I need solution for it.
Titration of a strong acid with a strong base
a student tritrated 25.00ml of a 0.100m solution of hydrochloric acid, HCl with a solution of 0.100m NaOH
a) write out the balanced equation for this reaction.
b)what is the initial pH of hydrochloric acid solution?
c)what is the pH after the student added 12.50ml of the base to the flask?
d)what is the pH after student added 24.50ml of the base to the flask?
e) what is the pH after student added 25.00ml of the base to the flask?(what is the name of this point called?)
f)what is the pH after student added 25.10ml of the base to the flask?
acid base equilibrium?
2016-03-22 4:21 pm
回答 (1)
2016-03-22 4:52 pm
✔ 最佳答案
a)HCl + NaOH → NaCl + H₂O
b)
Initial [H⁺] = Initial [HCl] = 0.1 M
pH = -log[H⁺] = -log(0.1) = 1
c)
Mole ratio HCl : NaOH = 1 : 1
25.00 mL of 0.100 M HCl needs 25.00 mL of 0.100 M NaOH for complete neutralization.
Now, only 12.50 mL NaOH is added.
It is insufficient to react with all HCl, and thus some HCl is unreacted.
Unreacted [HCl] = (0.100 × 25.00 - 0.100 × 12.50) / (25.00 + 12.50) = 0.0333 M
[H⁺] = [HCl] = 0.0333 M
pH = -log[H⁺] = -log(0.0333) = 1.48
d)
24.50 mL NaOH is added.
It is insufficient to react with all HCl, and thus some HCl is unreacted.
Unreacted [HCl] = (0.100 × 25.00 - 0.100 × 24.50) / (25.00 + 24.50) = 0.00101 M
[H⁺] = [HC] = 0.00101 M
pH = -log[H⁺] = -log(0.00101) = 3.00
e)
25.00 mL NaOH is added.
Both HCl and NaOH just completely react, and the final solution is neutral.
pH = 7.00
f)
25.10 mL NaOH is added.
It is in excess, and thus some NaOH is unreacted.
Unreacted [NaOH] = (0.100 × 25.10 - 0.100 × 25.00) / (25.10 + 25.00) = 0.000200 M
[OH⁻] = [NaOH] = 0.000200 M
pOH = -log[OH⁻] = -log(0.000200) = 3.70
pH = 14 - pOH = 14 - 3.70 = 10.30
收錄日期: 2021-04-18 14:37:52
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