更新1:
What is the pH of a solution in which 15 mL of 0.83 M NaOH is added to 25 mL of 0.83 M HCl? pH =
What is the pH?
回答 (2)
2016-02-27 6:36 pm
NaOH + HCl → NaCl + H₂O
1 mole of NaOH reacts with 1 mole of HCl.
Initial number of moles of NaOH = 0.83 × (15/1000) = 0.01245 mol
Initial number of moles of HCl = 0.83 × (25/1000) = 0.02075 mol
Number of moles of HCl unreacted = 0.02075 - 0.01245 = 0.0083 mol
Volume of the final solution = (15 + 25)/1000 = 0.04 L
[HCl] = 0.0083/0.04 = 0.2075 M
HCl is a strong acid which completely dissociates to give hydronium ions.
[H₃O⁺] = 0.2075 M
pH = -log(0.2075) = 0.68
1 mole of NaOH reacts with 1 mole of HCl.
Initial number of moles of NaOH = 0.83 × (15/1000) = 0.01245 mol
Initial number of moles of HCl = 0.83 × (25/1000) = 0.02075 mol
Number of moles of HCl unreacted = 0.02075 - 0.01245 = 0.0083 mol
Volume of the final solution = (15 + 25)/1000 = 0.04 L
[HCl] = 0.0083/0.04 = 0.2075 M
HCl is a strong acid which completely dissociates to give hydronium ions.
[H₃O⁺] = 0.2075 M
pH = -log(0.2075) = 0.68
2016-02-27 6:27 pm
excess of acid: (25*0.83 - 15*0.83) / 40 M = 0.208 M
so
pH = 0.68
so
pH = 0.68
收錄日期: 2021-04-18 14:31:04
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https://hk.answers.yahoo.com/question/index?qid=20160227101903AAnldc2