### Molarity Problems#26 - 35

Problem #26: What is the concentration of each type of ion in solution after 23.69 mL of 3.611 M NaOH is added to 29.10 mL of 0.8921 M H2SO4? Assume that the final volume is the sum of the original volumes.

Solution:

The answer requires you to know how NaOH and H2SO4 react. Here is the chemical equation:

H2SO4 + 2NaOH ---> Na2SO4 + 2H2O

A key point is that two NaOH formula units are required for every one H2SO4

1) Calculate moles of NaOH and H2SO4:

moles NaOH ---> (3.611 mol/L) (0.02369 L) = 0.08554459 mol
moles H2SO4 ---> (0.8921 mol/L) (0.02910 L) = 0.02596011 mol

2) Determine how much NaOH remains after reacting with the H2SO4:

0.02596011 mol x 2 = 0.05192022 mol <--- moles of NaOH that react

0.08554459 mol - 0.05192022 mol = 0.03362437 mol <--- moles of NaOH that remain

3) The above was required to determine the hydroxide ion concentration:

0.03362437 mol / 0.05279 L = 0.6369 M

0.05279 L is the sum of the two solution volumes.

4) Determine the sodium ion concentration:

0.08554459 mol / 0.05279 L = 1.620 M

The sole source of sodium ion is from the NaOH.

5) Determine the sulfate ion concentration:

0.02596011 mol / 0.05279 L = 0.4918 M

The sole source of sulfate ion is from the H2SO4.

6) Determine the hydrogen ion concentration:

[H+] [OH¯] = 1.000 x 10-14

(x) (0.636945823) = 1.000 x 10-14

x = 1.570 x 10-14 M

Problem #27: Given 3.50 mL of sulfuric acid (98.0% w/w) calculate the number of mmols in the solution (density: 1.840 g/mL).

Solution:

3.50 mL times 1.840 g/mL = 6.44 g <--- mass of the 3.50 mL

6.44 g times 0.980 = 6.3112 g <--- mass of H2SO4 in the solution

6.3112 g / 98.0768 g/mol = 0.06434957 mol

0.06434957 mol times (1000 mmol / 1 mol) = 64.3 mmol (to three sig figs)

Problem #28: Given 8.00 g of HBr calculate the volume (mL) of a 48.0% (w/w) solution. (MW HBr: 80.9119 g/mol, density: 1.49 g/mL). Then, calculate the molarity.

Solution:

8.00 g divided by 0.48 = 16.6667 g <--- total mass of the solution in which the HBr is 48% by mass

16.6667 g divided by 1.49 g/mL = 11.18568 mL

to three sig figs, the volume of the solution is 11.2 mL

For the molarity, determine the moles of HBr:

8.00 g / 80.9119 g/mol = 0.098873 mol

0.098873 mol / 0.01118568 L = 8.84 M

Problem #29: A solution is made by dissolving 0.100 mol of NaCl in 4.90 mol of water. What is the mass % of NaCl?

Solution:

1) Convert moles to masses:

NaCl ---> 0.100 mol times 58.443 g/mol = 5.8443 g

H2O ---> 4.90 mol times 18.015 g/mol = 88.2735 g

2) Calculate mass percent of NaCl:

[5.8443 g / (5.8443 g + 88.2735 g)] * 100 = 6.21% (to three sig figs)

Problem #30: 2.00 L of HCl gas (measured at STP) is dissolved in water to give a total volume of 250. cm3 of solution. What is the molarity of this solution?

Solution using molar volume:

2.00 L divided by 22.414 L/mol = 0.0892299 mol of HCl

0.0892299 mol / 0.250 L = 0.357 M (to three sig figs)

Solution using Ideal Gas Law:

PV = nRT

(1.00 atm) (2.00 L) = (n) (0.08206 L atm / mol K) (273.15 K)

n = 0.0892272 mol

0.0892272 mol / 0.250 L = 0.357 M (to three sig figs)

If the volume of HCl gas was not at STP, you must use PV = nRT to calculate the moles. You cannot use molar volume since it is only true at STP.

Problem #31: You need to make an 80.0 g mixture of ethanol and water containing equal molar amounts of both, what mass of each substance would be required?

Solution:

1) Some preliminary information and discussion:

Ethanol has a molar mass of 46.0684 g/mol
Water has a molar mass of 18.0152 g/mol

Moles of ethanol = moles of water = x

(x) (46.0684) = mass of ethanol in the 80.0 g total
(x) (18.0152) = mass of water in the 80.0 g total

2) Solving for x:

(x) (46.0684) + (x) (18.0152) = 80.0

(x) (46.0684 + 18.015) = 80.0

(x) (64.0836) = 80.0

x = 80.0 / 64.0836

x = 1.24837 mol

1.25 moles of each. to three sig figs

3) Doing this for the second step works too:

46.0684x + 18.015x = 80.0

Problem #32: What mass of pure sulfuric acid must be made up to 250 cm3 of aqueous solution so that the resulting solution has the same concentration as a potassium hydroxide solution containing 2.00 g of the pure alkali in 100 cm3?

