Calculate the pH of a solution prepared by the addition of 0.1 ml of 0.5M NaOH to 50.0ml of "pure" water.
I've done:
0.0001L×0.5M = 5x10^-5mol of NaOH
5x10^-5mol/0.05L = 1x10^-3M
...what do I do from here to get pH? Thanks!
Calculate pH of a solution?
2016-03-02 3:14 pm
回答 (1)
2016-03-02 3:56 pm
✔ 最佳答案
No. of milli-moles of NaOH in the solution = 0.5 × 0.1 = 0.05 mmolMolarity of NaOH in the final solution = 0.05/(50.0+0.1) = 9.98 × 10⁻⁴ M
[OH⁻] = Molarity of NaOH in the final solution = 9.98 × 10⁻⁴ M
pOH = -log[OH⁻] = -log(9.98 × 10⁻⁴) = 3.0
Thus, pH = 14.0 - pH = 14.0 - 3.0 = 11.0
OR :
pH = -log[H⁺] = -log(Kw/[OH⁻]) = -log{(1.0 × 10⁻¹⁴)/(9.98 × 10⁻⁴)} = 11.0
收錄日期: 2021-04-18 14:34:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20160302071420AADIng2