Chemistry Help?!?
回答 (2)
2018-06-08 5:19 am
[OH⁻] = 7.5 × 10⁻³ M
pOH = -log[OH⁻] = -log(7.5 × 10⁻³) = 2.1
pH = pKw - pOH = 14.0 - 2.1 = 11.9
pOH = -log[OH⁻] = -log(7.5 × 10⁻³) = 2.1
pH = pKw - pOH = 14.0 - 2.1 = 11.9
2018-06-08 6:04 am
To determine the pH of a solution, you need to calculate the hydronium ion concentration. You can use the following equation
[H3O+1] * [OH-1] = 1 * 10^-14
[H3O+1] * 7.5 * 10^-3 = 1 * 10^-14
[H3O+1] = 1 * 10^-14 ÷ 7.5 * 10^-3
This is approximately 1.33 * 10^12. To determine the pH, take the negative log of the number.
pH = -1 * log (1 * 10^-14 ÷ 7.5 * 10^-3)
This is approximately 11.9.
[H3O+1] * [OH-1] = 1 * 10^-14
[H3O+1] * 7.5 * 10^-3 = 1 * 10^-14
[H3O+1] = 1 * 10^-14 ÷ 7.5 * 10^-3
This is approximately 1.33 * 10^12. To determine the pH, take the negative log of the number.
pH = -1 * log (1 * 10^-14 ÷ 7.5 * 10^-3)
This is approximately 11.9.
收錄日期: 2021-05-01 22:29:33
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