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Show that the concentration of chloride ions in the solution, in mol L^-1 is 0.061...

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xinchaoyo | Student, Grade 11 | Honors

Posted January 22, 2012 at 11:40 AM via web

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Show that the concentration of chloride ions in the solution, in mol L^-1 is 0.061 M.

Each 200 mL of an electrolyte solution designed for treating dehydration contains 0.47 g of sodium chloride (NaCl), 0.30 g of potassium chloride (KCl) and 3.56 g of glucose (C6H12O6).

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted January 22, 2012 at 2:00 PM (Answer #1)

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To solve this problem, you need to first separate important information from nonimportant information. We know we're looking for the concentration of chloride ions (Cl-) in the solution. This means we can completely ignore chemicals that do not contain Cl! You see where it says 3.56 g of Glucose? Just go ahead and white that part out, cover it up, don't worry about it, because it has NOTHING to do with the question.

Moving on, we need to calculate the amount of chloride we have. We are given the masses of NaCl and KCl added to the solution, and this will give us:

1) the moles of each salt

2) the moles of Chloride

Now, let's work with each salt individually.

We'll start with NaCl. To determine the number of moles of NaCl added, we must divide the mass of NaCl we added by the molar mass of NaCl. We find this number by adding the molar mass of Na and Cl that we can obtain from the periodic table (or the first two links below):

Moles NaCl = mass NaCl/(Molar Mass Na + Molar Mass Cl)

Moles NaCl = 0.47g/(23.0 g/mol + 35.5 g/mol)

Moles NaCl = 8.03 * 10^-3 moles

Now, for every mole of NaCl, we have one mole of Cl- (there is one chloride ion per molecule of salt), so we know the following fact:

Moles Cl- from NaCl = 8.03 * 10^-3 moles

Now, we'll continue by finding the number of moles of KCl in solution. We'll use the same technique as above, just substituting K for Na (you can find molar masses at the last two links):

Moles KCl = Mass KCl/(Molar Mass K + Molar Mass Cl)

Moles KCl = 0.30/(39.1 g/mol + 35.5 g/mol)

Moles KCl = 4.02 * 10^-3 moles

Again, for each mole of KCl in solution, we have one mole of Cl- (because, again, we have one chloride ion per molecule of salt). Therefore,

Moles Cl- from KCl = 4.02 * 10^-3 moles

Now, we can determine the total number of moles of Cl- is in our solution:

Total Moles Cl- = Moles Cl- from KCl + Moles Cl- from NaCl

Total Moles Cl- = 8.03 * 10^-3 moles + 4.02 * 10^-3 moles

Total Moles Cl- = 1.21 * 10^-2 moles

Now, we can find the concentration. Keep in mind our concentration units are moles per liter, so we'll need to convert volume units to liters. In our case, we'll need to divide 200 mL by 1000 to convert the volume to liters. To get the molar concentration, simply divide the number of moles by the volume ([Cl-] = concentration of Cl- in M = mol/L):

[Cl-] = moles Cl- / volume of solution

[Cl-] = 1.21 * 10^-2 / (200 mL/1000)

[Cl-] = 0.061 M

Side Note: There is a small problem here (REALLY small). I was only able to get the exact 0.061 M when I used 3 significant figures for the entire solution. However, in the problem, the values given have only 2 significant figures. When you use 2 significant figures in your solution, you end up with 0.060 M. Not a huge difference, but if you have quizzes on BlackBoard or some other online learning tool, this small issue can make a difference.

Hope that helps!

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