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harshit agarwal Grade: 9
        how find ph values
7 years ago

Answers : (1)

Anurag Kishore
37 Points
										

Hi

















































pH



pOH



pH is a measure of the hydrogen ion concentration, [H+]



pOH is a measure of the hydroxide ion concentration, [OH-]



pH is calculated using the following formula:
pH = -log10[H+]



pOH is calculated using the following formula:
pOH = -log10[OH-]



Example 1:


Find the pH of a 0.2mol L-1
(0.2M) solution of HCl



  • Write the balanced equation for the dissociation of the acid
    HCl -----> H+(aq) + Cl-(aq)

  • Use the equation to find the [H+]:
    0.2 mol L- HCl produces 0.2 mol L-1 H+ since HCl is a strong acid that fully dissociates

  • Calculate pH: pH = -log10[H+]
    pH = -log10[0.2] = 0.7



Example 1:


Find the pOH of a 0.1mol L-
(0.1M) solution of NaOH



  • Write the balanced equation for the dissociation of the alkali
    NaOH -----> OH-(aq) + Na+(aq)

  • Use the equation to find the [OH-]:
    0.1 mol L-1 NaOH produces 0.1 mol L-1 OH- since NaOH is a strong alkali that fully dissociates

  • Calculate pOH: pOH = -log10[OH-]
    pOH = -log10[0.1] = 1



Example 2:


Find the pH of a 0.2 mol L-1
(0.2M) solution of H2SO4



  • Write the balanced equation for the dissociation of the acid
    H2SO4 -----> 2H+(aq) + SO42-(aq)

  • Use the equation to find the [H+]:
    0.2 mol L-1 H2SO4 produces 2 x 0.2 = 0.4 mol L-1 H+ since H2SO4 is a strong acid that fully dissociates

  • Calculate pH: pH = -log10[H+]
    pH = -log10[0.4] = 0.4



Example 2:


Find the pOH of a 0.1mol L-1
(0.1M) solution of Ba(OH)2



  • Write the balanced equation for the dissociation of the alkali:
    Ba(OH)2 -----> 2OH-(aq) + Ba2+(aq)

  • Use the equation to find the [OH-]:
    0.1mol L-1 Ba(OH)2 produces 2 x 0.1 = 0.2 mol L-1 OH- since Ba(OH)2 is a strong alkali that fully dissociates

  • Calculate pOH: pOH = -log10[OH-]
    pOH = -log10[0.2] = 0.7



Hydrogen ion concentration, [H+], can be calculated using the following formula:
[H+] = 10-pH



Hydroxide ion concentration, [OH-], can be calculated using the following formula:
[OH-] = 10-pOH



Example:


Find the [H+] of a nitric acid solution with a pH of 3.0
pH= 3.0
[H+] = 10-pH
[H+] = 10-3.0 = 0.001mol L-1


You can check this answer by using the calculated value [H+] in the equation for pH to make sure you arrive at the original pH
pH = -log10[H+]
pH = -log10[0.001] = 3
We get the same value for pH using the calculated value for [H+], so the calculated value for [H+] is correct.



Example:


Find the [OH-] of a sodium hydroxide solution with a pOH of 1
pOH = 1
[OH-] = 10-pOH
[OH-] = 10-1 = 0.1 mol L-1


You can check this answer by using the calculated value [OH-] in the equation for pOH to make sure you arrive at the original pOH
pOH = -log10[OH-]
pOH = -log10[0.1] = 1
We get the same value for pOH using the calculated value for [OH-], so the calculated value for [OH-] is correct.



pH + pOH = 14



Example A(1):


Find the pH of a solution of sodium hydroxide
that has a pOH of 2


pH = 14 - pOH
pH = 14 - 2 = 12



Example B(1):


Find the pOH of a solution of hydrochloric acid
that has a pH of 3.4


pOH = 14 - pH
pOH = 14 - 3.4 = 10.6



Example A(2):


Find the [H+] in a solution of sodium hydroxide
that has a pOH of 1



  • Calculate the pH
    pH = 14 - pOH
    pH = 14 - 1 = 13

  • Calculate [H+]
    [H+] = 10-pH
    [H+] = 10-13 = 10-13mol L-1



Example B(2):


Find the [OH-] of a sulfuric acid solution
with a pH of 3



  • Calculate the pOH
    pOH = 14 - pH
    pOH = 14 - 3 = 11

  • Calculate [OH-]
    [OH-] = 10-pOH
    [OH-] = 10-11 = 10-11 mol L-1



Example A(3):


Find the pH of 0.2mol L-1 sodium hydroxide



  • Write the equation for the dissociation of NaOH:
    NaOH -----> Na+(aq) + OH-(aq)

  • Use the equation to find [OH-]:
    0.2mol L-1 NaOH produces 0.2mol L-1 OH- since NaOH is a strong base that fully dissociates

  • Calculate the pOH:
    pOH = -log10[OH-]
    pOH = -log10[0.2] = 0.7

  • Calculate pH:
    pH = 14 - pOH
    pH = 14 - 0.7 = 13.3



Example B(3):


Find the pOH of 0.2mol L-1 sulfuric acid



  • Write the equation for the dissociation of H2SO4:
    H2SO4 -----> 2H+(aq) + SO42-(aq)

  • Use the equation to find [H+]:
    0.2 mol L-1 H2SO4 produces 2 x 0.2 = 0.4 mol L-1 H+ since H2SO4 is a strong acid that fully dissociates

  • Calculate the pH:
    pH = -log10[H+]:
    pH = -log10[0.4] = 0.4

  • Calculate pOH:
    pOH = 14 - pH
    pOH = 14 - 0.4 = 13.6



 


 


Thanks


Anurag Kishore

7 years ago
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