Consider a solution containing .100 M fluoride ions and .126M hydrogen fluoride. The concentration of fluoride ions after the ad
dition of 5.00 mL of .0100 M HCl to 25.0 mL of this solution is ___M
1 answer:
Given:
Concentration of Fluoride ions = 0.100 M
Concentration of Hydrogen Fluoride = 0.126 M
Asked: Concentration of fluoride ions after the addition of 5ml of 0.0100 M HCl to 25 mL of the solution
Assume: 50:50 ratio of fluoride ions and HF
12.5ml*0.1mol/L *1L/1000mL + 12.5*0.126mol/L * 1L/1000mL = 2.825x10^-3 moles F-
5ml * 0.01 mol/L *1L/1000mL = 5x10^-5 moles
Assume: Volume additive
Final concentration = 2.825x10^-3 + 5x10^-5 moles/ 30 ml * 1000ml/L =0.0958 M
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Answer:
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Explanation:
Data
Pressure 1 = P1 = 5 atm
Volume 1 = V1 = 363 ml
Pressure 2 = P2 = ?
Volume 2 = 0.0002 ml
Process
To solve this problem use Boyle's law
P1V1 = P2V2
-Solve for P2
P2 = P1V1/V2
-Substitution
P2 = (5 x 363) / 0.0002
-Simplification
P2 = 1815 / 0.0002
-Result
P2 = 9075000 atm
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