分析化學題 - Ans.fyi 參考答案

分析化學題

2013-01-15 4:05 am

How many grams of NaHCO3 ( 84.007)
must be added to 5.0 g K2CO3( 138.206) to give a pH
of 10.8 in 500mL of water
(H2CO3 Ka1 4.46x10^-7, Ka2 4.69x 10^-11)

請問此題如何解弄了好久謝謝

回答 (1)

胡雪8°

2013-01-15 10:03 am

✔ 最佳答案

HCO3⁻(aq) ⇌ H⁺(aq) + CO3²⁻(aq) .. Ka2 = 4.69 x 10⁻¹¹

[CO3²⁻]o = [K2CO3]o= [5/138.206] / (500/1000) = 0.07235 M

pH = pKa2 - log([HCO3⁻]/[CO3²⁻])
pH ≈ pKa2 - log([HCO3⁻]o/[CO3²⁻]o)
10.8 = -log(4.69 x 10⁻¹¹) - log([HCO3⁻]o/0.07235)
10.8 = 10.3 - log([HCO3⁻]o/0.07235)
log([HCO3⁻]o/0.07235) =-0.5
([HCO3⁻]o/0.07235) = 0.3162
[HCO3⁻]o= 0.02288 M
[NaHCO3]o = 0.02288 M

No. of moles of NaHCO3 = 0.02288 x (500/1000) = 0.01144 mol
Mass of NaHCO3 = 84.007 x 0.01144 = 0.961 g

參考: 胡雪

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