AcidsLesson 9

Weak AcidspH Calculations

1. Calculate the pH of 0.45 M HCN. You need an ICE chart for weak acids or bases!

HCN ⇄ H+ + CN-

I 0.45 M 0 0C x x xE 0.45 - x x x

Ka = [H+][CN-] = 4.9 x 10-10

[HCN]

Ka = x2 = 4.9 x 10-10

0.45 - x

The Ka is small, so x is small

We will find that x = 0.000015

0.45 - 0.000015 = 0.45

This means we can make the approximation that 0.45 - x = 0.45

We can do this anytime the Ka has an exponent of 10-4 or less

Ka = x2 = 4.9 x 10-10

0.45 - x

x2 = 4.9 x 10-10

0.45

x = [H+] = 0.000014849 M

pH = -Log[0.000014849]

pH = 4.83

2 sig figs due to molarity and Ka

2. Calculate the pH of 0.60 M H3BO3

H3BO3 ⇄ H+ + H2BO3-

I 0.60 M 0 0C x x xE 0.60 - x x x

0 small ka

x2 = 7.3 x 10-10

0.60

x2 = 7.3 x 10-10

0.60

x = [H+] = 2.09 x 10-5 M

pH = -Log[2.09 x 10-5]

pH = 4.68

2 sig figs due to molarity and Ka

3. Calculate the pH of a 0.20 M diprotic acid with a Ka = 4.7 x 10-7

H2X ⇄ H+ + HX-

I 0.20 M 0 0C x x xE 0.20 - x x x

0 small ka

x2 = 4.7 x 10-7

0.20

x = [H+] = 3.066 x 10-4 M

pH = -Log[3.066 x 10-4]

pH = 3.51

2 sig figs due to molarity and Ka

4. Calculate the pH of a saturated solution of Mg(OH)2.

This is a solubility equilibrium- no ICE

Mg(OH)2(s) ⇄ Mg2+ + 2OH-

s s 2s

Ksp = [Mg2+]][OH-]2 = 5.6 x 10-12

[s][2s]2= 5.6 x 10-12

4s3 = 5.6 x 10-12

s = 1.119 x 10-4 M

2s = [OH-] = 2.237 x 10-4 M

pOH = 3.65

pH = 10.35