A buffer solution is made by dissolving 0.45
moles of a weak acid (HA) and 0.13 moles of
KOH into 830 mL of solution. What is the
pH of this buffer? Ka = 4 × 10−6
for HA.
Answer in units of pH.
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2018-05-09 8:45 am
回答 (1)
2018-05-09 8:58 am
HA + OH⁻ → A⁻ + H₂O
OH⁻ is the limiting reactant/reagent.
No. of moles of HA after reaction = (0.45 - 0.13) mol = 0.32 mol
No. of moles of A⁻ after reaction = 0.13 mol
Hence, in the final solution, [A⁻]/[HA] = 0.13/0.32
Henderson-Hasselbalch equation:
pH = pKa + log([A⁻]/[HA]) = -log(4 × 10⁻⁶) + log(0.13/0.32) = 5.0
OH⁻ is the limiting reactant/reagent.
No. of moles of HA after reaction = (0.45 - 0.13) mol = 0.32 mol
No. of moles of A⁻ after reaction = 0.13 mol
Hence, in the final solution, [A⁻]/[HA] = 0.13/0.32
Henderson-Hasselbalch equation:
pH = pKa + log([A⁻]/[HA]) = -log(4 × 10⁻⁶) + log(0.13/0.32) = 5.0
收錄日期: 2021-05-01 22:23:21
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