Valine ( HV ) is a diprotic amino acid with ?a1=5.18×10−3 and ?a2=1.91×10−10 . Determine the pH of each of the solutions.
A 0.182 M valine hydrochloride ( H2V+ Cl− ) solution.
pH=
A 0.182 M valine ( HV ) solution.
pH=
A 0.182 M sodium valinate ( Na+ V− ) solution.
pH=
A 0.182 M valine hydrochloride ( H2V+Cl- ) solution. pH = 1.55
A 0.182 M valine ( HV ) solution. pH = 5.23
A 0.182 M sodium valinate ( Na+V- ) solution. pH = 11.5 (exact pH = 11.486)
Valine ( HV ) is a diprotic amino acid with ?a1=5.18×10−3 and ?a2=1.91×10−10 . Determine the...
Valine (HV) is a diprotic amino acid with Kal = 5.18 x 10-3 and K 2 = 1.91 x 10-10. Determine the pH of each of the solutions. A 0.288 M valine hydrochloride (H, V+Cl-) solution. pH = 1.44 A 0.288 M valine (HV) solution. pH = 5.12979 A 0.288 M sodium valinate (Na+vº) solution. pH = 11.59
Alanine (HA) is a diprotic amino acid with K_a1 = 4.53 times 10^-3 and K_a2 = 1.36 times 10^-10. Determine the pH of the following solutions. A) 0.260 M alanine hydrochloride (H_2 A^+ Cl^-) pH = B) 0.260 M alanine (HA) pH = C) 0.260 M sodium alaninate(Na^+ A^-) pH =
A diprotic acid, H 2 A , has acid dissociation constants of K a1 = 3.21 × 10 − 4 and K a2 = 5.67 × 10 − 12 . Calculate the pH and molar concentrations of H 2 A , HA − , and A 2 − at equilibrium for each of the solutions. A 0.130 M solution of H 2 A . pH = [ H 2 A ] = ? [ HA − ] = ? [...
For the diprotic weak acid H2A, ?a1=3.0×10−6 and ?a2=7.3×10−9 . What is the pH of a 0.0500 M solution of H2A ? pH= What are the equilibrium concentrations of H2AH2A and A2−A2− in this solution? [H2A]= [A2−]=
A diprotic acid, H2A,H2A, has acid dissociation constants of ?a1=4.15×10−4Ka1=4.15×10−4 and ?a2=3.73×10−12.Ka2=3.73×10−12. Calculate the pH and molar concentrations of H2A,H2A, HA−,HA−, and A2−A2− at equilibrium for each of the solutions. A 0.176 M0.176 M solution of H2A.H2A. pH = [H2A]=[H2A]= MM [HA−]=[HA−]= MM [A2−]=[A2−]= MM A 0.176 M0.176 M solution of NaHA.NaHA. pH= [H2A]=[H2A]= MM [HA−]=[HA−]= MM [A2−]=[A2−]= MM A 0.176 M0.176 M solution of Na2A.Na2A. pH= [H2A]=[H2A]= MM [HA−]=[HA−]= MM [A2−]=[A2−]= M
For the diprotic weak acid H2A, ?a1=3.5×10−6 and ?a2=7.8×10−9 What is the pH of a 0.07000 M solution of H2A? pH= What are the equilibrium concentrations of H2A and A2− in this solution? [H2A]= [A2−]=
Picolinic acid is a diprotic acid with ionization constants ?a1=9.80×10−2 and ?a2=4.10×10−6 . Calculate the pH of a 0.217 M potassium hydrogen picolinate ( KHP ) solution. PH=
For the diprotic weak acid H2A, ?a1=2.9×10−6 and ?a2=8.4×10−9 What is the pH of a 0.07000.0700 M solution of H2A? What are the equilibrium concentrations of H2A and A2− in this solution?
A diprotic acid, H2A, has acid dissociation constants of Ka1=1.01×10−4 and Ka2=4.08×10−12. Calculate the pH and molar concentrations of H2A, HA−, and A2−at equilibrium for each of the solutions. A diprotic acid, H, A, has acid dissociation constants of Kal = 1.01 x 104 and K22 = 4.08 x 10-12. Calculate the pH and molar concentrations of H, A, HA, and A? at equilibrium for each of the solutions. A 0.176 M solution of H, A. pH= pH = 1...
Given a diprotic acid, H2A , with two ionization constants of ?a1=2.0×10−4 and ?a2=3.1×10−12 , calculate the pH for a 0.120 M solution of NaHA.