24. Calculate the equilibrium constant for the reaction Glucose-1-phosphate + H2

Question

24. Calculate the equilibrium constant for the reaction Glucose-1-phosphate + H2O glucose+H,Por at pH7.0 and temp. 25 C. Gas constant is 8.314 J Kmol 0°,--20.9 kJ mol-I a. 1.2x 10 b. 2.5x 10 c.4.6x 10' d. 5.1x 10 e.7.3x 10 25. A solution is made by mixing 50 mL of 2.0M KJHPOs and 25 mL of 2.0 M KH,Po The solution is diluted to a final volume of 200 ml. What is the pH of the final solution? pK, for this acid is 6.82 a. 634 b. 5.45 .4.56 d.7.12 e. 7.85 26. Glycine hydrochloride CTH,N'CH COOH is a diprotic acid that contains a carboxylic acid group and an ammonium group and thus is an amino acid. A solution containing 0.01 M Glycine hydrochloride and 0.2 M of the monodissociated species (CTH N CH.COO) has pH-2.65. What is the pK of this dissociation ? a. 1.75 b. 2.15 c. 2.35 d.3.05 e.3.45 27. What is the pH of a solution prepared by dissolving 5.35 g of NH4CI in a. 7.58 b. 8.35 c. 9.56d. 10.46e 1 Liter of 0.2 M NHs ? N-14,H-1,C1-35.5 28. Calculate the pH of O.1 M solution of isolectric Alanine. pka -2.34,pK.2-9.69 29. Calculate the hydrolysis constant pKs for the acetic acid with pk.-4.75 30. Calculate the isolectric pH for Glutamic acid with pKa -2.10, pk,2-9.47, a 2.34 b. 4.68 c. 1.67 d. 1.17 e. 6.02 a. 2.25 b. 3.75 c. 5.45 d. 7.85 e. 9.25 and pK,-4.07 a. 6.77 b. 5.8 c. 3.08 d. 4.07 e. 4.08 31. At pH 7, aspartic acid (pKs are -carboxylate 1 .99, a-amino 99, -carboxylate 3.9 ) would be charged as follows: A) 0 -carboxylate, +1 -amino, 0 -carboxylate, +1 net charge B)-1 -carboxylate, +1 -amino,-1-carboxylate,-1 net charge C) 0 -carboxylate, -1 -amino, 0 -carboxylate,-1 net charge D) +1 -carboxylate, -I -amino, +1 -carboxy late, +1 net charge +1 -carboxylate, +1 -amino, +1 -carboxylate, +3 net charge

Explanation / Answer

Q24

given th eequilbirium

dG = -RT*ln(K)

K = exp(-dG/(RT))

K = exp(20900/(8.314*298))

K = 4608

K = 4.61*10^3

best answer is C

Q25

this is a buffer, since acid / base wil be present

mmol of K2HPO4 = MV = 50*2 = 100

mmol of KH2PO4 = MV = 25*2 = 50

then

pH = pKa + log(K2HPO4 /KH2PO4 )

pH = 6.82 + log(100/50)

pH = 7.12

choose D

choose a

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