The Ka of propionic acid, C2H5COOH is 1.3 x 10-5. What is [OH-] in a 0.839 M sol
ID: 606508 • Letter: T
Question
The Ka of propionic acid, C2H5COOH is 1.3 x 10-5. What is [OH-] in a 0.839 M solution of C2H5COOH? Use E notation and 3 significant figures.Explanation / Answer
C2H5COOH ----------> H+ + C2H5COO- ........................ initial concentrations of C2H5COOH = 0.839 M , H+ = 0 M and C2H5COO- = 0 M..................... now let the Concentration of H+ at equilibrium be 'x' then the concentration of C2H5COO- also 'x' and the concentration of C2H5COOH is (0.839-x) M................ Now as propanoic acid is a weak acid so 'x' is very small when compared to 0.839 M...............so we can write (0.839-x) as 0.839 M only.............................Now Ka = [H+][C2H5COO-] / [C2H5COOH].......................substituting values we get................. 1.3 *10-5 = x * x / 0.839.................which gives x = 3.303*10-3 M .............therefore [H+] = 3.303*10-3 M ......................we have [H+][OH-] = 10^-14..........plugging the value of [H+] we get [OH-] = 3.028 * 10^-12 M
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