# how to determine molar concentration of an ion?

Submitted by Anonymous (not verified) on Wed, 06/22/2011 - 02:43

1.

What is the molar concentration of an NaOH (aq) solution that contains 12.0 grams of NaOH (s) in 600. mL of solution?

2.

What mass of K3PO4 (s) is required to prepare 1500. mL of 0.20 M solution?

3.

Describe how you would prepare 250. mL of 0.100 M NaCl (aq). Make sure you show the calculation of the mass of the NaCl (s) that is required and a description of the process of making the solution.

4.

What volume of 0.020 M KOH (aq) solution contains 11.2 grams of KOH (s)?

5.

What is the resulting concentration of KBr (aq) when 100. mL of water are added to 400. mL of 0.350 M KBr (aq)? Assume the volumes are additive.

6.

How much water must be added to 500. mL of 1.00 M NaCl (aq) in order to make the final [NaCl (aq)] be 0.750 M?

7.

Given the equation:

2 NaI (aq) + Cl2 (g) 2 NaCl (aq) + I2 (s)

What volume of 0.100 M NaI (aq) must completely react in order to produce 508 grams of I2 (s)?

If you are given the mass, the number of moles = mass given/molar mass

Molar mass of NaCl = 23 + 35.5 = 58.5g

No of moles in a solution = volume in L x molarity

Volume in L = no.of moles/molarity
Molarity = no. of moles/volume in L

If you had a 0.05M solution of Na2SO4 or (Na+)2SO42- this contains 0.05 moles of sodium sulphate in one litre or 0.025 moles in 0.5L

Also as each mole contain 2 mole of Na+ ions and 1 mole of SO4 2- ions, a 0.05M solution is 0.10M with respect to Na+ and 0.05M with respect to SO42- ions

cute.not.smart. wrote:

3.  Describe how you would prepare 250. mL of 0.100 M NaCl (aq). Make sure you show the calculation of the mass of the NaCl (s) that is required and a description of the process of making the solution.

You are given a target concentration with 3 significant figures, so you are expected to be accurate and precise. A 0.1004 M solution would be OK, because it still rounds to 0.100 M, but you will need good measuring equipment and technique.
To measure the final volume of solution as 250 mL, it would be a good idea to use a volumetric flask, for accuracy.
You could weigh as close to the amount of NaCl you calculate, which will be more than 1 gram. You should be able to get much closer than 1% with any analytical balance.
You could weigh into a tared weighing boat, transfer into the flask and rinse any residue into the flask with water.
You could fill the flask with water half way, and swirl until dissolved. (NaCl dissolves easily, and the less complicated the procedure, the lesser your chance of losing some of the NaCl).
After dissolved, if it did not get cold otr hot (you would check), you would dilute to the mark with water, stopper, and mix by inverting and shaking a few times.
Then you would label it with composition, date, your initials or name, and maybe storage conditions and expiration date.

cute.not.smart. wrote:

5.
What is the resulting concentration of KBr (aq) when 100. mL of water are added to 400. mL of 0.350 M KBr (aq)? Assume the volumes are additive.

6.
How much water must be added to 500. mL of 1.00 M NaCl (aq) in order to make the final [NaCl (aq)] be 0.750 M?

When you have a substance (solute) in solution in a solvent (like water),
amount of solute = volume X concentration,
amount could be in moles, or grams, or milligrams, or ....
volume could be in liters, or milliliters, or ....
concentration could also be in various units, matching the units of amount and volume.

When adding solvent (water) to your solution, the amount of solute dissolved does not change, so
amount = initial volume X initial concentration =  final volume X final concentration
As long as you use the same units for initial and final, the units for amount do not matter; you could make up some,  like "gallon x moles/L"
For example in problem 3, to find the final volume (V) in mL, you solve the equation:
(500 mL)(1.00 M) = (V)(0.750 M)
(The units are mL and M on both sides).

The problem tells you to assume that volumes are additive, because some times, you add 100 mL of water to 100 mL solution and the final volume is a little different from 200 mL. Volume does not change much on mixing when the iniital solution wass not very concentrated, so in those cases volumes are additive.

cute.not.smart. wrote:

7.

Given the equation:

2 NaI (aq) + Cl2 (g) 2 NaCl (aq) + I2 (s)

What volume of 0.100 M NaI (aq) must completely react in order to produce 508 grams of I2 (s)?

Step 1 _ Find the molecular weight or molar mass of I2
Step 2 - Find how many moles of I2 are in 508 grams I2
Step 3 _ Make sure your reaction is balanced and finde the ratio of moles of NaI and I2. Do you need 2 moles NaI to make one mole of I2?
Step 4 _ Find how many moles of NaI you will need to make that many moles of I2.
Step 5 _ Find what volume of 0.100 M NaI solution contains the needed number of moles of NaI.

Of course you could dazzle the teacher by writing all the calculations for steps 2 to 5 on one line of paper and calculating the result in one step, like:
Volume=(508 g I2)/(253.80 g I2/mole I2)(2 moles NaI/mole I2)/(0.100 mole NaI/L NaI solution)