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Solutions and Buffers
- 1. Practical No. 1 Preparationof Standard Solutions & Buffers Aim: To prepare standard solutions & buffers. Relevant Information: Standard solution is the solution whose concentration is exactly known. Standard solutions are classified as- Percent, Molar, Normal & Part solution. Percent solution is the most common way to express the concentration or strength of solution. A one percent solution is defined as 1 gram of solute per 100 ml final volume. For example, 1 gram of sodium chloride, brought to a final volume of 100 ml with distilled water, is a 1% NaCl solution. Molarity is another standard expression of solution concentration. Molar solutions use the gram molecular weight of a solute in calculating molar concentration in a 1 liter ofsolution. The gram molecular weight (GMW) of a substance (sometimes called the "formula weight") is the sum of the combined atomic weights of all atoms in the molecule expressed in grams. For example, the GMW of NaCl is equal to the atomic weight (these atomic weights may be found on a periodic table or as a formula weight on the bottle of substance) of Na (22.99) and the atomic weight of Cl (35.45) for a total of 58.44 g. A 1 molar (M) solution will contain 1.0 GMW of a substance dissolved in water to make 1 liter of final solution. Hence, a 1M solution of NaCl contains 58.44 g.Normality (N) is another way to quantify solution concentration. A 1N solution contains 1 gram-equivalent weight of solute per liter of solution. Expressing gram- equivalent weight includes the consideration of the solute's valence. The valence is a reflection of the combining power of an element often as measured by the number of hydrogen atoms it can displace or combine with. For example: Normality = GMW/valence (the valence for NaCl is 1.0). 58.44 g/1.0 = 58.44 g = 1.0 gram-equivalent weight of NaCl = 1N solution of NaCl. Inthis situation, because NaCl has a valence of one, the molarity and normality of the solution are the same. When solute & solvent are added together in parts it’s called Part solution. For example: 1:2 solutions where one part of solute is dissolved in two parts of solvent. A buffer solution (more precisely, pH buffer or hydrogen ion buffer) is an aqueous solution consisting of a mixture of a weak acid and its conjugate base, or vice versa. Buffer solutions are used as a means of keeping pH at a nearly constant value in a widevariety of chemical applications. In nature, there are many systems that use buffering for pH regulation. Buffer solutions resist pH change because of an equilibrium between the weak acid HA and its conjugate base A−: HA ⇌ H+ + A−. A mixture containing citric acid, monopotassium phosphate, boric acid, and diethyl barbituric acid can be made to cover the pH range 2.6 to 12. Requirements: Beaker, glass rod, weighing balance, Volumetric flask, pipettes & chemicals like NaOH, KOH, HCL etc.
- 2. Procedure: For Solids 1. Weighsolid chemicals according to chart and dissolve in small amount of distilled water in beaker. 2. Transfer this solution in volumetric flask & then add distilled water upto the mark. 3. Label the flask with name of solution & put date. For Liquids 1. Pipette out required volume of solute as per chart and transfer it to volumetric flask. 2. Rinse the pipette with distilled water and transfer the washings to the volumetric flask. 3. now add distilled water upto the mark. Label the flask with name & date. Calculations: 1. NaOH Molecular Weight of NaOH= Na+O+H = 23+16+1 = 40 Equivalent Weight of NaOH= GMW / no. of replaceable OH + ions = 40/1 = 40 1) Preparation of 500 ml of 5 % NaOH solution For 100 ml solution, dissolve 5 gm of NaOH to small amount of D/W & make thevolume upto 100ml with D/W in volumetric flask. If 100 ml = 5 gm, then for 500 ml = 25 gm. Thus for 500 ml of NaOH solution, add 25 gm of NaOH to small amount of D/W &make the volume upto 500ml with D/W in volumetric flask. 2) Preparation of 1000 ml of 1 M NaOH solution Add 40 gm of NaOH to small amount of D/W & make the volume upto 1000 ml with D/W in volumetric flask. 3) Preparation of 1000 ml of 0.1 M NaOH solution If for 1 M = 40 gm in 1000 ml of D/W Then for 0.1 M= 4 gm in 1000 ml of D/W 4) Preparation of 1000 ml of 0.5 M NaOH solution If for 0.1 M= 4 gm in 1000 ml of D/W Then for 0.5 M= 20 gm in 1000 ml of D/W 5) Preparation of 100 ml of 1 N NaOH solution If 40 gm NaOH + 1000 ml D/W = 1 N NaOH Then 4 gm NaOH + 100 ml D/W = 1 N NaOH
- 3. 6) Preparation of100 ml of 0.1 N NaOH solution If 1 N NaOH= 4gm NaOH + 100 ml D/W Then 0.1 N NaOH = 0.4 gm NaOH + 100 ml D/W 2. KOH Molecular Weight of KOH= Equivalent Weight of KOH= GMW / no. of replaceable OH + ions = 1) Preparation of 500 ml of 2 % KOH solution 2) Preparation of 1000 ml of 1 M KOH solution 3) Preparation of 1000 ml of 0.1 M KOH solution 4) Preparation of 1000 ml of 0.5 M KOH solution 5) Preparation of 100 ml of 1 N KOH solution 6) Preparation of 100 ml of 0.2 N KOH solution
- 4. 3. HCL Molecular Weightof HCL= Equivalent Weight of HCL= GMW / no. of replaceable OH + ions = 1) Preparation of 500 ml of 10 % HCL solution 2) Preparation of 1000 ml of 1 M HCL solution 3) Preparation of 1000 ml of 0.1 M HCL solution 4) Preparation of 1000 ml of 0.5 M HCL solution 5) Preparation of 100 ml of 1 N HCL solution 6) Preparation of 100 ml of 0.4 N HCL solution
- 5. 4. H2SO4 Molecular Weightof H2SO4 Equivalent Weight of H2SO4 = GMW / no. of replaceable OH + ions = 1) Preparation of 500 ml of 5 % H2SO4 solution 2) Preparation of 1000 ml of 1 M H2SO4 solution 3) Preparation of 1000 ml of 0.1 M H2SO4 solution 4) Preparation of 1000 ml of 0.5 M H2SO4 solution 5) Preparation of 100 ml of 1 N H2SO4 solution 6) Preparation of 100 ml of 0.1 N H2SO4 solution Result: