#26 - 35 (Incomplete)

Go to Molarity Problems #11-25

**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 H_{2}SO_{4}? Assume that the final volume is the sum of the original volumes.

**Solution:**

The answer requires you to know how NaOH and H_{2}SO_{4} react. Here is the chemical equation:

H_{2}SO_{4}+ 2NaOH ---> Na_{2}SO_{4}+ 2H_{2}O

A key point is that two NaOH formula units are required for every one H_{2}SO_{4}

1) Calculate moles of NaOH and H_{2}SO_{4}:

moles NaOH ---> (3.611 mol/L) (0.02369 L) = 0.08554459 mol

moles H_{2}SO_{4}---> (0.8921 mol/L) (0.02910 L) = 0.02596011 mol

2) Determine how much NaOH remains after reacting with the H_{2}SO_{4}:

0.02596011 mol x 2 = 0.05192022 mol <--- moles of NaOH that react0.08554459 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 M0.05279 L is the sum of the two solution volumes.

4) Determine the sodium ion concentration:

0.08554459 mol / 0.05279 L = 1.620 MThe sole source of sodium ion is from the NaOH.

5) Determine the sulfate ion concentration:

0.02596011 mol / 0.05279 L = 0.4918 MThe sole source of sulfate ion is from the H

_{2}SO_{4}.

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 mL6.44 g times 0.980 = 6.3112 g <--- mass of H

_{2}SO_{4}in the solution6.3112 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 mass16.6667 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 gH

_{2}O ---> 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. cm^{3} 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 HCl0.0892299 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/molMoles 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 cm^{3} 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 cm^{3}?

**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

3) This problem was answered on an "answers" website like this:

(2.00 g KOH) / (56.10564 g KOH/mol) / (100 cm4) You must recognize that the value connecting the KOH and H^{3}) x (250 cm^{3}) x (98.0791 g H_{2}SO_{4}/mol) = 8.74 g H_{2}SO_{4}No additional comments were made. The above is a technique known as dimensional analysis. It is presented as one line of calculations and, typically, has no explanation added. The above answer inspired the follow comments.

mass _{1}MV _{1}=––––– MM _{1}<--- MM means molar mass

mass _{1}M = ––––––––––– (MM _{1}) (V_{1})

5) Here is the H_{2}SO_{4} set up:

mass _{2}<--- mass _{2}is the answer to the questionMV _{2}=––––– MM _{2}

mass _{2}M = ––––––––––– (MM _{2}) (V_{2})

6) Since M = M, we have this:

mass _{1}mass _{2}––––––––––– = ––––––––––– (MM _{1}) (V_{1})(MM _{2}) (V_{2})

7) Cross-multiply:

(mass_{1}) (MM_{2}) (V_{2}) = (mass_{2}) (MM_{1}) (V_{1})

8) Solve for mass_{2}:

(mass _{1}) (MM_{2}) (V_{2})mass _{2}=––––––––––––––––– (MM _{1}) (V_{1})

9) Insert numbers and solve:

(2.00 g) (98.0791 g/mol) (250 cm ^{3})mass _{2}=––––––––––––––––––––––––––––– (56.10564 g/mol) (100 cm ^{3})

Up above, I converted cm^{3} 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):

M_{1}V_{1}= M_{2}V_{2}(1.2419 mol/L) (12.00 mL) = (x) (71.20 mL) x = 0.209 M

3) Determine concentration of the ions:

(NH_{4})_{2}SO_{4}(aq) ---> 2NH_{4}^{+}(aq) + SO_{4}^{2}¯(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:

[NH_{4}^{+}] = 0.418 M

[SO_{4}^{2}¯] = 0.209 M

**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 M0.0097724 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 L15.1 mL (to three sig figs)