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Lab #2 info

# Lab #2 info - Laboratory II Topic Preparation of Buffers...

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Laboratory II Topic: Preparation of Buffers and Principles of Light Absorption. Activity 1: Preparing a Phosphate buffer A buffer works best when the pH is close to the pKa and at concentrations between 0.1 – 10 M concentrations Phosphoric acid [H 3 PO 4 ] has three pKa values; 2.15, 6.86 and 12.32. Phosphate buffers range between pH 4.8 and pH 8.8 and prepared using salts of phosphoric acid. Acid : Monosodium phosphate: Na + H 2 PO 4 - H + + Na + + HPO 4 2- Base : Disodium sodium phosphate 2Na + HPO 4 -2 + H + H 2 PO 4 - + 2Na+ Monosodium phosphate is also known as monobasic sodium phosphate. Disodium phosphate is also known as dibasic sodium phosphate. The phosphate salts are available in several hydrated forms (Monosodium phosphate anhydrous, Monosodium phosphate monohydrate and disodium phosphate anhydrous, disodium phosphate heptahydrate). Always check labels before you begin weighing. The purpose of today’s experiment is to apply Henderson-Hasselbalch equation in the preparation of buffers. The procedure involves calculating the quantities of the phosphate salts required to prepare the buffer and to verify the pH using a pH meter. Buffer Specifications Buffer Name: Phosphate buffer pH : 7 pKa: 6.86 Molarity: 0.1 M Volume : 250 ml [Acid] : Monosodium phosphate anhydrous Mol wt =120 gms [Base] : Disodium sodium phosphate anhydrous Mol wt =142 gms Calculate the ratio of Acid to Base pH = pKa + log [Base/Acid] log ([Base]/[Acid]) = pH - pKa log ([Base]/[Acid]) = 7 - 6.86 log ([Base]/[Acid]) = 0.14 ([Base]/[Acid] = 1.38 ............................. ………………………………………….(1) Calculate moles of [Acid] Molarity of the buffer, 0.1 M is the sum of molarities of the [Acid] + [Base] components. Therefore [Base] = 0.1 - [Acid] ………………………………………………………………(2) Substituting for [Base] in equation (1): [0.1 - [Acid] / [Acid] = 1.38 [0.1 - [Acid] = 1.38 [Acid] 0.1 = 1.38 [ Acid] + [Acid] = [ Acid] (1.38 + 1) = 2.38 [Acid] [Acid] = 0.1/2.38 = 0.042 moles of Monosodium phosphate per Litre …………… (3a) Calculate moles of [Base] using equation (2) [Base] = 0.1 - 0.042 = 0.058 moles of Disodium sodium phosphate per liter ………….(3b) Calculate the weight of [Acid] and [Base] powder required for 250 ml of buffer - 1 -

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= Mol wt (g) x Moles x volume (L) (substitute the value for moles of [Acid] obtained from (3a) and of [Base] from (3b) Molecular weight of Monosodium phosphate anhydrous 120 g [ Acid] = 120 x 0.042 x 0.25 L = 1.26 g Molecular weight of Disodium phosphate anhydrous 142 g [ Base] = 142 x 0.058 x 0.25 L = 2.058 g Activity 2: Absorption of light and complimentary colors White light is a mixture of all the colors of light of different wavelengths ranging form 400-700 nm. Table 1 below shows the wavelength of the visible spectrum Color Wavelength in nm (~range) Color Wavelength in nm (~range) Violet: 400 – 420 nm Yellow: 570 - 585nm Indigo : 420 – 440 nm Orange : 585 – 620 nm Blue: 440 – 490 nm Red: 620 – 780 nm Green: 490 – 570 nm When light passes through a colored solution, certain colors (wavelengths) are absorbed while the remaining pass (transmitted) through the solution.
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