Acid Base Calculator

Weak acids are compounds that partially dissociate to produce an equilibrium concentration of H⁺. An example equilibrium is acetic acid in water :

CH₃COOH + H₂O <--> CH₃COO^- + H₃O⁺

Weak bases partially react with water to produce an equilibrium concentration of OH^-. An example equilibrium is ammonia in water:

NH₃ + H₂O <--> NH₄⁺ + OH^-

Sample Weak Acid Problem (Exact Solution)

What is the pH of a 0.200 M solution of acetic acid (K_a = 1.75E-5)

These are the important equations:

CH₃COOH + H₂O <--> CH₃COO^- + H₃O⁺

( [H₃O⁺] [CH₃COO^-] ) / [CH₃COOH] = 1.75 x 10^-5

Searching for the [H₃O⁺]

[H₃O⁺] = [CH₃COO^-] = x

The [CH₃COOH] started at 0.2 M and went down as CH₃COOH molecules dissociated.

0.2 - x

We now have our equation:

1.75E-5 = x² / (0.2 - x)

-x² - 1.75E-5x + 3.5E-6 = 0

Solving this equation, we get:

x = 0.00186 M

pH -log 0.00186 = 2.73

Sample Weak Acid Problem (Approximation)

What is the pH of a 0.200 M solution of acetic acid (K_a = 1.75E-5)

These are the important equations:

CH₃COOH + H₂O <--> CH₃COO^- + H₃O⁺

( [H₃O⁺] [CH₃COO^-] ) / [CH₃COOH] = 1.75 x 10^-5

Searching for the [H₃O⁺]

[H₃O⁺] = [CH₃COO^-] = x

The [CH₃COOH] started at 0.2 M and went down as CH₃COOH molecules dissociated. In this approximation we simplify by ignoring that molecules dissociate from the original conc. substituting (0.2-x) by 0.2, since x is rather small.

This new equation is easy to solve:

1.75E-5 = x² / 0.2

(x²)^1/2 = 3.5E-6^1/2

Solving this equation, we get:

x = 0.00187 M

pH = -log 0.00187 = 2.73

Sample Weak Acid Problem - Acid Base Calculator Solution