Question

Be sure to answer all parts. The pH of a 0.051 M weak monoprotic acid is 3.41. Calculate the K, of the acid. x 10 K = Enter your answer in scientific notation.

Answer 1

Ka = 2.99 * 10^-6

or approx value of Ka = 3.00 * 10^-6

PH = -log[ht 0 3.41 = -log (²) log x= -3.41 x= antilog (-3:41) antilog (0.59-4) x= 3.89 x 10-47 log (3.89) = 0.59 .: antilog (0.59) - 3.89 log (10-4) = -4 antilog (-4) = 10 H+ + A HA weak acid 0.051 M initially change equlibrium (0.051-x) Ka = [H +] . [A-] [HA] (3.89 x 10 9 2 0.051 - 3.89*10-4 (0.051-x) 15.1321 *10-8 0.050611 ka = 15.1321 x 10-8 0.051 0.000389 Ka = 298.99 x 10-8 = 2.99x10-6