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, has acid dissociation constants of ?a1=3.98×10−4 and ?a2=4.11×10−11.Calculate the pH and molar...
A diprotic acid, H2A, has acid dissociation constants of ?a1=3.98×10−4 and ?a2=4.11×10−11.Calculate the pH and molar concentrations of [H2A], [HA−],and [A2−] at equilibrium for each of the solutions. 1) A 0.199 M solution of H2A. 2) A 0.199 M solution of NaHA. 3) A 0.199 M solution of Na2A.
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
A diprotic acid, H2A, has acid dissociation constants of ?a1=3.69×10−4 and ?a2=4.08×10−12. Calculate the pH and...
A diprotic acid, H2A, has acid dissociation constants of ?a1=3.69×10−4 and ?a2=4.08×10−12. Calculate the pH and molar concentrations of H2A, HA−, and A2− at equilibrium for each of the solutions. A 0.102 M solution of H2A. A 0.102 M solution of NaHA. A 0.102 M solution of Na2A
Polyprotic acids contain more than one dissociable proton. Each dissociation step has its own acid-dissociation constant,...
Polyprotic acids contain more than one dissociable proton. Each dissociation step has its own acid-dissociation constant, Ka1, Ka2, etc. For example, a diprotic acid H2A reacts as follows: H2A(aq)+H2O(l)⇌H3O+(aq)+HA−(aq) Ka1=[H3O+][HA−][H2A] HA−(aq)+H2O(l)⇌H3O+(aq)+A2−(aq) Ka2=[H3O+][A2−][HA−] In general, Ka2 = [A2−] for a solution of a weak diprotic acid because [H3O+]≈[HA−]. Many household cleaning products contain oxalic acid, H2C2O4, a diprotic acid with the following dissociation constants: Ka1=5.9×10−2, Ka2=6.4×10−5 Part A) Calculate the equilibrium concentration of H3O+ in a 0.20 M solution of oxalic...
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) +...
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) +...
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