Dilution Problems
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:
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:
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:
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:
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:
Substituting:
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:
1) Calculate moles HCl in 0.445 M solution:
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:
Problem #8: If volumes are additive and 95.0 mL of 0.55 M KBr is mixed with 165.0 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:
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:
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 CaCl2solution?
(Warning: there's a complication in the solution. It has to do with the CaCl2.)
Solution #1:
1) Get total moles of chloride:
Suppose you really wanted to use this equation:
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:
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:
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:
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:
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:
Solution #2:
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:
1) Let x = the volume (in liters) of 40% acid. Therefore:
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:
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:
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:
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:
Problem #21: Determine the percentage concentration of the solution prepared by mixing 150 g of 20% solution and 250 g of 40%.
Solution #1:
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:
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:
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:
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:
Bonus Problem: 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:
How many milliliters of 2.00 M copper(II) sulfate solution must be added to 165 mL of water to achieve a 0.300 M copper(II) sulfate solution?
Solution:
Problem #27: Calculate the volume of solution prepared by diluting 6.929 mL of 3.555 M solution to 0.8229 M.
Solution:
Problem #28: Calculate the concentration of formaldehyde (CH2O) in a solution prepared by mixing 125 mL of 6.13 M CH2O and 175 mL of 4.34 M CH2O and diluting the mixture to 500.0 mL.
Solution:
1) Determine the total moles of CH2O in the two solutions being mixed:
Problem #29: Calculate the following quantity: volume of 2.48 M calcium chloride that must be diluted with water to prepare 356.0 mL of a 0.0586 chloride ion solution. (give answer in mL)
Solution:
Problem #30: The concentration of muriatic acid is 11.7 M. A diluted solution of 3.50 M is prepared. How many milliliters of 3.50 M muriatic acid solution contains 35.7 g of HCl? (give answer in mL)
Solution:
Problem #31: Determine the mass (g) of calcium nitrate in each milliliter of a solution prepared by diluting 56.0 mL of 0.705 M calcium nitrate to a final volume of 0.100 L
Solution:
1) use M1V1 = M2V2:
Problem #32: Concentrated sulfuric acid is 98.0% H2SO4 by mass and has a density of 1.84 g/mL. Determine the volume of acid required to make 1.00 L of 0.100 M H2SO4 solution.
Solution:
Problem #33a: What is the [NO3¯] in 200. mL of 0.350 M Al(NO3)3?
Problem #33b: What is the [NO3¯] in the solution above after adding 200.0 mL of 0.150 M Ca(NO3)2
Solution:
Problem #34: What volume of a 15.0% by mass NaOH solution, has a density of 1.116 g/mL, should be used to make 5.30 L of an NaOH solution with a pH of 11.00?
Solution:
1) A pH of 11.00 means a pOH of 3.00 which means a hydroxide concentration of 0.0010 M. The total moles of hydroxide is this:
Problem #35: If a solution of MgCl2 is 1/8 M, what will its concentration be if it is diluted by 27%?
Solution:
Bonus Problem: 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 4.30 L of an HCl solution with a pH of 1.86?
Solution:
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)Note that 1000 mL was used rather than 1.0 L. Remember to keep the volume units consistent.
x = 0.28 M
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)Please note how I use the molarity unit, mol/L, in the calculation rather than the molarity symbol, M.
x = 4.8 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:
M1V1 = M2V2(5.00 mol/L) (x) = (0.3 mol/L) (160 + x)The solution to this problem assumes that the volumes are additive. That's the '160 + x' that is V2.
5x = 48 + 0.3x
4.7x = 48
x = 10. mL (to two sig figs)
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)Here is the second way to solve this problem:
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
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 = M3V3There 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)Solution #2:
x = 0.125 M
1) Calculate moles HCl in 0.445 M solution:
(0.445 mol/L) (2.00 L) = 0.890 moles2) Set up expression for moles of HCl in second solution:
(x) (3.88 L) = moles HCl in unknown solution3) Calculate moles of HCl in final solution:
(0.974 mol/L) (5.88 L) = 5.73 moles4) 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 CuCl22) 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 M3) Use the dilution formula:
M1V1 = M2V2(7.90 M) (133 mL) = (0.648 M) (V2)You should dilute the 133 mL of an 7.90 M CuCl2 solution to 1620 mL.
V2 = 1620 mL
Problem #8: If volumes are additive and 95.0 mL of 0.55 M KBr is mixed with 165.0 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 mol2) Calculate moles in 180 mL of resulting solution:
3.00 mol in 1000 mL so 3 x (180/1000) = 0.54 mol in 180 mL3) 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 CaCl2solution?
