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Calculation of molarity of chloride ion
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Topic: Calculation of molarity of chloride ion (Read 1680 times)
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sanguine
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Calculation of molarity of chloride ion
«
on:
December 31, 2018, 10:55:16 AM »
Hi, just trying to self-teach from an old book and I figured I'd join just to check my work if that's okay.
25 ml of NaCl required 17.9 ml of 0.100 mol/L AgNO3 for complete reaction.
What is the molarity of the chloride ion?
Balanced equation: AgNO3 + NaCl = AgCl + NaNO3.
Taking 1/1 ratio:
0.100 mol / 0.025 L = 4 M
Finding the molarity of ion:
NaCl = 4 M
Cl x 1 = 4 M
Answer:
4 M (correct?)
Thanks in advance
«
Last Edit: December 31, 2018, 11:10:22 AM by sanguine
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AWK
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Re: Calculation of molarity of chloride ion
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Reply #1 on:
December 31, 2018, 11:13:31 AM »
Wrong answer. Concentration of chloride should be lower than silver nitrate.
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Borek
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Re: Calculation of molarity of chloride ion
«
Reply #2 on:
December 31, 2018, 01:44:05 PM »
Quote from: sanguine on December 31, 2018, 10:55:16 AM
0.100 mol / 0.025 L = 4 M
0.100 is not the number of moles of AgNO
3
(expressed in moles), it is concentration of the AgNO
3
solution (expressed in moles per liter).
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ChemBuddy
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sanguine
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Re: Calculation of molarity of chloride ion
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Reply #3 on:
January 01, 2019, 09:48:51 PM »
Fix:
(17.9 mL)•(0.100 mol/L) = (0.0179 L)•(0.100 mol/L) = 0.00179 mol
mol AgNO3 = mol NaCl = CV. Solving for C, C = (mol NaCl)/V = (0.00179 mol)/(0.025 L) = 0.0716 mol/L. Since [NaCl] = [Cl–], the concentration of chloride ion is 0.0716 mol/L.
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Borek
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Re: Calculation of molarity of chloride ion
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Reply #4 on:
January 02, 2019, 02:35:23 AM »
Looks OK.
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ChemBuddy
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Calculation of molarity of chloride ion