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oliviaclerk5
10.03.2020 •
Chemistry
An analytical chemist is titrating 242.5 mL of a 1.200 M solution of hydrazoic acid HN3 with a 0.3400 M solution of NaOH. The pKa of hydrazoic acid is 4.72.
Calculate the pH of the acid solution after the chemist has added 1006.mL of the NaOH solution to it. Note for advanced students: you may assume the final volume equals the initial volume of the solution plus the volume of NaOH solution added. Round your answer to 2 decimal places.
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Ответ:
The pH of the solution is 12.61
Explanation:
Step 1: Data given
Molarity of hydrazoic acid solution = 1.200 M
Volume of solution = 242.5 mL
Molarity of NaOH solution = 0.3400 M
pKa = 4.72
Step 2: The balanced equation
HN3 + NaOH → NaN3 + H2O
Step 3: Calculate moles hydrazoic acid
Moles hydrazoic acid = molarity * volume
Moles hydra
zoic acid = 1.200 M * 0.2425 L
Moles hydrazoic acid = 0.291 moles
Step 4: Calculate moles NaOH
Moles NaOH = 0.3400 M * 1.006 L
Moles NaOH = 0.342 moles
Step 5: Calculate the limiting reactant
HN3 is the limiting reactant. It will completely be consumed (0.291 moles)
NaOH is in excess. There reacts 0.29 moles. There will remain 0.342 - 0.291 = 0.051 moles NaOH
Step 6: Calculate total volume
Total volume = 242.5 mL + 1006 mL = 1248.5 mL = 1.2485 L
Step 7: Calculate molarity of NaOH
Molarity NaOH = 0.051 moles / 1.2485 L
Molarity NaOH = 0.0408 M
Step 8: Calculate pOH
pOH = -log [OH-] = -log(0.0408)
pOH = 1.39
Step 9: Calculate pH
pH = 14 - pOH
pH = 14 - 1.39
pH = 12.61
The pH of the solution is 12.61
Ответ:
H2SiO3:
STOCK: Ácido trioxosilícico
SISTEMATICA: Trioxosilicato de hidrógeno
TRADICIONAL: Ácido metasilícico