(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.2) Get chloride molarity:
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
0.350 mol / 0.150 L = 2.33 MSolution #2:
Suppose you really wanted to use this equation:
M1V1 + M2V2 = M3V3Set 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
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 mol2) Determine nitrate concentration:
0.0524832 mol + 0.224756 mol = 0.2772392 mol (of nitrate)
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 mol2) Determine nitrate concentration:
0.031396 mol + 0.03924 mol = 0.070636 mol (of nitrate)
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 molsulfate: (0.309 mol/L) (0.0150 L) (1 sulfate / formula unit) = 0.004635 mol2) new molarities
chloride: (0.200 mol/L) (0.0356 L) (1 chloride / formula unit) = 0.00712 mol
sodium: 0.00927 mol / 0.0506 L = 0.183 Msulfate: 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 molfrom CaCl2: (0.484 mol/L) (0.0415 L) (2 chloride / formula unit) = 0.040172 mol2) determine [Cl-]:
0.0598874 mol + 0.040172 mol = 0.1000594 mol
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 AlCl3dissolved.Use M1V1 = M2V2: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.
(4.5 mol/L) (x) = (0.21 mol/L) (70.0 mL)x = 3.27 mL
Solution #2:
(1.5 mol/L) (x) = (0.21 mol/L) (70.0 mL)x = 9.8 mLHowever, 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 - x3) 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 acid5) We will solve this:
L acid from 40% solution + L acid from 75% solution = total acid in 50% solution6) Substitute and solve:
0.40x + (0.75)(140 - x) = 700.40x + 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 = C2V2where 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% acidx + 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/mol2) Determine volume required for dilution:
x = 0.1770161 M (I'll carry some guard digits.)
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 times 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 soluteSolution #2:
250 g solution x 40% solute = 100 g solute400 g solution containing 130 g of solute
% solute = (g solute / g solution) x 100 = (130 / 400) x 100 = 32.5%
C1M1 + C2M2 = C3M3where 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 mol3) From the chemical equation, one mole of AgClO3 dissolving yields one mole of ClO3- in solution.
moles ClO3- = 0.458883 mol4) 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 mol6) 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 mol7) total moles ClO3- in solution:
0.510234 mol + 0.458883 mol = 0.969117 mol8) total volume of solution:
0.921 L + 0.603 L = 1.524 L9) 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 molSince 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 molSince two nitrate ions are produced for every one Mg(NO3)2 that dissolves,3) Total moles of nitrate produced:
0.138966 mol x 2 = 0.277932 molof nitrate ion is produced.
0.138966 mol + 0.277932 mol = 0.416898 mol4) Total volume of solution:
265 mL + 318 mL + 774 mL = 1357 mL = 1.357 L5) 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: 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 g3) Determine volume of concentrated HCl to be diluted:
1179 g times 0.36 = 424.44 g of HCl
(424.44 g/36.46 g/mole) / 1.00 L = 11.64 M.
M1V1 = M2V2(11.64 mol/L) (x) = (0.02511 mol/L) (4.75 L)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:
x = 0.010247 L
x = 10.25 mL
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 #26:
How many milliliters of 2.00 M copper(II) sulfate solution must be added to 165 mL of water to achieve a 0.300 M copper(II) sulfate solution?
Solution:
We will assume that volumes are additive.M1V1 = M2V2
(2.00 mol/L) (x) = (0.300 mol/L) (165 + x)
2x = 49.5 + 0.3x
1.7x = 49.5
x = 29.1 mL
Problem #27: Calculate the volume of solution prepared by diluting 6.929 mL of 3.555 M solution to 0.8229 M.
Solution:
Use this equation: M1V1 = M2V2(3.555 mol/L) (6.929 mL) = (0.8229 mol/L) (x)
x = 29.93388 mL
To four sig figs, this is 29.93 mL
Problem #28: Calculate the concentration of formaldehyde (CH2O) in a solution prepared by mixing 125 mL of 6.13 M CH2O and 175 mL of 4.34 M CH2O and diluting the mixture to 500.0 mL.
Solution:
1) Determine the total moles of CH2O in the two solutions being mixed:
(6.13 mol/L) (0.125 L) = 0.76625 mol2) The total moles are diluted to a final volume of 500.0 mL:
(4.34 mol/L) (0.175 L) = 0.7595 mol0.76625 mol + 0.7595 mol = 1.52575 mol
1.52575 mol / 0.5000 L = 3.05 M (to three sig figs)Although you are not asked, you can easily determine the molarity of the two formaldehyde solutions after they are mixed, but before they are diluted to the final volume of 500.0 mL:
1.52575 mol / 0.300 L = 5.08 M (to three sig figs)You could have also used this:
M1V1 + M2V2 = M3V3and solved for M3.
Problem #29: Calculate the following quantity: volume of 2.48 M calcium chloride that must be diluted with water to prepare 356.0 mL of a 0.0586 chloride ion solution. (give answer in mL)
Solution:
The key to solving this problem is to know the formula of calcium chloride is CaCl2. That means that, while the solution is 2.48 M in CaCl2, it is 4.96 M from the perspective of just the chloride.We use M1V1 = M2V2 to solve this problem.
