pOH and Dissociation Constant

[Soaring Crane]

1. A bottle of strong monoprotic acid was labelled as having a concentration of 2.040 x 10-1 mol/L. Given that KW = 1.00 x 10-14, determine the p0H of the acid solution.

[H3O+][OH-]= K_w

[OH-] = k_W/[H3O+] = (1.00*10^-14)/(2.040*10^-1 M) = 4.90196078E-14

-log[OH-] = pOH
-log[4.90196E-14] = pOH = 13.3096 = 13.3

2. The pH of a 0.2700 molar solution of unknown monoprotic acid was measured and found to be 5.75. Calculate the Ka of this acid.

pH = -log[H3O+]
5.75 = -log[H3O+]
antilog[-5.75] = [H3O+] = 0.000001778 = X

K_A = [H3O+][A-]/[HA] = X^2/[0.2700 - X] = (0.000001778)/(0.2700 - 0.000001778) = 1.17122166E-11 = 1.17E-11

Thanks.

[Haxx0rm4ster]

For the first one you could have just found the pH and added that number from 14 to get pOH. But anyway, stick to your way since you probably memorized it already.


K_A = [H3O+][A-]/[HA] = X^2/[0.2700 - X] = (0.000001778)^2/(0.2700 - 0.000001778) = 1.17122166E-11 = 1.17E-11


Though your answer is right (you just posted what you did after finding right solution, right?), you forgot to put square 1.778E-6 in the message.