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Dilution Problems
#1 - 10
Problem #1: If you dilute 175 mL of a 1.6 M solution of LiCl to 1.0 L, determine the new concentration of the solution.
Solution:
M1V1 = M2V2(1.6 mol/L) (175 mL) = (x) (1000 mL)
x = 0.28 M
Note that 1000 mL was used rather than 1.0 L. Remember to keep the volume units consistent.
Problem #2: You need to make 10.0 L of 1.2 M KNO3. What molarity would the potassium nitrate solution need to be if you were to use only 2.5 L of it?
Solution:
M1V1 = M2V2(x) (2.5 L) = (1.2 mol/L) (10.0 L)
x = 4.8 M
Please note how I use the molarity unit, mol/L, in the calculation rather than the molarity symbol, M.
Problem #3: How many milliliters of 5.0 M copper(II) sulfate solution must be added to 160 mL of water to achieve a 0.30 M copper(II) sulfate solution?
Solution:
The Final Take 1 0 2
M1V1 = M2V2(5.00 mol/L) (x) = (0.3 mol/L) (160 + x)
5x = 48 + 0.3x
4.7x = 48
x = 10. mL (to two sig figs)
The solution to this problem assumes that the volumes are additive. That's the '160 + x' that is V2.
Problem #4: What volume of 4.50 M HCl can be made by mixing 5.65 M HCl with 250.0 mL of 3.55 M HCl?
Solution:
Here is the first way to solve this problem:
M1V1 + M2V2 = M3V3(3.55) (0.250) + (5.65) (x) = (4.50) (0.250 + x)
Where x is volume of 5.65 M HCl that is added
(0.250 + x) is total resultant volume
0.8875 + 5.65x = 1.125 + 4.50 x
1.15x = 0.2375
x= 0.2065 L
Total amount of 4.50 M HCl is then (0.250 + 0.2065) = 0.4565 L
Total amount = 456.5 mL
Here is the second way to solve this problem:
Since the amount of 5.65 M added is not asked for, there is no need to solve for it.M1V1 + M2V2 = M3V3
(3.55) (250) + (5.65) (x − 250) = (4.50) (x)
That way, x is the answer you want, the final volume of the solution, rather than x being the amount of 5.65 M solution that is added.
Problem #5: A 40.0 mL volume of 1.80 M Fe(NO3)3 is mixed with 21.5 mL of 0.808M Fe(NO3)3 solution. Calculate the molar concentration of the final solution.
Solution:
Let's use a slightly different way to write the subscripts:
M1V1 + M2V2 = M3V3
There is no standard way to write the subscripts in problems of this type.
Substituting:
(1.80) (40.0) + (0.808) (21.5) = (M3) (40.0 + 21.5)M3 = 1.45 M
Problem #6: To 2.00 L of 0.445 M HCl, you add 3.88 L of a second HCl solution of an unknown concentration. The resulting solution is 0.974 M. Assuming the volumes are additive, calculate the molarity of the second HCl solution.
Solution #1:
M1V1 + M2V2 = M3V3(0.445) (2.00) + (x) (3.88) = (0.974) (2.00 + 3.88)
0.890 + 3.88x = 5.72712
3.88x = 4.83712
x = 1.25 M (to three sig figs)
Solution #2:
1) Calculate moles HCl in 0.445 M solution:
(0.445 mol/L) (2.00 L) = 0.890 moles
2) Set up expression for moles of HCl in second solution:
(x) (3.88 L) = moles HCl in unknown solution
3) Calculate moles of HCl in final solution:
(0.974 mol/L) (5.88 L) = 5.73 moles
4) Moles of HCl in two mixed solutions = moles of HCl in final solution:
0.890 moles + [(x) (3.88 L)] = 5.73 molesx = 1.25 M (to three sig figs)
Problem #7: To what volume should you dilute 133 mL of an 7.90 M CuCl2 solution so that 51.5 mL of the diluted solution contains 4.49 g CuCl2?
