Question

. Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:

Answer 1

A. HCl:
pH= -log [H3O+]
PH=-log (0.200)
= 0.699

poH= 14-0.699
= 13.301

b. NaOH:
PoH= -log [OH-]
= -log (0.0143)
= 1.845
pH= 14-poH
= 14- 1.845
= 12.16

c. HNO3:
PH= -log[H3O+]
=-log(3.0)
= -0.4771
poH= 14-pH
= 14-9-0.4771
= 14.4771

pH= -0.4771, poH= 14.4771

d. [Ca(OH)2] = 0.0031M
[OH-]= 2X0.0031
[OH-] = 0.0062M

PoH= - log[OH-]
=-log(0.0062)
=-log(6.2x10-3)
=-(-2.21)
= 2.21
PH=14-poH
=14-2.21
=11.79
POH=2.21, PH= 11.79