Question

1. Calculate pH of 10^-8 M H₂SO₄ 

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

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

Answer 1

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

[H₃O⁺] = 2 x 10^-8 (from H₂SO₄) + 10^-7(from water) 

= 10^-8 (2+ 10)

= 12 x 10^-8 = 1.2 x 10^-7

pH = – log [H₃O⁺] 

= – log₁₀ (1.2 x 10^-7) 

= 7 – log₁₀1.2 

= 7 – 0.0791 = 6.9209

2. pH of the solution = 5.4 

[H₃O⁺] = antilog of (- pH)

3. No. of moles of HCl = 0.2 x 50 x 10^-3 = 10 x 10^-3 

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

No. of moles of HCl after mixing = 10 x 10^-3 – 5 x 10^-3 = 5 x 10^-3

after mixing total volume = 100 mL

[H₃O⁺] = 5 x 10^-2 M

pH = – log ( 5 x 10^-2) 

= 2 – log 5 

= 2 – 0.6990 

= 1.30

Comments

In the first question, the concentration of H3O+ ions from sulphuric acid is taken as 2×10^-8. However, in the dissociation of sulphuric acid, only one mole of H3O+ is produced. The equation is as follow:

 H2SO4 + H2O —> HSO4- + H3O+(all are in aqueous state)

This will make the concentration of H3O+ 1×10^-8. If this is not the case, please let me know the reason for using 2H3O+.

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∵ H2SO4 is a dibasic acid and Here we assume all H2SO4 completely ionized as
H2SO4 ⇌ 2H^+  + SO4^-2