The pH scale

The pH scale

We usually deal with acid / base solution in the concentration range 10^-1 to 10^-7M . To express the strength of such low concentrations, Sorensen introduced a logarithmic scale known as the pH scale. The term pH is derived from the French word ‘Purissance de hydrogene’ meaning, the power of hydrogen. pH of a solution is defined as the negative logarithm of base 10 of the molar concentration of the hydronium ions present in the solution.

pH = - log ₁₀[H₃O⁺ ]      .....(8.5)

The concentration of H₃O⁺ in a solution of known pH can be calculated using the following expression.

 [H₃O⁺] = 10^-pH (or) [H₃O⁺] = antilog of (-pH)   .....(8.6)

Similarly, pOH can also be defined as follows

pOH = - log ₁₀[OH^-]     .....(8.7)

As discussed earlier, in neutral solutions, the concentration of [H₃O⁺ ] as well as [OH⁺] is equal to 1 × 10^-7 M at 25ºC . The pH of a neutral solution can be calculated by substituting this H₃O⁺ concentration in the expression (8.5)

pH = - log₁₀ [H₃O⁺]

= - log ₁₀ 10^-7

= (-7)(-1) log₁₀10= +7 (1)= 7    { [ log₁₀¹⁰ =1] }

Similary, we can calculate the pOH of a neutral solution using the expression (8.7), it is also equal to 7.

The negative sign in the expression (8.5) indicates that when the concentration of [H₃O⁺ ] increases the pH value decreases. For example, if the [H₃O⁺ ] increases from to 10^-7 to10^-5 M , the pH value of the solution decreases from 7 to 5. We know that in acidic solution, [H₃O⁺ ] > [OH^- ] , i.e., [H₃O⁺ ] > 10^-7 . Similarly in basic solution [H₃O⁺ ] < 10^-7. So, we can conclude that acidic solution should have pH value less than 7 and basic solution should have pH value greater than 7.

Relation between pH and pOH

A relation between pH and pOH can be established using their following definitions

pH = -log₁₀[H₃O⁺]    .....(8.5)

pOH = - log₁₀[OH^-]    .....(8.7)

Adding equation (8.5) and (8.7)

pH + pOH = - log₁₀[H₃O⁺] - log₁₀[OH^-]

= - ( log₁₀[H₃O⁺] + log₁₀[OH^-])

pH+pOH = -log₁₀[H₃O⁺][OH^-]       [log a+logb = logab]

We know that [H₃O⁺][OH^-]=K_w

⇒ pH + pOH = -log₁₀ K_w

⇒ pH+pOH = pK_w        [pK_w = -log₁₀ K_w]

at 25º C, the ionic product of water, K_w =1 × 10^-14

pK_w = - log₁₀10^-14 = 14 log₁₀10

 = 14

∴ (8.7) ⇒ ∴ At 25º C, pH + pOH= 14

Example 8.2

Calculate the pH of 0.001M HCl solution

H₃O⁺ from the auto ionisation of H₂O (10^-7M) is negligible when compared to the H₃O⁺ from 10^-3M HCl.

Hence [H₃O⁺ ]= 0.001 mol dm^–3

pH = -log₁₀ [H₃O⁺]

= -log₁₀ (0.001)

= -log₁₀ (10^-3 ) = 3

Note: If the concentration of the acid or base is less than 10^–6 M, the concentration of H₃O⁺ produced due to the auto ionisation of water cannot be negleted and in such cases

[H₃O⁺] = 10^-7(from water) + [H₃O⁺] (from the acid)

similarly, [OH^-] = 10^-7 M (from water) + [OH^-] (from the base)

Example 8.3

Calculate pH of 10^-7 M HCl

If we do not consider [H₃O⁺ ] from the ionisation of H₂O,

then [H₃O⁺ ] = [HCl] = 10^-7M

i.e., pH = 7, which is a pH of a neutral solution. We know that HCl solution is acidic whatever may be the concentration of HCl i.e, the pH value should be less than 7. In this case the concentration of the acid is very low (10^-7M) Hence, the H₃O⁺ (10^-7M) formed due to the auto ionisation of water cannot be neglected.

so, in this case we should consider [H₃O⁺ ] from ionisation of H ₂O

[H₃O⁺ ] = 10^-7 (from HCl) + 10^-7 (from water)

= 10^-7 (1+1)

= 2 ×10^-7

pH = -log₁₀[H₃O⁺ ]

= - log 2 - (-7).log₁₀ 10

= 7-log 2

= 7-0.3010 = 6.6990

= 6.70

Evaluate yourself – 6

a. Calculate pH of 10^-8M H₂ SO₄

b. Calculate the concentration of hydrogen ion in moles per litre of a solution whose pH is 5.4

c. Calculate the pH of an aqueous solution obtained by mixing 50ml of 0.2 M HCl with 50ml 0.1 M NaOH

a) Answer

 In this case the concentration of H₂SO₄ is very low and hence [H₃O⁺ ] from water cannot be neglected

 âˆ´[H₃O⁺] = 2 ×10^-8 (from H₂SO₄ ) + 10^-7

(from water)

= 10^-8 (2+10)

= 12×10^-8 = 1.2 ×10^-7

pH = - log₁₀[H₃O⁺]

 = - log₁₀ (1.2 × 10^-7)

= 7 - log₁₀1.2

= 7 -0.0791

= 6.9209

b) Answer

pH of the solution = 5.4

[H₃O⁺] = antilog of (-pH)

 = anitlog of (-5.4)

  = antilog of (-6 + 0.6) = 6.6

 = 3.981×10^-6

 i.e., 3.98 × 10^-6 mol dm^-3

c) Answer

No of moles of HCl = 0.2× 50 × 10^-3 = 10 × 10^-3

No of moles of NaOH = 0.1 × 50 × 10^-3 = 5 × 10^-3

No of moles of HCl after mixing = 10 × 10^-3 - 5 × 10^-3

 = 5 ×10^-3

after mixing total volume = 100mL

∴ Concentration of HCl in moles per litre =  5×10^-3mole /  100×10^-3L

[H₃O⁺] = 5 ×10^-2 M

pH = - log (5 × 10^-2 )

 = 2 - log 5

= 2 - 0.6990

= 1.30