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Acid-Base Chemistryintermediate

pH of a Weak Acid Solution

Calculate the pH of a weak acid solution using the Ka value and equilibrium expressions. Essential for understanding acid-base chemistry.

Problem Scenario

Calculate the pH of a 0.15 M solution of acetic acid (CH3COOH). The Ka of acetic acid is 1.8 x 10^-5.

Given Data

Initial concentration0.15 M
Ka of acetic acid1.8 x 10^-5
Dissociation equationCH3COOH โ‡Œ H+ + CH3COO-

Requirements

  1. Write the dissociation equation
  2. Set up the ICE table
  3. Write the Ka expression
  4. Solve for [H+]
  5. Calculate pH

Solution

Step 1:

Write the dissociation equation: CH3COOH โ‡Œ H+ + CH3COO-

Step 2:

Set up ICE table: Initial: [CH3COOH] = 0.15 M, [H+] = 0, [CH3COO-] = 0. Change: -x, +x, +x. Equilibrium: (0.15 - x), x, x

Step 3:

Write Ka expression: Ka = [H+][CH3COO-]/[CH3COOH] = (x)(x)/(0.15 - x) = 1.8 x 10^-5

Step 4:

Check if approximation is valid: Since Ka << 0.15 (specifically, 0.15/Ka = 8333 > 100), we can assume x << 0.15, so (0.15 - x) โ‰ˆ 0.15. Then x^2/0.15 = 1.8 x 10^-5, so x^2 = 2.7 x 10^-6

Step 5:

Solve for x: x = sqrt(2.7 x 10^-6) = 1.64 x 10^-3 M = [H+]

Step 6:

Verify approximation: x/0.15 = 1.64 x 10^-3 / 0.15 = 0.011 = 1.1%, which is < 5%, so approximation is valid.

Step 7:

Calculate pH: pH = -log[H+] = -log(1.64 x 10^-3) = 2.79

Final Answer

The pH of the 0.15 M acetic acid solution is 2.79. This is acidic but much less so than a strong acid at the same concentration (which would have pH = 0.82) because acetic acid only partially dissociates.

Key Takeaways

  • โœ“Weak acids partially dissociate, requiring equilibrium calculations
  • โœ“The 5% rule determines if approximations are valid
  • โœ“Ka values indicate acid strength - smaller Ka means weaker acid
  • โœ“Always verify approximations after solving

Common Errors to Avoid

  • โœ—Treating weak acids like strong acids ([H+] โ‰  initial concentration)
  • โœ—Forgetting to check if the approximation is valid
  • โœ—Taking the log of x instead of -log
  • โœ—Using Kb instead of Ka for acids

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FAQs

Common questions about this problem type

When x is more than 5% of the initial concentration, the approximation fails and you need the quadratic formula. This happens when the acid is stronger (larger Ka) or more dilute.

Strong acids completely dissociate, so [H+] equals the initial acid concentration. Weak acids only partially dissociate, so we must use equilibrium calculations.

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