Question

Values for Kw at 0, 50, and 100

Answer 1

(a)

Given, 4.30 x 10^-2 M NaOH solution.

So, [OH-] = 4.30 x 10^-2 M

We know, [H+][OH-] = Kw

[H+] [ 4.30 x 10^-2 M] = 1.14 x 10^-15 (since, it is given, value of Kw = 1.14 x 10^-15 at 0 oC)

[H+] = 2.65 x 10^-14 M

pH = -log[H+]

pH = -log [2.65 x 10^-14]

pH = 13.6

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(b)

Given, 4.30 x 10^-2 M NaOH solution.

So, [OH-] = 4.30 x 10^-2 M

We know, [H+][OH-] = Kw

[H+] [ 4.30 x 10^-2 M] = 5.47 x 10^-14 (since, it is given, value of Kw = 5.47 x 10^-14 at 50 oC)

[H+] = 1.27 x 10^-12 M

pH = -log[H+]

pH = -log [1.27 x 10^-12]

pH = 11.9 = 12

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(c)

Given, 4.30 x 10^-2 M NaOH solution.

So, [OH-] = 4.30 x 10^-2 M

We know, [H+][OH-] = Kw

[H+] [ 4.30 x 10^-2 M] = 4.9 x 10^-13 (since, it is given, value of Kw = 4.9 x 10^-13  at 100 oC)

[H+] = 1.14 x 10^-11 M

pH = -log[H+]

pH = -log [1.14 x 10^-11]

pH = 10.6 = 11