Solution:
1) Find moles:
(4.49g CuCl2) (1 mole CuCl2 / 134.45 grams) = 0.033395 moles CuCl2
2) Find the molarity of the 51.5 mL of the diluted solution that contains 4.49g CuCl2:
(0.033395 moles CuCl2) / (0.0515 liters) = 0.648 M
3) Use the dilution formula:
M1V1 = M2V2(7.90 M) (133 mL) = (0.648 M) (V2)
V2 = 1620 mL
You should dilute the 133 mL of an 7.90 M CuCl2 solution to 1620 mL.
Problem #8: If volumes are additive and 95.0 mL of 0.55 M KBr is mixed with 165.0 mL of a BaBr2 solution to give a new solution in which [Br¯] is 0.65 M, what is the concentration of the BaBr2 used to make the new solution?
Solution:
moles of Br¯ from KBr: (0.55 mol/L) (0.095 L) = 0.05225 molmoles of Br¯ in final solution: (0.65 mol/L) (0.260 L) = 0.169 mol
moles Br¯ provided by the BaBr2 solution: 0.169 − 0.05225 = 0.11675 mol
BaBr2 provides two Br¯ per formula unit so (0.11675 divided by 2) moles of BaBr2 are required for 0.11675 moles of Br¯ in the solution.
molarity of BaBr2 solution: 0.058375 mol / 0.165 L = 0.35 M
Problem #9: 1.00 L of a solution is prepared by dissolving 125.6 g of NaF in it. If I took 180 mL of that solution and diluted it to 500 mL, determine the molarity of the resulting solution.
Solution:
1) Calculate moles of NaF:
125.6 g / 41.9 g/mol = 3.00 mol
2) Calculate moles in 180 mL of resulting solution:
The Final Take 1 0 Xanax
3.00 mol in 1000 mL so 3 x (180/1000) = 0.54 mol in 180 mL
3) Calculate molarity of diluted solution:
0.54 mol / 0.50 L = 1.08 mol/L = 1.08 M
Problem #10: What is the molar concentration of chloride ions in a solution prepared by mixing 100.0 mL of 2.0 M KCl with 50.0 mL of a 1.50 M CaCl2 solution?
(Warning: there's a complication in the solution. It has to do with the CaCl2.)
Solution #1:
1) Get total moles of chloride:
KCl ⇒ (2.00 mol/L) (0.100 L) = 0.200 mol of chloride ionCaCl2 ⇒ (1.50 mol/L) (0.0500 L) (2 ions / 1 formula unit) = 0.150 mol of chloride ion.
The '2 ions / 1 formula unit' is the problem child. The solution is 1.50 M in calcium chloride, but 3.00 M in just chloride ion.
total moles = 0.200 mol + 0.150 mol = 0.350 mol
2) Get chloride molarity:
0.350 mol / 0.150 L = 2.33 M
Solution #2:
Suppose you really wanted to use this equation:
M1V1 + M2V2 = M3V3
Set it up like this:
(2.00 mol/L) (0.100 L) + (3.00 mol/L) (0.0500 L) = (M3) (0.150 L)Note that the CaCl2 molarity is 3.00 because that is the molarity of the solution from the point-of-view of the chloride ion.
Bonus Problem: What volume of a 30.% (w/v) hydrogen peroxide solution is required to prepare 425 mL of a 6.0% (w/v) solution?
Solution:
1) 6.0% (w/v) means 6 g per 100 ml of solution:
| 6 g | x | |
| ––––––– | = | ––––––– |
| 100 mL | 425 mL |
x = 25.5 g
We must use sufficient 30.% (w/v) solution to provide 25.5 g of H2O2.
2) 30.% (w/v) means 30 g of solute per 100 mL of solution:
| 30 g | 25.5 g | |
| ––––––– | = | ––––––– |
| 100 mL | x |
x = 85 mL
3) This question can be written so as to ask for the mass of 30.% (w/v) required. To do this, we follow the above steps, then this:
(85 mL) (1.10 g/mL) = 93.5 mLThe density of the solution is required, which necessitates some Internet searching. Here is an example of a site which gives a value for the density.