Worksheet - Dilution - Problems 11 - 20


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Problem #11: 46.2-mL of a 0.568 M Ca(NO3)2 solution is mixed with 80.5 mL of 1.396 M calcium nitrate solution. What is the nitrate ion concentration?

Solution:

1) Determine total moles of nitrate in solution:

(0.568 mol/L) (0.0462 L) (2 nitrate / formula unit) = 0.0524832 mol

(1.396 mol/L) (0.0805 L) (2 nitrate / formula unit) = 0.224756 mol

0.0524832 mol + 0.224756 mol = 0.2772392 mol (of nitrate)

2) Determine nitrate concentration:

0.2772392 mol / 0.1267 L = 2.19 M (to three sig figs)

Problem #12: 66.8 mL of a 0.235 M solution of Ca(NO3)2 is mixed with 87.2 mL of a 0.450 solution of KNO3. What is the nitrate ion concentration?

Solution:

1) Determine total moles of nitrate in solution:

(0.235 mol/L) (0.0668 L) (2 nitrate / formula unit) = 0.031396 mol

(0.450 mol/L) (0.0872 L) (1 nitrate / formula unit) = 0.03924 mol

0.031396 mol + 0.03924 mol = 0.070636 mol (of nitrate)

2) Determine nitrate concentration:

0.070636 mol / 0.154 L = 0.459 M (to three sig figs)

Problem #13: 15.0 mL of 0.309 M Na2SO4 and 35.6 mL of 0.200 M KCl are mixed. Determine the following concentrations: [Na+]; [SO42-]; [Cl-]

Solution:

Comment: notice that no ion is in both solutions. That means that this problem is essentially three dilutions.

1) moles of each ion:

sodium: (0.309 mol/L) (0.0150 L) (2 sodium / formula unit) = 0.00927 mol

sulfate: (0.309 mol/L) (0.0150 L) (1 sulfate / formula unit) = 0.004635 mol

chloride: (0.200 mol/L) (0.0356 L) (1 chloride / formula unit) = 0.00712 mol

2) new molarities

sodium: 0.00927 mol / 0.0506 L = 0.183 M

sulfate: 0.004635 mol / 0.0506 L = 0.0916 M

chloride: 0.00712 mol / 0.0506 L = 0.141 M


Problem #14: 3.50 g of NaCl is dissolved in 41.5 mL of 0.484 M CaCl2 solution. Determine the chloride ion concentration.

Solution:

1) total moles of chloride:

from NaCl: (3.50 g / 58.443 g/mol (1 chloride / formula unit) = 0.0598874 mol

from CaCl2: (0.484 mol/L) (0.0415 L) (2 chloride / formula unit) = 0.040172 mol

0.0598874 mol + 0.040172 mol = 0.1000594 mol

2) determine [Cl-]:

0.1000594 mol / 0.0415 L = 2.41 M (to three sig figs)

Notice that I assumed the addition of the solid NaCl caused no volume change. There probably was some volume change, but, in the context of the problem, it was negligible.


Problem #15: How many milliliters of 1.5 M AlCl3 must be used to make 70.0 mL of a solution that has a concentration of 0.21 M Cl¯?

Solution #1:

Think of the 1.5 M solution of AlCl3 as being 4.5 M in chloride ion. This is because there are three chlorides in solution for every one AlCl3 dissolved.

Use M1V1 = M2V2:

(4.5 mol/L) (x) = (0.21 mol/L) (70.0 mL)

x = 3.27 mL

Notice how the mol/L cancels out and mL is left as the unit on the answer. You do not have to convert to liters for this particular type of problem, you just need to have the volumes be the same unit.

Solution #2:

(1.5 mol/L) (x) = (0.21 mol/L) (70.0 mL)

x = 9.8 mL

However, you MUST realize how this answer is incorrect. We are being asked only for the chloride ion concentration. The 9.8 mL answer would give us a solution that is 0.21 M in aluminum chloride, AlCl3. The concentration of ONLY the chloride ion is three times larger than the AlCl3 concentration.

To obtain the correct answer, we must divide the 9.8 mL by three to get 3.27 mL. Be careful on this because you might think you should mltiply by three. Oh no. In the solution, the three winds up in the denominator (note where the 4.5 goes in solution #1. 4.5 is 1.5 times 3).

