 # Calc. the pH of 10^-8 M acid sol.

11 years ago

Dear student,

Find the dissociation constant first and then write log[H+]

pH=-log[H+]

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Sagar Singh

B.Tech, IIT Delhi

11 years ago

10-8M acid = 10-8M H+ ion concentration...

in solution H+ from water is 10-7 M

total H+ = 10-7+10-8 = 1.01*10-7

pH = -log[H+] =-log[1.01*10-7]

= 6.85       approx

11 years ago

Molarity = 10^-8 .

n/V = 10^-8

n = 10^-8 * V

pH = -log [H+] .

Water normally dissociates: H2O <===> H+ + OH-

The dissociation constant is Kw = 10^-7

So you would have 10^-7 moles of H+ in pure water. Add the 10^-8 moles of acid, and the resulting concentration of H+ is [H+] = 1.1 x 10^-7

the negative log of 1.1 x 10^-7 = 6.96, which is the pH value.

therefore pH of the solution is 6.96

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