✔ 最佳答案
1.a)
Mixing the solutions:
No. of moles of CH3COOH added = 0.1 x (25/1000) = 0.0025 mol
No. of moles of CH3COO- added = 0.1 x (50/1000) = 0.005 mol
Volume of the final solution = (25 + 50) cm3 = 75 cm3 = 0.075 dm3
[CH3COOH]o before dissociation = 0.0025/0.075 = 0.0333 M
[CH3COO-]o before dissociation = 0.005/0.075 = 0.0667 M
eqm CH3COOH(aq) ≒ CH3COO-(aq) + H+(aq)
eqm (0.0333-y) M ≒ 3(0.0667+y) M (a+ y M
Assume that 0.0333 >>y.
Then (0.0333 - y) M ≈ 0.0333 M, and (0.0667 + y) M ≈ 0.0667 M
Ka = (0.0667+y)y/(0.0333-y) = 1.8 x 10-5
0.0667y/0.0333 = 1.8 x 10-5 (Take approximation)
y = 9 x 10-6
pH = -log(9 x 10-6) = 5.05
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2.
H+(aq) + CH3COO-(aq) → CH3COOH(aq)
No. of moles of H+(aq) added = 0.442 x (184/1000) = 0.0813 mol
No. of moles of CH3COO-(aq) added = 0.4 x 0.5 = 0.2 mol
H+(aq) is the limiting reactant.
No. of moles of CH3COOH(aq) formed = 0.0813 mol
No. of moles of CH3COO-(aq) unreacted = 0.2 - 0.0813 = 0.1187 mol
Final volume = (184 mL) + (0.5 L) = 0.684 L
[CH3COOH]o = 0.0813/0.684 = 0.119 M
[CH3COO-]o = 0.1187/0.684 = 0.174 M
eqm CH3COOH(aq) ≒ CH3COO-(aq) + H+(aq)
eqm (0.119-y) Maq ≒ 3(0.174+y) M (a+ y M
Assume that 0.119 >>y.
Then (0.119 - y) M ≈ 0.119 M, and (0.174 + y) M ≈ 0.174 M
Ka = (0.174+y)y/(0.119-y) = 1.8 x 10-5
0.174y/0.119 = 1.8 x 10-5 (Take approximation)
y = 1.23 x 10-5
pH = -log(1.23 x 10-5) = 4.91
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