Weak Acid equilibria

Unit 8: Khan academy. Lesson 3

Write the standard equation for a strong acid reacting with water.

Include the K_a

HA (aq) + H₂O (l) → H₃O⁺ (aq) + A^- (aq)

K_a = [H₃O⁺][A^-] / [HA]

For a strong acid is K_a < 1 or K_a > 1?

Explain.

Because it is a strong acid, K_a > 1 because there would be more products and less reactants. Strong acids completely ionize in water.

For weak acids is K_a < 1 or K_a > 1?

Explain.

Because it is a weak acid, K_a < 1 because there would be less products and more reactants. Weak acids partially ionize in water.

What is the equation for percent ionization?

([H₃O⁺]_eq / [HA]_init.) x 100%

How would I solve a problem given the M of the solution and pH while trying to look for K_a?

1. Use pH = -log(H₃O⁺)

   1. Plug in the pH value and solve for [H₃O⁺]

2. Create an ice table.

   1. Once you solve for [H₃O⁺] in the previous step, take this value and plug it into the equilibrium of H₃O⁺

   2. Solve for the rest of the ice table accordingly

3. Solve for K_a

   1. Once you have all the values for E in the ice table, plug them into the K_a equation

   2. Solve for K_a

Determine which of these are acidic, nuetral, or basic.

K_a < K_b

K_a > K_b

K_a ~ K_b

Acidic: K_a > K_b

Neutral: K_a ~ K_b

Basic: K_a < K_b