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Reactivity 3.1.2—A pair of species differing by a single proton is called a conjugate acid–base pair. Deduce the formula of the conjugate acid or base of any Br.nsted–Lowry base or acid.
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[N/A]Directly related questions
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19M.1A.SL.TZ1.27:
Which has the strongest conjugate base?
A. HCOOH (Ka = 1.8 × 10−4)
B. HNO2 (Ka = 7.2 × 10−4)
C. HCN (Ka = 6.2 × 10−10)
D. HIO3 (Ka = 1.7 × 10−1)
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19M.1A.SL.TZ1.27:
Which has the strongest conjugate base?
A. HCOOH (Ka = 1.8 × 10−4)
B. HNO2 (Ka = 7.2 × 10−4)
C. HCN (Ka = 6.2 × 10−10)
D. HIO3 (Ka = 1.7 × 10−1)
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19M.1A.SL.TZ1.27:
Which has the strongest conjugate base?
A. HCOOH (Ka = 1.8 × 10−4)
B. HNO2 (Ka = 7.2 × 10−4)
C. HCN (Ka = 6.2 × 10−10)
D. HIO3 (Ka = 1.7 × 10−1)
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19M.1A.SL.TZ1.27:
Which has the strongest conjugate base?
A. HCOOH (Ka = 1.8 × 10−4)
B. HNO2 (Ka = 7.2 × 10−4)
C. HCN (Ka = 6.2 × 10−10)
D. HIO3 (Ka = 1.7 × 10−1)
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19M.2.SL.TZ2.5a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
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19M.2.SL.TZ2.5a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
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19M.2.SL.TZ2.a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
-
19M.2.SL.TZ2.5a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
-
19M.2.SL.TZ2.5a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
-
19M.2.SL.TZ2.a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
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19M.2.HL.TZ1.2a:
Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
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19M.2.HL.TZ1.2a:
Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
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19M.2.HL.TZ1.a:
Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
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19M.2.HL.TZ1.2a:
Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
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19M.2.HL.TZ1.2a:
Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
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19M.2.HL.TZ1.a:
Draw the structure of the conjugate base of benzoic acid showing all the atoms and all the bonds.
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19M.2.SL.TZ2.5a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
-
19M.2.SL.TZ2.5a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
-
19M.2.SL.TZ2.a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
-
19M.2.SL.TZ2.5a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
-
19M.2.SL.TZ2.5a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
-
19M.2.SL.TZ2.a(ii):
The hydrogencarbonate ion, produced in Equilibrium (2), can also act as an acid.
State the formula of its conjugate base.
- 19N.2.SL.TZ0.4a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.4a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.4a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.4a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.4a(ii): The value of Ka at 298 K for the first dissociation is 5.01 × 10−4. State, giving a reason, the...
- 19N.2.SL.TZ0.4a(ii): The value of Ka at 298 K for the first dissociation is 5.01 × 10−4. State, giving a reason, the...
- 19N.2.SL.TZ0.a(ii): The value of Ka at 298 K for the first dissociation is 5.01 × 10−4. State, giving a reason, the...
- 19N.2.SL.TZ0.4a(ii): The value of Ka at 298 K for the first dissociation is 5.01 × 10−4. State, giving a reason, the...
- 19N.2.SL.TZ0.4a(ii): The value of Ka at 298 K for the first dissociation is 5.01 × 10−4. State, giving a reason, the...
- 19N.2.SL.TZ0.a(ii): The value of Ka at 298 K for the first dissociation is 5.01 × 10−4. State, giving a reason, the...
- 19N.2.SL.TZ0.4a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.4a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.4a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.4a(i): Identify a conjugate acid–base pair in the equation.
- 19N.2.SL.TZ0.a(i): Identify a conjugate acid–base pair in the equation.
- 21M.2.SL.TZ1.2b(i): State the formula of its conjugate base.
- 21M.2.SL.TZ1.2b(i): State the formula of its conjugate base.
- 21M.2.SL.TZ1.b(i): State the formula of its conjugate base.
- 21M.2.SL.TZ1.2b(i): State the formula of its conjugate base.
- 21M.2.SL.TZ1.2b(i): State the formula of its conjugate base.
- 21M.2.SL.TZ1.b(i): State the formula of its conjugate base.
- 21N.1A.SL.TZ0.21: What is the conjugate acid of HS−? A. H2S B. S2− C. H2SO3 D. H2SO4
- 21N.1A.SL.TZ0.21: What is the conjugate acid of HS−? A. H2S B. S2− C. H2SO3 D. H2SO4
- 21N.1A.SL.TZ0.21: What is the conjugate acid of HS−? A. H2S B. S2− C. H2SO3 D. H2SO4
- 21N.1A.SL.TZ0.21: What is the conjugate acid of HS−? A. H2S B. S2− C. H2SO3 D. H2SO4