Chemical Forums
Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: sanguine on December 31, 2018, 10:55:16 AM
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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
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Wrong answer. Concentration of chloride should be lower than silver nitrate.
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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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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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Looks OK.