A solution is prepared by diluting concentrated HNO3to2.0M, 0.30Mand 0.0063M
HNO3
at25°C.Calculate [H3O+
],[OH–
],pH and pOH of the three solutions.
To find the [H3O+], [OH-], pH, and pOH of the three solutions, we can use the formula:
pH = -log[H3O+]
pOH = -log[OH-]
pH + pOH = 14
First, let's calculate the [H3O+], [OH-], pH, and pOH of the 2.0 M HNO3 solution:
[H3O+] = 2.0 M
[OH-] = 1.0 x 10^-14 / [H3O+] = 1.0 x 10^-14 / 2.0 = 5.0 x 10^-15 M
pH = -log(2.0) = -0.30
pOH = -log(5.0 x 10^-15) = 14 + log(5.0 x 10^-15) = 14 + 14.30 = 28.30
Next, let's calculate the [H3O+], [OH-], pH, and pOH of the 0.30 M HNO3 solution:
[H3O+] = 0.30 M
[OH-] = 1.0 x 10^-14 / [H3O+] = 1.0 x 10^-14 / 0.30 = 3.3 x 10^-14 M
pH = -log(0.30) = 0.52
pOH = -log(3.3 x 10^-14) = 14 + log(3.3 x 10^-14) = 14 + 13.48 = 27.48
Finally, let's calculate the [H3O+], [OH-], pH, and pOH of the 0.0063 M HNO3 solution:
[H3O+] = 0.0063 M
[OH-] = 1.0 x 10^-14 / [H3O+] = 1.0 x 10^-14 / 0.0063 = 1.6 x 10^-12 M
pH = -log(0.0063) = 2.2
pOH = -log(1.6 x 10^-12) = 14 + log(1.6 x 10^-12) = 14 + 11.80 = 25.80
Therefore, the [H3O+], [OH-], pH, and pOH of the three solutions are as follows:
2.0 M HNO3:
[H3O+] = 2.0 M
[OH-] = 5.0 x 10^-15 M
pH = -0.30
pOH = 28.30
0.30 M HNO3:
[H3O+] = 0.30 M
[OH-] = 3.3 x 10^-14 M
pH = 0.52
pOH = 27.48
0.0063 M HNO3:
[H3O+] = 0.0063 M
[OH-] = 1.6 x 10^-12 M
pH = 2.2
pOH = 25.80