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Finding the pH of Weak Acids

Finding the pH of Weak Acids. Strengths of Acids and Bases. “ Strength ” refers to how much an acid or base ionizes in a solution. Weak acids will generally have a higher pH than a strong acid because the [H + ] is lower.

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Finding the pH of Weak Acids

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  1. Finding the pH of Weak Acids

  2. Strengths of Acids and Bases • “Strength” refers to how much an acid or base ionizes in a solution.

  3. Weak acids will generally have a higher pH than a strong acid because the [H+] is lower. • However, you can make a weak acid more concentrated and that will lower the pH. • A 1 Molar solution of a strong acid will have a pH of 0. • This will not be the case for a 1M solution of a weak acid if only 5% dissociates. • The [H+] will be less, so pH will be higher.

  4. Concentration vs- StrengthHA ↔ H+ + A-

  5. Finding the pH of Weak Acids • Weak acids do not dissociate 100%. • Therefore, we cannot assume that the concentration of the acid equals the concentration of H+ ions like we did with the strong acids.

  6. We must use equilibrium expressions to help us solve for the pH. • Assume HA is a weak acid - • HA  H+ + A- • Then • Ka = [H+] [A-] • [HA]

  7. Generic Example- • What is the pH of a 1 Molar solution of a weak acid HA? • HA  H+ + A- • Initial 1 M 0 0 • At 1 - x x x Equilibrium • * at equilibrium “some” of the HA will dissociate, which we call “x” . Therefore, [HA] will decrease by “x” and [H+] = [A-] = x

  8. With a known Ka value and plugging in our equilibrium values we get • Ka = (x)(x) = x2 • 1 – x 1 – x • Because the Ka is small, we can assume that “x” is very small in comparison to 1 M, so we can ignore the “x” in the (1 – x) value. So, • Ka = x2 x = [H+] • 1 • from this we can get pH!!!!!

  9. Real Example Problem • What is the pH of a 0.5 Molar solution of benzoic acid , C6H5COOH, Ka = 6.6 x 10-5? • Write the equilibrium reaction and expression C6H5COOH  H+ + C6H5COO- • Initial 0.5 M 0 0 • At Eq. 0.5 - x x x • Ka = [H+] [C6H5COO-] [C6H5COOH]

  10. Ka = [H+] [C6H5COO-] • [C6H5COOH] 6.6 x 10-5 = x2 0.5 - x • We can ignore the ‘x’ as being very small compared to 0.5 in the denominator, so the equation becomes • 6.6 x 10-5 = x2 • 0.5 • Solve for ‘x’ = [H+] = 5.7 x 10-3 • Remember - solve for x2 and then take square root of both sides to solve for x!!!!

  11. Once you have the [H+], you can solve for pH pH = - log [H+] = - log (5.7 x 10-3) = 2.24 Ka = _x2 [HA] x = [H+]

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