Simple organic chem question?
2018-03-08 4:33 pm
What volume of 0.200M k2co3 must be added to 0.1L of 0.1M khco3 to prepare a buffer solution with a pH of 11.0?
回答 (2)
2018-03-08 4:53 pm
Refer to: http://www4.ncsu.edu/~franzen/public_html/CH201/data/Acid_Base_Table.pdf
Second dissociation constant for H₂CO₃ = 4.7 × 10⁻¹¹
Let V L be the volume of K₂CO₃ added.
[CO₃²⁻]/[HCO₃⁻] = (No. of moles of K₂CO₃)/(No. of moles of KHCO₃) = 0.200V/(0.1×0.1) = 20V
HCO₃⁻(aq) ⇌ CO₃²⁻(aq) + H⁺(aq) …… Ka = 4.7 × 10⁻¹¹
pH = pKa + log([CO₃²⁻]/[HCO₃⁻])
11.0 = -log(4.7 × 10⁻¹¹) + log(20V)
log(20V) = 11.0 + log(4.7 × 10⁻¹¹)
20V = 10^[11.0 + log(4.7 × 10⁻¹¹)]
V = 10^[11.0 + log(4.7 × 10⁻¹¹)] / 20
V = 0.235
Volume of K₂CO₃ added = 0.235 L
Second dissociation constant for H₂CO₃ = 4.7 × 10⁻¹¹
Let V L be the volume of K₂CO₃ added.
[CO₃²⁻]/[HCO₃⁻] = (No. of moles of K₂CO₃)/(No. of moles of KHCO₃) = 0.200V/(0.1×0.1) = 20V
HCO₃⁻(aq) ⇌ CO₃²⁻(aq) + H⁺(aq) …… Ka = 4.7 × 10⁻¹¹
pH = pKa + log([CO₃²⁻]/[HCO₃⁻])
11.0 = -log(4.7 × 10⁻¹¹) + log(20V)
log(20V) = 11.0 + log(4.7 × 10⁻¹¹)
20V = 10^[11.0 + log(4.7 × 10⁻¹¹)]
V = 10^[11.0 + log(4.7 × 10⁻¹¹)] / 20
V = 0.235
Volume of K₂CO₃ added = 0.235 L
2018-03-08 4:51 pm
0.3 mL of QUIT gets added to 0.200 mol of CHEMISTRY. The solution is... BURN YOUR TEXTBOOK.
收錄日期: 2021-05-01 22:22:55
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