### Dilution Problems#11 - 25

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: A 74.31 g sample of Ba(OH)2 is dissolved in enough water to make 2.450 L of solution. How many mL of this solution must diluted with water in order to make 1.000 L of 0.1000 molar Ba(OH)2.

Solution:

1) Determine molarity of first solution:

MV = mass / molar mass

(x) (2.450 L) = 74.31 g / 171.3438 g/mol

x = 0.1770161 M (I'll carry some guard digits.)

2) Determine volume required for dilution:

M1V1 = M2V2

(0.1770161 mol/L) (x) = (0.1000 mol/L) (1000 mL)

x = 564.9 mL (to four sig figs)

Problem #20: How many milligrams of MgI2 must be added to 234.0 mL of 0.0720 M KI to produce a solution with [I¯] = 0.1000 M?

Solution:

moles I¯ already dissolved ---> (0.0720 mol/L) (0.2340 L) = 0.016848 mol (KI produces iodide ion in a 1:1 molar ratio)

moles I¯ desired in solution ---> (0.1000 mol/L) (0.2340 L) = 0.02340 mol

moles I¯ needed ---> 0.02340 − 0.016848 = 0.006552 mol

Every MgI2 dissolved yields two iodine ions, so:

moles MgI2 required ---> 0.006552 / 2 = 0.003276 mol

mass MgI2 required ---> (0.003276 mol) (278.105 g/mol) = 0.911072 g

convert to mg ---> 0.911072 g = 911 mg (to 3 sig figs, based on the 0.0720 M)

Problem #21: Determine the percentage concentration of the solution prepared by mixing 150 g of 20% solution and 250 g of 40%.

Solution #1:

150 g solution x 20% solute = 30 g solute
250 g solution x 40% solute = 100 g solute

400 g solution containing 130 g of solute

% solute = (g solute / g solution) x 100 = (130 / 400) x 100 = 32.5%

Solution #2:

C1M1 + C2M2 = C3M3

where C = concentration and M = mass of solution

(20) (150) + (40) (250) = (x) (400)

x = 32.5%

Problem #22: What volume of 0.416 M Mg(NO3)2 should be added to 255 mL of 0.102 M KNO3 to produce a solution with a concentration of 0.278 M NO3¯ ions? Assume volumes are additive.

Solution:

We need to consider the 0.416M Mg(NO3)2 solution for the standpoint of just the nitrate ion, in which case the concentration is twice the 0.416 value. So, [NO3¯] in the Mg(NO3)2 solution is 0.832 M.

In like manner, the [NO3¯] in 0.102 M KNO3 is 0.102 M.

M1V1 + M2V2 = M3V3

(0.832 mol/L) (x) + (0.102 mol/L) (255 mL) = (0.278 mol/L) (255 + x)

0.832x + 26.01 = 70.89 + 0.278x

0.554x = 44.88

x = 81.01083 mL

to three sig figs, the answer is 81.0 mL

Problem #23: What concentration of ClO3¯ results when 603 mL of 0.761 M AgClO3 is mixed with 921 mL of 0.277 M Mn(ClO3)2?

Solution:

1) AgClO3 ionizes as follows:

AgClO3 ---> Ag+ + ClO3¯

2) moles AgClO3 dissolved:

(0.761 mol/L) (0.603 L) = 0.458883 mol

3) From the chemical equation, one mole of AgClO3 dissolving yields one mole of ClO3¯ in solution.

moles ClO3¯ = 0.458883 mol

4) Mn(ClO3)2 dissolves as follows:

Mn(ClO3)2 ---> Mn2+ + 2ClO3¯

5) moles Mn(ClO3)2 dissolved:

(0.277 mol/L) (0.921 L) = 0.255117 mol

6) From the chemical equation, one mole of Mn(ClO3)2 dissolving yields two moles of ClO3¯ in solution.

moles ClO3¯ = 0.255117 mol x 2 = 0.510234 mol

7) total moles ClO3¯ in solution:

0.510234 mol + 0.458883 mol = 0.969117 mol

8) total volume of solution:

0.921 L + 0.603 L = 1.524 L

9) molarity of ClO3¯:

0.969117 mol / 1.524 L = 0.636 M (to three sig figs)

Problem #24: Assuming the volumes are additive, what is the [NO3¯] in a solution obtained by mixing 265 mL of 0.290 M KNO3, 318 mL of 0.437 M Mg(NO3)2, and 774 mL of H2O

Solution:

1) Moles nitrate ion from KNO3:

(0.290 mol/L) (0.265 L) = 0.07685 mol

Since one nitrate ion is produced for every one KNO3 that dissolves, 0.07685 mol of nitrate ion is produced.

2) Moles nitrate ion from Mg(NO3)2:

(0.437 mol/L) (0.318 L) = 0.138966 mol

Since two nitrate ions are produced for every one Mg(NO3)2 that dissolves,

0.138966 mol x 2 = 0.277932 mol

of nitrate ion is produced.

3) Total moles of nitrate produced:

0.138966 mol + 0.277932 mol = 0.416898 mol

4) Total volume of solution:

265 mL + 318 mL + 774 mL = 1357 mL = 1.357 L

5) Molarity of nitrate ion:

0.416898 mol / 1.357 L = 0.307 M (to three sig figs)

Problem #25: Describe the preparation of 900.0 mL of 3.00 M HNO3 from the commercial reagent that is 70.5% HNO3 (w/w) and has a specific gravity of 1.42.

Solution:

Let us start with 1.00 kg of the 70.5% reagent. From the 70.5%, we know that 705 g of HNO3 are present. From the specific gravity of 1.42, we know the density to be 1.42 g/mL, so the volume of the 1.00 kg of solution is this:

1000 g / 1.42 g/mL = 704.2 mL

moles of HNO3 ---> 705 g / 63.0119 g/mol = 11.19 mol

molarity of the 70.5% solution ---> 11.19 mol / 0.7042 L = 15.89 M

Now, use M1V1 = M2V2

(3.00 mol/L) (900 mL) = (15.89 mol/L) (x)

x = 169.92 mL

To three sig figs, this is 170. mL

Diluting 170. mL of the 70.5% solution to a final volume of 900.0 mL will create the 3.00 molar solution we desire.

Bonus Problem: A vat contains 40 liters of a 5% salt solution. How much of a 20% salt solution should be added to get a 10% solution?

Solution:

1) We can write the following equation:

C1V1 + C2V2 = C3V3

(5%) (40 L) + (20%) (x) = (10%) (y)

2) Since we have two variables we need to find another equation. We assume that all the volumes are additive, so:

40 L + x = y

3) We can now write one equation in one unknown:

(5%) (40 L) + (20%) (x) = (10%) (40 + x)

4) A bit of algebra (ignore all units):

200 + 20x = 400 + 10x

10x = 200

x = 20

So you need to add 20 L of the 20% salt solution to the 40 L of 5%. The result is 60 L of 10% solution.