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2 Answers
- RameshwarLv 710 years agoFavorite Answer
[ H+ ] = 3x10^-8 + 10^-7 [ 10^-7 that of water [H+] ]
= 10^-8 [ 3 + 10 ]
= 13x 10^-8
pH = - log[13x10^-8]
= 8 - log 13
= 8 - 1.1139
=6. 886 answer
[ note --- when solution is very very dilute then [H+] of water is taken into consideration ]
- Anonymous10 years ago
pH=-log(H+)
Since HCl is strong enough to completely deprotonate, you can simply plug 3.0x10^-8 into the above equation.
pH=7.52