Solution:

1) Determine the concentration of the KOH solution:

MV = mass / molar mass

(x) (0.1 L) = 2.00 g / 56.10564 g/mol

x = 0.35647 M

2) Determine the mass of sulfuric acid needed to make a 0.35647 M solution:

MV = mass / molar mass

(0.35647 mol/L) (0.25 L) = y / 98.0791 g/mol

y = 8.74 g

(2.00 g KOH) / (56.10564 g KOH/mol) / (100 cm3) x (250 cm3) x (98.0791 g H2SO4/mol) = 8.74 g H2SO4

4) You must recognize that the value connecting the KOH and H2SO4 solutions is that they have the same molarity. Here is the KOH set up:
 mass1 MV1 = ––––– MM1 <--- MM means molar mass

 mass1 M = ––––––––––– (MM1) (V1)

5) Here is the H2SO4 set up:

 mass2 <--- mass2 is the answer to the question MV2 = ––––– MM2

 mass2 M = ––––––––––– (MM2) (V2)

6) Since M = M, we have this:

 mass1 mass2 ––––––––––– = ––––––––––– (MM1) (V1) (MM2) (V2)

7) Cross-multiply:

(mass1) (MM2) (V2) = (mass2) (MM1) (V1)

8) Solve for mass2:

 (mass1) (MM2) (V2) mass2 = ––––––––––––––––– (MM1) (V1)

9) Insert numbers and solve:

 (2.00 g) (98.0791 g/mol) (250 cm3) mass2 = ––––––––––––––––––––––––––––– (56.10564 g/mol) (100 cm3)

Up above, I converted cm3 to L, but I did not have to. In retrospect, that can be seen in the set up just above. If you compare the set up just above to the dimensional analysis set up, you will see that they are comparable.

Problem #33: A solution is prepared by dissolving 19.20 g ammonium sulfate in enough water to make 117.0 mL of stock solution. A 12.00 mL sample of this stock solution is added to 59.20 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Solution:

1) Determine the molarity of the stock solution:

MV = mass / molar nass

(x) (0.1170 L) = 19.20 g / 132.1382 g/mol

x = 1.2419 M

2) Determine molarity of diluted solution (assume volumes are additive):

M1V1 = M2V2

(1.2419 mol/L) (12.00 mL) = (x) (71.20 mL) x = 0.209 M

3) Determine concentration of the ions:

(NH4)2SO4(aq) ---> 2NH4+(aq) + SO42¯(aq)

For every one mole of ammonium sulfate that dissolves, two moles of ammonium ion are present in solution as well as one mole of sulfate ions.

Therefore:

[NH4+] = 0.418 M
[SO42¯] = 0.209 M

Problem #34: A solution was prepared by adding pure KCl to deionized water. When the water in 1.00 mL of the solution was evaporated completely, the residue remaining weighed 90.0 mg. What was the molarity of the KCl solution.

Solution:

MV = mass / molar mass

(x) (1.00 mL) = 90.0 mg / 74.551 mg/mmole

(x) (1.00 mL) = 1.21 mmol

x = 1.21 M (to three sig figs)

Note the use of milligrams, milliliters and millimoles.

Problem #35: An aqueous solution is 0.556 M in NaCl and 0.245 M in KCl. If 100.0 mL of the solution is left uncovered and all of the water evaporates away, calculate the mass of the residue.

Solution #1:

1) Moles of each solute:

NaCl ---> (0.556 mol/L) (0.1000 L) = 0.0556 mol
KCl ---> (0.245 mol/L) (0.1000 L) = 0.0245 mol

2) Mass of each solute:

NaCl ---> (0.0556 mol) (58.443 g/mol) = 3.24943 g
KCl ---> (0.0245 mol) (74.551 g/mol) = 1.82650 g

3.24943 g + 1.82650 g = 5.07593

Use the 'rounding by 5' rule to get the final answer of 5.08 g

Solution #2: 1) Use MV = mass/molar mass to solve for the mass of NaCl.

2) Use MV = mass/molar mass to solve for the mass of KCl.

3) Add the two masses and round off to three sig figs.

Bonus Problem: How many milliliters of concentrated hydrochloric acid solution (36.0% HCl by mass, density = 1.18 g/mL) are required to produce 18.0 L of a solution that has a pH of 2.01?

Solution:

1) Get moles of hydrogen ion needed for the 18.0 L:

[H+] = 10-pH = 10-2.01 = 0.0097724 M

0.0097724 mol/L) (18.0 L) = 0.1759032 mol of HCl required

Remember, HCl is a strong acid, dissociating 100% in solution

2) Determine molarity of 36.0% HCl:

Assume 100. g of solution present.

36.0 g of that is HCl

100. g / 1.18 g/mL = 84.745763 mL

Use MV = mass / molar mass

(x) (0.084745763 L) = 36.0 g / 36.4609 g/mol

x = 11.6508 M

3) Volume of 11.6508 M acid needed to deliver 0.1759032 mol:

0.1759032 mol / 11.6508 mol/L = 0.01509795 L

15.1 mL (to three sig figs)