Brønsted-Lowry Acid: A substance that donates a proton (H+$$H^+$$).
Brønsted-Lowry Base: A substance that accepts a proton (H+$$H^+$$).
Acetic Acid + Water:
Acetic acid acts as the acid, donating a proton to water.
Water acts as the base, accepting a proton from acetic acid.
Acetate is the conjugate base (formed from the acid).
Hydronium ion (H3O+$$H_3O^+$$) is the conjugate acid (formed from the base).
$$CH3COOH + H2O \rightleftharpoons CH3COO^- + H3O^+$$
Ethylamine + Water:
Ethylamine acts as the base, accepting a proton from water.
Water acts as the acid, donating a proton to ethylamine.
Ethylammonium is the conjugate acid (formed from the base).
Hydroxide ion (OH−$$OH^−$$) is the conjugate base (formed from the acid).
$$CH3CH2NH2 + H2O \rightleftharpoons CH3CH2NH_3^+ + OH^-$$
Definition: A substance that can act as both an acid and a base. Water is a common example.
Acid Dissociation Constant (KA):
$$KA = \frac{[A^-]{eq}[H3O^+]{eq}}{[HA]_{eq}}$$
* Where:
* $$[A^-]_{eq}$$ is the equilibrium concentration of the conjugate base.
* $$[H_3O^+]_{eq}$$ is the equilibrium concentration of the hydronium ion.
* $$[HA]_{eq}$$ is the equilibrium concentration of the acid.
Base Dissociation Constant (KB):
$$KB = \frac{[BH^+]{eq}[OH^-]{eq}}{[B]{eq}}$$
* Where:
* $$[BH^+]_{eq}$$ is the equilibrium concentration of the conjugate acid.
* $$[OH^-]_{eq}$$ is the equilibrium concentration of the hydroxide ion.
* $$[B]_{eq}$$ is the equilibrium concentration of the base.
Strength Correlation:
Large KA$$K_A$$ value = Strong acid.
Large KB$$K_B$$ value = Strong base.
Example Values (from the slides):
KA=1.8×10−5$$K_A = 1.8 \times 10^{-5}$$
KB=5.0×10−4$$K_B = 5.0 \times 10^{-4}$$
Use arrows to show bond breaking and formation during the reaction with water.
Identify the acid, base, conjugate acid, and conjugate base in the reaction.
Classify the acid or base as weak or strong.
Write the corresponding $$KAor$$ or $$KB$$ expression.
Hydrochloric Acid (HCl):
KA=1.3×106$$ K_A = 1.3 \times 10^6$$
$$HCl + H2O \rightarrow H3O^+ + Cl^-$$
Acid | Hydrochloric Acid |
Base | Water |
Conjugate Base | Chloride |
Conjugate Acid | Hydronium Ion |
Acid Strength | Strong Acid |
Benzoate ($$C6H5COO^−$$):
KB=1.5×10−10$$ K_B = 1.5 \times 10^{-10}$$
$$C6H5COO^- + H2O \rightleftharpoons C6H_5COOH + OH^-$$
Acid | Water |
Base | Benzoate |
Conjugate Base | Hydroxide |
Conjugate Acid | Benzoic Acid |
Base Strength | Weak Base |
Ammonium (NH4+$$NH_4^+$$):
KA=5.6×10−10$$ K_A = 5.6 \times 10^{-10}$$
$$NH4^+ + H2O \rightleftharpoons H3O^+ + NH3$$
Acid | Ammonium |
Base | Water |
Conjugate Base | Ammonia |
Conjugate Acid | Hydronium |
Acid Strength | Weak Acid |
KA>1$$K_A > 1$$: Strong acid.
KB<1$$K_B < 1$$: Weak base.
KA<1$$K_A < 1$$: Weak acid.
Acids and Bases Notes
CH3COOH+H2O⇌CH3COO−+H3O+
CH3CH2NH2+H2O⇌CH3CH2NH3++OH−
KA=[HA]eq[A−]eq[H3O+]eq
* Where:
* [A−]eq is the equilibrium concentration of the conjugate base.
* [H3O+]eq is the equilibrium concentration of the hydronium ion.
* [HA]eq is the equilibrium concentration of the acid.
KB=[B]eq[BH+]eq[OH−]eq
* Where:
* [BH+]eq is the equilibrium concentration of the conjugate acid.
* [OH−]eq is the equilibrium concentration of the hydroxide ion.
* [B]eq is the equilibrium concentration of the base.
KA=1.3×106
HCl+H2O→H3O++Cl−
Acid | Hydrochloric Acid |
Base | Water |
Conjugate Base | Chloride |
Conjugate Acid | Hydronium Ion |
Acid Strength | Strong Acid |
KB=1.5×10−10
C6H5COO−+H2O⇌C6H5COOH+OH−
Acid | Water |
Base | Benzoate |
Conjugate Base | Hydroxide |
Conjugate Acid | Benzoic Acid |
Base Strength | Weak Base |
KA=5.6×10−10
NH4++H2O⇌H3O++NH3
Acid | Ammonium |
Base | Water |
Conjugate Base | Ammonia |
Conjugate Acid | Hydronium |
Acid Strength | Weak Acid |