Homework Answers
FROM THE ABOVE DISSOCIATION REACTION ITS CLEAR THAT ion A2- HAVE VALENCY TWO AND FORM HYDRIDE H2A so it can be either from second group (Be, Ca ,Sr)or grpoub 6 (O,S,Se,Te)
because it can donate proton so it must belong to group 6 like (H2O,H2S)
AND ACCORDING TO THIS KHA(KOH,KSH) Should be basic in nature
while pKa value of both reaction its clear that H2A Is more acidic than HA-
THAT ALSO indicate that it belong to Oxygen family
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Question 10 Consider the acid dissociation reactions and for the diprotic acid H2A: H2A(aq) + H2O(1)...
Consider the acid dissociation reactions and for the diprotic acid H2A: H2A(aq) + H2O(1) =HA (aq)...
Consider the acid dissociation reactions and for the diprotic acid H2A: H2A(aq) + H2O(1) =HA (aq) + H2O+(aq) pKQ1 = 6 HA (aq) + H20(I) = A2-(aq) + H20+ (aq) pRaz = 10 Would a salt solution of KHA be acidic, basic or neutral? neutral basic O acidic
Consider the acid dissociation reactions and for the diprotic acid H2A: H2A(aq) + H2O (l) --><--...
Consider the acid dissociation reactions and for the diprotic acid H2A: H2A(aq) + H2O (l) --><-- HA-(aq) + H3O+(aq) pKa1= 3 H2A(aq) + H2O (l) --><-- A2- (aq) + H3O+(aq) pKa2= 8 Would a salt solution of KHA be acidic, basic or neutral? a) Acidic b) Basic c) Neural
Consider the acid dissociation reactions and for the diprotic acid H2A: Would a salt solution of...
Consider the acid
dissociation reactions and for the diprotic acid
H2A:
Would a salt solution
of KHA be acidic, basic or neutral?
A diprotic acid, H2A, has acid dissociation constants of Kai = 3.52 x 10-4 and Ka2...
A diprotic acid, H2A, has acid dissociation constants of Kai = 3.52 x 10-4 and Ka2 = 2.03 × 10-11 . Calculate the pH and molar concentrations of H2A, HA, and A2- at equilibrium for each of the solutions. A 0.206 M solution of H,A. pH = H2A] HA1 A 0.206 M solution of NaHA pH- [H2A] = [HA-] = A 0.206 M solution of Na,A. pH- [H2A] EA T [A21
A diprotic acid, H2A, has acid dissociation constants of Ka1=1.01×10−4 and Ka2=4.08×10−12. Calculate the pH and...
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...
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=3.73×10−1...
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
Question 4 (1 point) You will determine the concentration of an unknown diprotic acid using your standardized bas...
Question 4 (1 point) You will determine the concentration of an unknown diprotic acid using your standardized base. The chemical equation for this reaction can be expressed as: Vous déterminez la concentration d'un acide inconnu diprotique par le titrage avec une base étalonnée. La réaction chimique s'exprime comme: O a) H2A (aq) + 2 NaOH (aq) → Na, A (aq) + 2 H20 (1) Ob) HA (aq) + NaOH (aq) → NaA (aq) + H2O (1) Oc) H2A (aq) +...
Chat Consider the following reactions. Adiprotic acid H2A will react with water as shown in the...
Chat Consider the following reactions. Adiprotic acid H2A will react with water as shown in the following reactions. What is the value of K2? H2A(aq) + H2O(1) <=> H30+(aq) + HA (aq) HA-(aq) + H20(1) <=> H30*(aq) + A2+ (aq) H2A(aq) + 2 H2O(l) <=> 2 H30*(aq) + A+ (aq) o bbw WOO ES K EN WS 1.3E-13 - 2.3E-7 1.0E-7 1.3E-27 ZZE12 < Previne Consider the following reaction which has AH = -1311 k). If the following changes are...
The diprotic acid, H2A, has Ka1 i.e. (K1) = 1.00 X 10-4 and K2 = 1.00...
The diprotic acid, H2A, has Ka1 i.e. (K1) = 1.00 X 10-4 and K2 = 1.00 X 10-8. a) Consider a solution of 0.100 M H2A. Calculate the pH, and calculate the following concentrations: [H2A], [HA- ] and [A2- ]. b) Consider a solution of 0.100 M NaHA. Calculate the pH, and calculate the following concentrations: [H2A], [HA- ] and [A2- ].
(10) 1. The diprotic acid, H2A, has Kai i.e. (K1) = 1.00 X 10 and K2...
(10) 1. The diprotic acid, H2A, has Kai i.e. (K1) = 1.00 X 10 and K2 = 1.00 X 108. a) Consider a solution of 0.100 M H2A. Calculate the pH, and calculate the following concentrations: (H2A), (HA) and (A2). b) Consider a solution of 0.100 M NaHA. Calculate the pH, and calculate the following concentrations: (H2A), CHA') and (AP).
Consider the acid dissociation reactions and for the diprotic acid H2A: H2A(aq) + H2O(1) =HA (aq) + H2O+(aq) pKQ1 = 6 HA (aq) + H20(I) = A2-(aq) + H20+ (aq) pRaz = 10 Would a salt solution of KHA be acidic, basic or neutral? neutral basic O acidic
Consider the acid
dissociation reactions and for the diprotic acid
H2A:
Would a salt solution
of KHA be acidic, basic or neutral?
A diprotic acid, H2A, has acid dissociation constants of Kai = 3.52 x 10-4 and Ka2 = 2.03 × 10-11 . Calculate the pH and molar concentrations of H2A, HA, and A2- at equilibrium for each of the solutions. A 0.206 M solution of H,A. pH = H2A] HA1 A 0.206 M solution of NaHA pH- [H2A] = [HA-] = A 0.206 M solution of Na,A. pH- [H2A] EA T [A21
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...
Question 4 (1 point) You will determine the concentration of an unknown diprotic acid using your standardized base. The chemical equation for this reaction can be expressed as: Vous déterminez la concentration d'un acide inconnu diprotique par le titrage avec une base étalonnée. La réaction chimique s'exprime comme: O a) H2A (aq) + 2 NaOH (aq) → Na, A (aq) + 2 H20 (1) Ob) HA (aq) + NaOH (aq) → NaA (aq) + H2O (1) Oc) H2A (aq) +...
Chat Consider the following reactions. Adiprotic acid H2A will react with water as shown in the following reactions. What is the value of K2? H2A(aq) + H2O(1) <=> H30+(aq) + HA (aq) HA-(aq) + H20(1) <=> H30*(aq) + A2+ (aq) H2A(aq) + 2 H2O(l) <=> 2 H30*(aq) + A+ (aq) o bbw WOO ES K EN WS 1.3E-13 - 2.3E-7 1.0E-7 1.3E-27 ZZE12 < Previne Consider the following reaction which has AH = -1311 k). If the following changes are...
(10) 1. The diprotic acid, H2A, has Kai i.e. (K1) = 1.00 X 10 and K2 = 1.00 X 108. a) Consider a solution of 0.100 M H2A. Calculate the pH, and calculate the following concentrations: (H2A), (HA) and (A2). b) Consider a solution of 0.100 M NaHA. Calculate the pH, and calculate the following concentrations: (H2A), CHA') and (AP).
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