(16-14) Given that Kw for water is 2.4 x 10^-14 at 37 degree C, compute the...

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(16-14)

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Answer 1

Answer #1

Kw = 2.4 x 10^-14

Kw = [H+][OH-]

in neutral solution [H+] = [OH-]

Kw = [H+]^2

2.4 x 10^-14 = [H+]^2

[H+] = 1.549 x 10^-7

pH = -log [H+] = -log(1.549 x 10^-7) = 6.809

pH= 6.809 -----------------------------------------------> answer

now the pH scale is re-constructed at 37⁰C. at this temperature pH scale contain 0 to 13.619 .

below 6.809 acidic above 6.809 basic

now answer for the second question

at pH =7 solution is basic -----------------------------------------> answer

Answer 2

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