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`        how find ph values`
7 years ago

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 alkaliNaOH -----> 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 3.0pH= 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 pHpH = -log10[H+] pH = -log10[0.001] = 3We 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 1pOH = 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 pOHpOH = -log10[OH-]pOH = -log10[0.1] = 1We 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 - pOHpH = 14 - 2 = 12

Example B(1):
Find the pOH of a solution of hydrochloric acid that has a pH of 3.4
pOH = 14 - pHpOH = 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 pHpH = 14 - pOHpH = 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 pOHpOH = 14 - pHpOH = 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 - pOHpH = 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 - pHpOH = 14 - 0.4 = 13.6

Thanks
Anurag Kishore
```
7 years ago
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