Answer.
(a) (first part)
We all know at 250C , pH + pOH = 14, pH =-log[H3O+] and pOH =-log[OH-]
pH = 8.61
so, pOH = (14 -8.61) =5.39
[H3O+] = 10(-8.61) = 2.45 x 10-9 (M)
[OH-] = 10(-5.39) = 4.07 x 10-6 (M)
(b) (second part)
We all know at 250C , pH + pOH = 14, pH =-log[H3O+] and pOH =-log[OH-]
pH = 11.23
so, pOH = (14 -11.23) =2.77
[H3O+] = 10(-11.23) = 5.88 x 10-12 (M)
[OH-] = 10(-2.77) = 1.69 x 10-3 (M)
(c) (third part)
We all know at 250C , pH + pOH = 14, pH =-log[H3O+] and pOH =-log[OH-]
pH = 2.94
so, pOH = (14 -2.94) =11.06
[H3O+] = 10(-2.94) = 1.14 x 10-3 (M)
[OH-] = 10(-11.06) = 8.70 x 10-12 (M)
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