A lab technician adds 2.50 mL of a 0.00450 M stock solution of barium chloride to a test
tube, then adds other solutions so that the total volume is 10.00 mL. In the absence of reactions, find a) the molarity of barium ion and b) the molarity of chloride ion.
If you can explain your steps, I would greatly appreciate it! Thank you!
Equilibrium Question.?
2016-05-09 10:17 pm
回答 (1)
2016-05-09 10:39 pm
Number of milli-moles of BaCl₂ = (0.00450 mmol/mL) × (2.5 mL) = 0.01125 mmol
1 mole of BaCl₂ contains 1 mole of Ba²⁺ ions and 2 moles of Cl⁻ ions.
Number of moles of Ba²⁺ ions = 0.01125 mmol
Number of moles of Cl⁻ ions = (0.01125 mmol) × 2 = 0.0225 mmol
When the volume of the solution changes to 10.00 mL, the amount of BaCl₂ does not change.
(a) Molarity of Ba²⁺ ions = (0.01125 mmol) / (10 mL) = 0.001125 mol/L
(b) Molarity of Cl⁻ ions = (0.0225 mmol) / (10 mL) = 0.00225 mol/L
1 mole of BaCl₂ contains 1 mole of Ba²⁺ ions and 2 moles of Cl⁻ ions.
Number of moles of Ba²⁺ ions = 0.01125 mmol
Number of moles of Cl⁻ ions = (0.01125 mmol) × 2 = 0.0225 mmol
When the volume of the solution changes to 10.00 mL, the amount of BaCl₂ does not change.
(a) Molarity of Ba²⁺ ions = (0.01125 mmol) / (10 mL) = 0.001125 mol/L
(b) Molarity of Cl⁻ ions = (0.0225 mmol) / (10 mL) = 0.00225 mol/L
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