖𝐭×𝐯𝐨𝐥𝐮𝐦𝐞(𝐦𝐥) 𝟏𝟎×𝟏𝟎𝟎𝟎 𝐌= = 0.4 M 𝟗𝟖×𝟐𝟓𝟎 n= 2 𝑵=𝒏𝒙𝑴 𝑵 = 𝟐 𝒙 𝟎. 𝟒 = 𝟎. 𝟖 𝑵 Dr. ahmed jaber  𝐰𝐞𝐢𝐠𝐡𝐭×𝟏𝟎𝟎𝟎 (b) 𝐌 = 𝐌.𝐖𝐭×𝐯𝐨𝐥𝐮𝐦𝐞(𝐦𝐥) 𝟔×𝟏𝟎𝟎𝟎 𝐌= = 0.3 M 𝟒𝟎×𝟓𝟎𝟎 n= 1 𝑵=𝒏𝒙𝑴 𝑵 = 𝟏 𝒙 𝟎. 𝟑 = 𝟎. 𝟑 𝑵 𝐰𝐞𝐢𝐠𝐡𝐭×𝟏𝟎𝟎𝟎 (c) 𝐌 = 𝐌.𝐖𝐭×𝐯𝐨𝐥𝐮𝐦𝐞(𝐦𝐥) 𝟐𝟓×𝟏𝟎𝟎𝟎 𝐌= = 0.14 M 𝟏𝟕𝟎×𝟏𝟎𝟎𝟎 n= 1 𝑵=𝒏𝒙𝑴 𝑵 = 𝟏 𝒙 𝟎. 𝟏𝟒 = 𝟎. 𝟏𝟒 𝑵 Dr. ahmed jaber  Homework Q1:- prepare a standard solution of a Sodium Chloride (NaCl), 0.2 N in 250 ml of water. (atomic weight : Na=23 , Cl=35.5 ) ? Q2 :- prepare a standard solution of a potassium Chloride (KCl), 0.3 N in 500 ml of water. (atomic weight : K=19 , Cl=35.5 ) ? Dr. ahmed jaber Q3:- prepare a standard solution of a Sodium Chloride (CaCL2), 222000 PPM in 100 ml of water. And then find molarity (M) and normality(N) for solution ? (atomic weight : Ca=40 , CL=35 ) Q4:- How many millilitres of concentrated sulphuric acid, 99.0% , density 1.831 g/cm3, are required to prepare 0.6 liter of a 0.2 M solution? Note ( M.wt of sulphuric acid = 98.1 g/mol) Q5:- prepare 100 mL of 0.3 N (AgNO3) solution from a 0.5 N solution? Dr. ahmed jaber THANK YOU! ?? ANY QUESTIONS PLEASE ASK Dr. ahmed jaber