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Compute the pH of a solution at 25 degree centigrade which is twice as alkaline as pure water
ph of water = 7
concentration of [OH-] = 10-7
concentration of [OH-] in the solution = 2*10-7 (2 times alkaline than water)
total OH- in sol is 10-7 + 2*10-7 =3*10-7
[H+] = 10-14/3*10-7 = 3.33*10-8
PH = -log3.33*10-8 =8 - log3.33 = 7.5 approx
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