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Author Topic: Calculation of molarity of chloride ion  (Read 666 times)

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sanguine

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Calculation of molarity of chloride ion
« on: December 31, 2018, 04:55:16 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, 05:10:22 AM by sanguine »
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AWK

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Re: Calculation of molarity of chloride ion
« Reply #1 on: December 31, 2018, 05:13:31 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, 07:44:05 AM »

0.100 mol / 0.025 L = 4 M

0.100 is not the number of moles of AgNO3 (expressed in moles), it is concentration of the AgNO3 solution (expressed in moles per liter).
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sanguine

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Re: Calculation of molarity of chloride ion
« Reply #3 on: January 01, 2019, 03:48:51 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
« Reply #4 on: January 01, 2019, 08:35:23 PM »

Looks OK.
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