BCH 312 [Buffers – Tutorial] Q1: Describe how you would prepare a 1 L of the following buffer 0.025M formic acid (mwt = 178.93 g/mol) / sodium formate (mwt = 68.01 g/mol) buffer, pH = 4.0, containing 0.05M glucose (mwt= 180.156g/mol) . (pKa =3.75) pH = pKa + log [!"#$%& !"#!"#$ (!!)] [!"#$%& !"#$ (!")] ; Assume [A - ] = y , [HA] = 0.025 –y 4 = 3.75 + log ! !.!"#!! 1.77 = ! !.!"#!! (after antilog) [A - ] = y = 0.016M ; [HA] = 0.025-0.016 = 9 x 10 -3 M no. of moles of A - = M x V (L) = 0.016 x 1 = 0.016 moles no. of moles of HA = M x V (L) = 9 x 10 -3 x 1 = 9 x 10 -3 moles wt. of A - = no. of moles x mwt = 0.016 x 68.01 = 1.088 g wt of HA= no. of moles x mwt = 9 x 10 -3 x 178.93 = 1.610 g Note: any additions of salts or sugars (that does not release H + or OH - ) such as NaCl, EDTA, CaCl 2 to buffers will not affect pH of solution. ** This buffer contains glucose, its calculated weight is added to the buffer components where it not effect the pH of buffer since it is a sugar that does not dissociate to H + or OH - ** no. of moles of glucose = M x V (L) = 0.05 x 1= 0.05 moles wt of glucose= no. of moles x mwt = 0.05 x 180.156 = 9.0078 g To prepare 1L this buffer à dissolve 1.088g of sodium formate, 1.61g of formic acid and 9.0078g of glucose in some water, then complete to the final volume of 1 liter and check pH. Components of 0.025M formate buffer, pH =4
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BCH 312 Buffers–Tutorial]fac.ksu.edu.sa/sites/default/files/tutorial-buffer.pdfBCH 312 [Buffers–Tutorial] Q4: An enzyme –catalyzed reaction was carried out in a solution containing
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BCH312 [Buffers–Tutorial]
Q1: Describe how you would prepare a 1 L of the following buffer 0.025M formic acid (mwt =