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

Dear student,

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

pH=-log[H+]

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

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

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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