Solution Preparation (2) Solution A solution is a homogeneous mixture composed of two or more substances. In such a mixture, a solute is dissolved in another substance, known as a solvent. Solute The substance which dissolves in a solution Solvent The substance which dissolves another to form a solution Saturation Saturation is the point at which a solution of a substance can dissolve no more of that substance and additional amounts of it will appear as a precipitate. SupersaturationSolution Preparation (3) It refers to a solution that contains more of the dissolved material than could be dissolved by the solvent under normal circumstances. ‘Types of solutions. * Percentage solution * Molar solution * Normal solution ° Weight/ Weight solution % (wiw) This type of solution is rarely if ever prepared in the laboratory since it is easier to measure volumes of liquids rather than weigh the liquid on an analytical balance. This type of percent solution is usually expressed as (w/w), where “w" denotes weight (usually grams) in both cases. An example of a correct designation for this type of solution is as follows: 10 g/100 g (w/w), which indicates that there are 10 grams of solute for every 100. grams total * Weight/volume solution % (w/v) Weight-volume percentage, (sometimes referred to as mass-volume percentage and often abbreviated as % m/v or % wiv) describes the mass of the solute in g per 100 ml of the resulting solution.Solution Preparation (4) € (required Conc.) X V of solvent (ml) 100 ) =Wt of solute (g 010% (W/V) NaCI solution has 10 grams of sodium chloride dissolved in 100 ml of solution. Procedure: * Weigh 10g of sodium chloride. * Pour it into a graduated cylinder containing about 80ml of water. npletely add water to bring the Bo not simply measureegA@Oigyftwater and add 10g of som chloride. This wil gk se Ge" ASS Uni AbacuReaTh BPANI® the (NG GREIRFAKON WHIRH SFiS HOW), RRAG'URalts of a certain substance are present in one litre of liquid. g/t = 1000 mg/l. = 100 mg/dl.= 1000000 ug/l = al volume of ppm or part per million, is a unit of concentration often used and denotes one part per 1,000,000 parts, one part in 10°, and a value of 1 x 10°.(5) * Volume /volume solution % (viv) Volume-volume percentage (abbreviated as % v/v) describes the volume of the salute in ml per 100 mi of the resulting solution. C2 (required Conc.) X V2 of solvent (ml) .JC1 (original solute Cone )=V1 of solute (ml So 30% (V/V) sulfuric acid has 30 mi of sulfuric acid dissolved in 70 ml of water. Procedure: * Calculate the required volume of solute * Subtract the volume of solute from the total solution volume © Dissolve 30 ml sulfuric acid in a 70 ml of water to bring final volume of solution up to 100ml.Solution Preparation (6) When you mix concentrated sulfuric acid and water, you must ic Sulfuric acid reacts very vigorously with water, in a highly exothermic reaction. Water is less dense than sulfuric acid, so if you pour water on the acid, the reaction occurs on top of the liquid. If you add the acid to the water, it sinks and any wild reactions have to get through the Ice and Water water, Molecular weight It is the sum of the atomic weights of all the atoms in a molecule. Also called formula weight Mole A mole is also called gram-molecular weight and defined as number of particles whose total mass in grams was numerically equivalent to the molecular weight or it is the amount of substance containing the Avogadro number (6.022 x 10) Molarity Molarity or molar concentration denotes the number of moles of a given substance per liter of solution. Molar solution.Solution Preparation (7) Itis a solution that contains 1 mole of solute in each liter of solution. Units of molarity The units for molar concentration are mol/L. These units are often denoted by a Capital letter M (pronounced "molar"). Name Abbreviation Concentration Millimolar mM 10° molar Micromolar uM 10° molar Nanomolar nM 10°9 molar Picomolar pM 40°? molar Femtomolar iM 10" molar )=Wt of solute (gM (mole/1) X M.wt (g/mole) X V (L) = M: the required molarity = M.wt: molecular weight of the salute = V: total volume in liters Procedure;Solution Preparation (8) Dissolve 58.5 g of NaCl in a 1000 mf (1 liter) of water to prepare 1M NaCl solution. © Molarity of liquid solute To prepare 1M solutions make the following steps: * Calculate the molecular weight of the solute * Calculate the molarity of the solute using the following formula: % X density X 10 M.wt )=Molarity (M * Calculate the volume of the solute needed to prepare 1M solution using the following formula: M2 (required molarity) X V2 (required volume) (ml) )=V1 of solute(mi )M1 (original molarity * Subtract the volume of solute from the total solution volume * Mix both volumes of solute and solvent to reach the total required volume HCL molarity = 10.345 Volume required from HCL = 96.66 ml Complete to 1 liter with 903.33 ml distilled water H2SOx molarity = 18.3Solution Preparation (9) Volume required from H,SO, = 54.64 ml Complete to 1 liter with 945,36 ml distilled water HO, molarity = 12.70 Volume required from HO. = 78.74 ml Complete to 1 liter with 921.25 ml distilled water Equivalent weight An equivalent weight is equal to the molecular weight divided by the valence (replaceable H ions). Normal A normal is one gram equivalent of a solute per liter of solution. Normality Normality is the total no of gram equivalents of the solute present per liter of the solution Normal solutions | The definition of a normal solution is a solution that contains 1 gram equivalent weight (gEW) per liter solution. Units of Normality(10) The units for normal concentration are Eq/L. These units are often denoted by a capital letter N (pronounced "normal"). M.wt of acid or base No of H+ in acids or OH- in bases * HCL the MW= 36.5 the EW = 36.5 * H-SO; the MW = 98 the EW = 49 © HPO, the MW = 98 the EW = 32.7 * NaOH the MW = 40 the EW = 40 * Ca(OH), the MW = 74 the EW = 37 M.wt of salt No of cations X valency or No of anions X valencySolution Preparation (11) * KCL the MW= 74.55 the EW = (74.55 / 1X 1) = 74.55 * CaCl, the MW= 110,98 the EW = (110.98 / 1 X 2) or (110.98 / 2 X 1) = 55.49 © NayCOs the MW= 105.98 the EW = (105.98 / 1 X 2) or (105.98 / 2 X 1) = 52.99 )=Wt of solute (g N (Eq/I) X Eq.wt XV (L) = N: the required normality = Eg.wt: equivalent weight of the solute = V: tofal volume in liters Procedure: * Dissolve 40 g of NaOH in a 1000 mI (1 liter) of water to prepare 1N NaOH solution. * Dissolve 37 g of Ca (OH), in a 1000 mi (1 liter) of water to prepare 1N Ca (OH), solution.Solution Preparation (12) © Dissolve 74.55 g of KCl in a 1000 mi (1 liter) of water to prepare 1N KCl solution. * Normality of liquid solute To prepare 1N salutions make the following steps: * Calculate the equivalent weight of the solute * Calculate the normality of the solute using the following formula: % X density X 10 )=Narmality (N Eq.wt * Calculate the volume of the solute needed to prepare 1N solution using the following formula: N2 (required normality) X V2 (required volume) )=V1 of solute(ml = JNi (origif@l normality * Subtract the volume of soluite from the total solution volume * Mix both volumes of solute and solvent to reach the total required volume HCL normality = 10,345 Volume required from HCL = 96.66 mi Complete to 1 liter with 903.33 mi distilled waterH2SO, normality = 36.6 Volume required from H2SO, = 27.32 ml Complete to 1 liter with 972.67 ml distilled water * Simple Dilution A simple dilution is one in which a unit volume of a liquid material of interest is combined with an appropriate volume of a solvent liquid to achieve the desired concentration, The dilution factor is the total number of unit volumes in which your material will be dissolved. The diluted material must then be thoroughly mixed to achieve the true dilution. Dilute a serum sample 4:5 dilution as follow: Combining 1 unit volume of serum (for example 200 | |) + 4 unit volumes of the saline (for example 800 _ |) In this case the dilution factor is 5 (1+ 4 = * Serial Dilution = dilution factor).Solution Preparation (1a) A serial dilution is simply a series of simple dilutions which amplifies the-dilution factor quickly beginning with a small initial quantity of material. The source of dilution material for each step comes from the diluted material of the previous. In-a serial dilution the total dilution factor at any point is the product of the individual dilution factors in each step up to it Final dilution factor (DF) = DF1 * DF2* DF3 etc The initial step combines 1 unit volume of stock solution (100 mi) with 9 unit volumes of distilled water (900 ml) = 1:10 dilution. Prepare another 6 tubes each one contains 500 mi distilled water Combines 1 unit volume of the first tube (500 ml) with the 500 ml distilled water in the second tube Combines 1 unit volume of the second tube (500 ml) with the 500 mi distilled water in the third tube and:so on until the last tube The total dilution would be: 1:10 X2X2X2K2X2X2=1:640Solution Preparation (15) FOOL 500uL 500uL 500uL S00uL 500 ub CNINTNIONON 200 12.5 6.25 = 3.12 mol/L fea ime ae nmol/L mol/L mol/L. 100 ul std. * Specific Dilution Itis used when we need to make a specific volume of known concentration from stock solutions To do this we use the following formula: V1C1=V2C2 « V1: the volume of stock we start with. (Unknown) * (C1: the concentration of stock solution « V2; total volume needed at the new concentration * C2: the new concentrationSolution Preparation (16) Suppose we have 3 ml of a stock solution of 100 mg/ml and we want to_ make 200 ml of solution having 25 mg/ ml, V4 = (V2 x C2)/C1 V1 = (0.2 ml x 25 mg/ml) / 100 mg/ml V1 = 0.05 mi, or 50 mi So, we would take 0.05 ml stock solution and dilute it with 150 ml of solvent to get the 200 ml of 25 mg/ mi solution needed “X* units Stock solutions of stable compounds are routinely maintained in labs as more concentrated solutions that can be diluted to working strength when used in typical applications. The usual working concentration is denoted as 1X. A solution 20 times more concentrated would be denoted as 20X and would require a 1:20 dilution to restore the typical working concentration. Molarity (M) = (% X 10)/Molecular wt * Gonversion of molarity to Wt% Percent Wt% Percent = (M X Molecular wt)/ 10 * Conversion of Wt% Percent to normality Normality (N) = (% X 10)/Equivalent wt * Conversion of normality to Wt% PercentSolution Preparation (17) Wt% Percent = (N X Equivalent wt)/ 10 * Conversion of Normality to Molarity Normality (N) = Molarity (M) Xn * Conversion of Molarity to Normality Molarity (M) = Normality (N) /n Where n= number of (H') in acids or (OH) in bases or valency in salts * Conversion of Wt% Percent to ail Wt% Percent = g/I / 10 * Conversion of g/l to Wt% Percent g/l = Wt% Percent X 10 * Gonversion of Molarity to g/l g/l = M X Molecular wt * Conversion of g/l to Molarity M = Molecular wt / g/I © Conversion of Normality to g/l g/l = M X Equivalent wt(18) * Conversion of g/lto Normality M = Equivalent wt / g/l [tawcxtom || aracaoa, | aratcaion —] Easncnaco, —I[_-sa9% Naxc0, 15299 gnc]