Answer :
pH = – log [H+]
for 10⁻⁷ M HCl
[H+] HCl = 1.0 x 10⁻⁷
but HCl is dissolve in water so
[H+] total = [H+] acid + [H+] water
= 1.0 x 10⁻⁷ + 1.0 x 10⁻⁷ { [H+] [OH-] = 1.0 x 10⁻¹⁴ }
=10⁻⁷ [ 1+1 ]
[H+] total = 2.0 x 10⁻⁷
pH = – log [H⁺]
= – log [ 2.0 x 10⁻⁷ ]
= – log 2.0 +(– log 10⁻⁷ )
= – 0.3010 +7.0000
= 6.699
pH ≈ 6.7
for 10⁻⁷ M HCl
[H+] HCl = 1.0 x 10⁻⁷
but HCl is dissolve in water so
[H+] total = [H+] acid + [H+] water
= 1.0 x 10⁻⁷ + 1.0 x 10⁻⁷ { [H+] [OH-] = 1.0 x 10⁻¹⁴ }
=10⁻⁷ [ 1+1 ]
[H+] total = 2.0 x 10⁻⁷
pH = – log [H⁺]
= – log [ 2.0 x 10⁻⁷ ]
= – log 2.0 +(– log 10⁻⁷ )
= – 0.3010 +7.0000
= 6.699
pH ≈ 6.7