Calculate the [OH−] and the pH of a solution with an [H+]=0.0071 M at 25 °C .?
回答 (2)
2020-08-04 12:35 am
[OH⁻] = Kw/[H⁺] = (1 × 10⁻¹⁴)/0.0071 M = 1.41 × 10⁻¹² M
pH = -log[H⁺] = -log(0.0071) = 2.15
pH = -log[H⁺] = -log(0.0071) = 2.15
2020-08-04 12:11 am
Given,
[H+] = 0.0071 M
Autoionization constant of water, Kw = 10-14 at 25 oC
Now,
Kw = [H+][OH-] = 10-14
or, [OH-] = 10-14/(0.0071)
= 1.408 x 10^-12M
Hence, the concentration of OH- = 1.408 x 10^-12 M
Now,
pH = 14 - pOH
= 14 + log[OH-]
= 14 + log(1.408 x 10^-12)
= 14 - 11.85
= 2.14
[H+] = 0.0071 M
Autoionization constant of water, Kw = 10-14 at 25 oC
Now,
Kw = [H+][OH-] = 10-14
or, [OH-] = 10-14/(0.0071)
= 1.408 x 10^-12M
Hence, the concentration of OH- = 1.408 x 10^-12 M
Now,
pH = 14 - pOH
= 14 + log[OH-]
= 14 + log(1.408 x 10^-12)
= 14 - 11.85
= 2.14
收錄日期: 2021-04-18 18:35:24
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