Question

A 0.100 M solution of an unknown acid, HA, is 1.47% ionized. Complete the following table, but instead of using the variable x, calculate the number values of each quantity and report these values to the correct number of significant figures. Fill in all of the cells. Include signs in the Change column to indicate a gain or loss of concentration. Do not include any units in your answers, M is assumed. To enter values in scientific notation follow the convention: 0.00753 = 7.53E-3 What is the K_a of HA? Calculate your answer to three sig figs. K_a = Now calculate the pH of the solution of HA. pH=

Answer 1

HA dissociates as: - - - - - > H+ + A- o Initial Change Equilibrium HA 0.1 -X 0.1-x of x * *

1)

% dissociation = (x*100)/c

1.47= x*100/0.1

x = 0.0015

HA dissociates as:

HA -----> H+ + A-

Initial 0.1 0 0

Change -x +x +x

Equilibrium 0.1-x x x

Ka = [H+][A-]/[HA]

Ka = x*x/(c-x)

Ka = 1.47*10^-3*1.47*10^-3/(0.1-1.47*10^-3)

Ka = 2.193*10^-5

Answer: 2.19E-5

2)

[H+] = x = 1.47*10^-3 M

use:

pH = -log [H+]

= -log (1.47*10^-3)

= 2.8327

Answer: 2.83