Personally, the ChemTeam thinks solution #1 is the better one. However, you do see some presentations that use #2.


Problem #16: A 40% acid solution is mixed with a 75% acid solution to produce 140 liters of a 50% acid solution? How many liters of each acid solution was mixed?

Comment: this will play a role:

percent times volume gives amount of pure acid present.

Solution:

1) Let x = the volume (in liters) of 40% acid. Therefore:

the amount of pure acid in that solution = (0.40)(x)

2) Since L 40% acid + L 75% acid = 140, then

the volume of 75% acid = 140 - x

3) The amount of pure acid in this solution:

(0.75)(140 - x)

4) The last set up conerns the final solution of 50% acid:

(0.50)(140 L) = 70 L pure acid

5) We will solve this:

L acid from 40% solution + L acid from 75% solution = total acid in 50% solution

6) Substitute and solve:

0.40x + (0.75)(140 - x) = 70

0.40x + 105 - 0.75x = 70

-0.35x = -35

x = 100 L
140 - x = 40 L

100 L of the 40% solution is mixed with 40 L of the 75% solution to yield 140 L of the 50% solution.


Problem #17: Container A holds a solution that is 80% alcohol while container B holds a solution that is 20% alcohol. How many liters of the solution in container A are needed to produce 12 liters of a solution that is 60% alcohol?

Solution:

Let us use this equation:

C1V1 = C2V2

where C stands for concentration. It's a more general way to write the dilution equation.

(80) (x) + (20) (12 - x) = (60) (12)

80x + 240 - 20x = 720

60x = 480

x = 8 L

Notice that x is the volume of the 80% solution while 12 - x is the volume of the 20% solution.


Problem #18: What quantity of a 45% acid solution must be mixed with a 20% acid solution to produce 800 mL of a 29.375% solution?

Solution:

let x = mL of 45% acid
let y = mL of 20% acid

x + y must equal 800 mL, therefore y = 800 - x

(x mL) (% acidity) + (y mL) (% acidity) = (mixture mL) (mixture acidity)

(x) (0.45) + (800 - x) (0.20) = (800) (0.29875)

The rest is left to the reader. Notice how the above solution uses decimal equivalents (0.45) rather than percents (45%). It doesn't matter which you use, just as long as you are consistent. Here's the set up with percents:

(x) (45) + (800 - x) (20) = (800) (29.875)

You'll get the same answer as with decimal equivalents.


Problem #19: What volume of a concentrated HCl solution, which is 36.0% HCl by mass and has a density of 1.179 g/mL, should be used to make 4.75 L of an HCl solution with a pH of 1.6?

Solution:

1) pH 1.6 leads to this concentration:

10-1.6 = 0.02511 M H+

Since HCl is a strong acid, this is also the concentration of the HCl desired.

2) Determine molarity of concentrated HCl:

1.00 L of the HCl ---> 1000 mL times 1.179 g/mL = 1179 g
1179 g times 0.36 = 424.44 g of HCl
(424.44 g/36.46 g/mole) / 1.00 L = 11.64 M.

3) Determine volume of concentrated HCl to be diluted:

M1V1 = M2V2

(11.64 mol/L) (x) = (0.02511 mol/L) (4.75 L)

x = 0.010247 L

x = 10.25 mL

When I copied this problem from Yahoo Answers, the solution was stated differently than I put it above. I offer it here for your study:

And you should dilute this concentrated solution 11.64/0.02511 fold = 463.56 fold.

4750 mL / 463.56 = 10.25 mL. So use 10.25 mL of the concentrated HCl and fill up with water to 4.75 L.


Problem #20: To 15.00 mL of distilled water in a 50 mL beaker, you add exactly 27 drops of 0.100 M HCl and swirl to mix. If exactly 19 drops of HCl are required to make 1.000 mL, calculate the following: (a) molarity of the diluted HCl and (b) the pH of the diluted HCl.

Solution:

27 drops / 19 drops/mL = 1.421 mL

M1V1 = M2V2

(0.100 mol/L) (1.421 mL) = (x) (16.421 mL)

x = 0.00865 M

pH = -log 0.00865 = 2.063


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