#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.

**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)