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. Calculate the pH and the pOH of each of the following solutions at 25 °C for which the substances ionize completely:
(a) 0.200 M HCl

(b) 0.0143 M NaOH

(c) 3.0 M HNO3

(d) 0.0031 M Ca(OH)2

Respuesta :

lastking1765
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
mickymike92

Based on the molarity of the solutions;

  • For 0.200 M HCl; pH = 0.699, pOH = 13.301
  • For 0.0143 M NaOH; pOH = 1.845, pH = 12.16
  • For 3.0 M HNO3; pH = -0.4771, poH = 14.4771
  • For 0.0031 M Ca(OH)2; pOH = 2.21, pH = 11.79

What pHand pOH?

pH is the negative logarithm to base ten of the hydrogen ions concentration of a solution.

  • pH = -log[H+]

pOH is the negative logarithm to base ten of the hydroxide ions concentration.

  • pOH = -log[OH-]

Also;

  • pH + pOH = 14

For HCl:

pH = -log [H3O+]

pH =-log (0.200)

pH = 0.699

Then;

poH= 14-0.699

pOH = 13.301

For NaOH:

pOH= -log [OH-]

= -log (0.0143)

pOH = 1.845

Then;

pH= 14-poH

= 14- 1.845

pH = 12.16

For HNO3:

pH= -log[H3O+]

=-log(3.0)

= -0.4771

Then;

pOH = 14-9-0.4771

pOH = 14.4771

For [Ca(OH)2]

molarity = 0.0031M

2 moles of OH- are produced

[OH-]= 2 × 0.0031

[OH-] = 0.0062M

pOH = - log[OH-]

=-log(0.0062)

=-log(6.2x10-3)

=-(-2.21)

pOH = 2.21

Then;

pH =14-2.21

pH =11.79

Learn more about pH and pOH at: https://brainly.com/question/13557815

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