calculate the PI of lysine and glutamic acid pk1=2 pk2=8 pkp=4 pkp2=10
A 1000 mL aliquot of .1 M diprotic acid H2A (pK1=4; pK2=8) was titrated with 1 M NaOH. find pH values at Vb: 0, 3, 8, 10, 13, 19, 20, 22.
Given that pK1 = 3.0 and pK2 = 8.0 for an amino acid X. Total concentration of amino acid X in solution is 10 mM. Calculate the concentrations of protonated form (P), dipolar form (Z) and de-protonated form (D) of amino acid X at pH 2, and 10.
The amino acid Prolinehas a pk1 of 1.99, pk2 of 10.60. The amino acid Serine has pk1 of 2.21 and a of pk2 of 9.15. If these two amino acids bind together by peptide bond creating the peptide sequence Pro-Ser . What will be the overall charge of this peptide at the physiological pH of 7?
(b) The amino acid lysine has the following values of pK : PK1 =2.2, pK2 = 8.9, PKR = 10.5 (the pK, of the amino group in the side chain). At pH 7 it has the structure shown below (6). Draw the structure that you would expect for lysine at pH 9.7, indicating the charges as appropriate. HZN 0 HINA (2 marks) (c) Phenol (C6H5OH) and 2,4,6-trichlorophenol (C6H2C13OH) are both weak acids with pK, values of 9.89 and 6.21, respectively,...
What is the pI of glutamate? pK1= 2.2 pK2=3.5 pKr=8.5 Rule: pH > pKa deprotonated, pH < pKa protonated
Calculate the pH of a 0.340 M NaO2CCO2H solution. pK1 = 1.250 and pK2 = 4.266
A stream has a sulfuric acid concentration of 5000 mg/L. Assume pK1 = -3 and pK2 = 1.99. What is the expected pH of the stream?
A 100.0 mL aliquot of 0.100M diprotic acid H2A(pK1 = 4.00; pK2 = 8.00) was titrated with 1.00 M NaOH. Find the pH values at the following volumes of base added, Vb; Vb= 0 mL, PH= Vb= 2 mL, PH= Vb= 8 mL, PH= Vb= 10 mL, PH= Vb= 12 mL, PH= Vb= 18 mL, PH= Vb= 20 mL, PH= Vb= 22 mL, PH=
#1) a) The pK1, pK2, and pKR for the amino acid histidine are 1.8, 9.3, and 6.0, respectively. At pH 4.0 would be charged predominantly as follows: A a-carboxylate 0, a-amino +1, imidazole +1, net charge +2 B a-carboxylate +1, a-amino +1, imidazole −1, net charge +1 C a-carboxylate −1, a-amino +1, imidazole 0, net charge 0 D a-carboxylate +1, a-amino 0, imidazole −1, net charge 0 E a-carboxylate −1, a-amino +1, imidazole +1, net charge +1 b) The pK1,...
Calculate the pH of a 0.03 M NaO2CCO2H solution. pK1 = 1.250 and pK2 = 4.266. THE ANSWER IS NOT 2.89 IT IS 2.76 Please show your work. Thanks!