(4.96 mol/L) (x) = (0.0586 mol/L) (356.0 mL)
x = 4.205968 mL
to three sig figs, this is 4.20 mL
Problem #30: The concentration of muriatic acid is 11.7 M. A diluted solution of 3.50 M is prepared. How many milliliters of 3.50 M muriatic acid solution contains 35.7 g of HCl? (give answer in mL)
Solution:
Use this: MV = mass / molar mass(3.50 mol/L) (x) = 35.7 g / 36.4609 g/mol
x = 0.27975 L
to three sig figs, 280. mL
Problem #31: Determine the mass (g) of calcium nitrate in each milliliter of a solution prepared by diluting 56.0 mL of 0.705 M calcium nitrate to a final volume of 0.100 L
Solution:
1) use M1V1 = M2V2:
(0.705 mol/L) (0.0560 L) = (x) (0.100 L)x = 0.3948 M2) moles of Ca(NO3)2 in 1 mL:
0.3948 mol/L = 0.3948 mol / 1000 mL = 0.0003948 mol/mL3) Convert moles to grams:
0.0003948 mol/mL times 164.086 g/mol = 0.0648 g/mL
Problem #32: Concentrated sulfuric acid is 98.0% H2SO4 by mass and has a density of 1.84 g/mL. Determine the volume of acid required to make 1.00 L of 0.100 M H2SO4 solution.
Solution:
Assume 100. g of the solution is present. This means 98.0 g of H2SO4 are present.98.0 g / 98.0768 g/mol = 0.999217 mol
100. g / 1.84 g/mL = 54.34783 mL <--- volume of the 100. g of solution
molarity of the 98.0% solution:
0.999217 mol / 0.05434783 L = 18.3856 M
Now, use M1V1 = M2V2
(18.3856 mol/L) (x) = (0.1 mol/L) (1 L)
x = 0.005439 L = 5.44 mL
Problem #33a: What is the [NO3¯] in 200. mL of 0.350 M Al(NO3)3?
Problem #33b: What is the [NO3¯] in the solution above after adding 200.0 mL of 0.150 M Ca(NO3)2
Solution:
For every one formula unit of Al(NO3)3 that dissolves, three nitrate ions are released to the solution.0.350 M times 3 = 1.05 M in nitrate. The 200. mL is not needed.
For the second part, the 200. mL is needed.
(0.200 L) (1.05 mol/L) = 0.210 mol of nitrate
(0.200 L) (0.300 mol/L) = 0.060 mol of nitrate
0.210 + 0.060 = 0.270 mol of nitrate in 0.400 L of total volume
0.270 mol / 0.400 L = 0.675 M
I got the 0.300 by doing 0.150 times 2 since each Ca(NO3)2 delivers 2 nitrates to the solution.
Problem #34: What volume of a 15.0% by mass NaOH solution, has a density of 1.116 g/mL, should be used to make 5.30 L of an NaOH solution with a pH of 11.00?
Solution:
1) A pH of 11.00 means a pOH of 3.00 which means a hydroxide concentration of 0.0010 M. The total moles of hydroxide is this:
(0.0010 mol/L) (5.30 L) = 0.0053 mol2) The mass of NaOH needed is this:
40.0 g/mol times 0.0053 mol = 0.212 g3) The mass of 15/0% solution required is found by ratio and proportion:
0.212 g is to x as 15 is to 100x = 1.41333 g <--- kept a couple extra digits4) The volume of NaOH solution required is this:
1.41333 g / 1.116 g/mL = 1.27 mL1.27 mL of 15.0%(w/w) NaOH solution, when diluted to 5.30 L of solution, has a pH equal to 1.00.
Problem #35: If a solution of MgCl2 is 1/8 M, what will its concentration be if it is diluted by 27%?
Solution:
M1V1 = M2V2(0.125) (100 mL) = (x) (127 mL)
x = 0.098 M
Bonus Problem: 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 4.30 L of an HCl solution with a pH of 1.86?
Solution:
From the pH, you know that in the final solution, [H+] = 10-1.86 = 0.0138 M (Quite properly, the pH has only 2 significant figures. The "1" is a place-holder digit. So, formally, you should say that the solution will have [H+] = 0.014 M, but we'll keep the third digit anyway...)Since HCl is a strong acid, we know that the final [HCl] will equal 0.0138 M.
Now, the problem is to calculate the molarity of the original HCl solution, and then use M1V1 = M2V2 to get your final answer.
36.0% by mass means that 100 g of the HCl solution will contain 36.0 grams of HCl.
36.0 g HCl / 36.45 g/mol = 0.988 moles HCl in 100 g solution.
The volume of 100 g of solution is: 100 g / 1.179 g/mL = 84.8 mL or 0.0848 L. So, the molarity of the original HCl solution is:
0.988 mol / 0.0848 L = 11.6 M
Now, use M1V1 = M2V2:
(11.6 mol/L) (V1) = (0.0138 mol/L) (4.30 L)
V1 = 5.11 x 10-3 L = 5.11 mL
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