Question

1. Calculate the pKa of lactic acid, given that when the concentration of lactic acid is 0.010 M and the concentration of lactate is 0.087 M, the pH is 4.80 B. Calculate the ratio of the concentrations of acetate and acetic acid required in a buffer system of pH 5.30 C The pH of a 0.20 M sodium acetate buffer is 5.1. The pKa of acetic acid is 4.76. Explain in details (showing all calculation) how to prepare the buffer Ordinary household bleach is an ageous solution of sodium hypochlorite, NaCIO a) What is the molarity of bleach solution that contains 42.4 g of sodium hypochlorite, in a total volume of 425 ml? 2. b) express the concentration of the bleach solution in g c) express the concentration of the bleach solution in w/v 2B. A stock solution of protein, 8 mg/mi is to be used to prepare a 50g/mL solution. How would you prepare 10 ml of this diluted solution? The following data were obtained in a study of an enzyme known to follow Michaelis Menten kinetics: Determine Km and Vmax 3. adeted 217 325 433 sous 647 O00 4. Identify the unknown sugars X1, X2, and X3 using the following test: lodine, Benedict's test and Molisch's test.

Answer 1

1. Pka = ? lactic acid lactate ion PH = 4.8 = 0.01M. = 0.087. pH = pka + log (lactate] I [Lactic acid] Pka = PH - log (lactate] Lactic Acid] = 4.8 - log 0.087M DOIM. = 4.8 - log. 8.7 = 4.8 - 0.94. . [pka = 3.86 | Pka of acetic acid 4.74. PH= 5.3 -PH = pka + log (CH3C00] [CH₂COOH) 5.3 - 4.76 = log (CH₂600) (CH₃Coor] 0.54 = log [CH₃Coo] CH₃ Cook] antilog 0.54 (CH3Coo) [CH₂ cool = 3.5 [CH₂COOH] [CH₂COOH

( C PH = 5.1 Pka = 4.76. 5.1 = 4.76 + - log 0.20m CH₃COOH) 0.34 = log (0-20m) [CH2Cook antilog (0.34) = 0.20M/Chycoon. CH₂ Cool - 0.2 - 0.1M or oil mel/ weight of Chylook per litre = 0.1 may & 600 = 69%

This buffer is prepared by dissolving 6g of acetic acid in 1L of solvent.