common ion effect
2007-03-27 23:08:53 · 3 answers · asked by fallingfaster1 1 in Science & Mathematics ➔ Chemistry
Lancenigo di Villorba (TV), Italy
pH = 4.9.
Your question interests an aqueous solution of A WEAK ACID AND ONE ITS SALT. This solution constitutes a system named as BUFFER which results able to minimize pH's effects related to ACID/BASE's addictions.
Your case is related to Acid Base and Its Sodium Salt, a solution obeying to Acid-Base's Chemical Equilibrium as measured in aqueous solutions
CH3COOH(aq) <---> CH3COO-(aq) + H+(aq)
1.8E-5 = Ka = |CH3COO-| * |H+| / |CH3COOH|
The liquid amount of Acetic Acid's solution undergoes dilution when it mix with Sodium Acetate's one : nonetheless, I am interested to Molarity's Ratio instead the Molarities, so I refer to Molarity's Ratio or the Molar Amount's Ratio.
I can overcome to the pH's level of this BUFFER
|H+| = Ka * |CH3COOH| / |CH3COO-| =
= 1.8E-5 * (0.15 * 50E-3) / (0.20 * 50E-3) = 1.35E-5 M
hence pH = 0 - LOG(|H+|) = 4.9
I hope this helps you.
2007-03-27 23:28:17 · answer #1 · answered by Zor Prime 7 · 1⤊ 0⤋
As mentioned, it's a buffer, not a common ion effect. You need to find the new concentrations when combining the two solutions, then plug into the Henderson-Hasselbach equation. Each concentration will be halved in the combined solution, so you have 0.075 M acetic acid and 0.10 M sodium acetate.
pH = pKa + log ([base]/[acid]) = 4.74 + log (0.1/0.075)
pH = 4.74 + 0.12 = 4.86
2007-03-28 08:23:01 · answer #2 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋
This is actually a buffer solution. To calculate the pH, use the Henderson-Hasselbach Equation.
pH = pKa (acid) + Log {[salt]/[acid]}
pKa = - Log Ka
2007-03-28 06:31:55 · answer #3 · answered by estheryltan 3 · 0⤊ 0⤋