Calculate the pH of a 1.00 L solution which is made up of 0.043 mol acetic acid and 0.020 mol sodium acetate. (Ka for acetic acid is 1.8x10-5.)
2007-11-17 09:29:51 · 3 answers · asked by Anonymous in Science & Mathematics ➔ Chemistry
it's a buffer solution.
[H+] = Ka * [A-]/[HA] where [H+] is the concentration of H+, [A-] is the concentration of the acetate, and [HA] is the concentration of the acetic acid.
[H+] = 1.8 * 10e-5 * (0.020/1) / (0.043/1) = 0.8372 * 10e-5 = 8.4 *10e-6
pH = log [H+] = - log (8.4 * 10e-6) = - (0.92 + (-6)) = 5.08
2007-11-17 09:44:33 · answer #1 · answered by j.investi 5 · 0⤊ 0⤋
You have buffer conditions and can use the Henderson-Hasselbach equation, or, if you prefer, substitute directly into the definition of the equilibrium constant Ka.
2007-11-17 09:44:48 · answer #2 · answered by Facts Matter 7 · 0⤊ 0⤋
9.2 is the answer
2007-11-17 09:35:34 · answer #3 · answered by courtneyboo2093 1 · 0⤊ 0⤋