#26 - 35

Go to dilution problems #1 - 10

Go to dilution problems #11 - 25

**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.M

_{1}V_{1}= M_{2}V_{2}(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: M_{1}V_{1}= M_{2}V_{2}(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 (CH_{2}O) in a solution prepared by mixing 125 mL of 6.13 M CH_{2}O and 175 mL of 4.34 M CH_{2}O and diluting the mixture to 500.0 mL.

**Solution:**

1) Determine the total moles of CH_{2}O in the two solutions being mixed:

(6.13 mol/L) (0.125 L) = 0.76625 mol

(4.34 mol/L) (0.175 L) = 0.7595 mol0.76625 mol + 0.7595 mol = 1.52575 mol

2) The total moles are diluted to a final volume of 500.0 mL:

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:

M_{1}V_{1}+ M_{2}V_{2}= M_{3}V_{3}

and solved for M_{3}.

**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 CaCl_{2}. That means that, while the solution is 2.48 M in CaCl_{2}, it is 4.96 M from the perspective of just the chloride.We use M

_{1}V_{1}= M_{2}V_{2}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 M_{1}V_{1} = M_{2}V_{2}:

(0.705 mol/L) (0.0560 L) = (x) (0.100 L)x = 0.3948 M

2) moles of Ca(NO_{3})_{2} in 1 mL:

0.3948 mol/L = 0.3948 mol / 1000 mL = 0.0003948 mol/mL

3) 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% H_{2}SO_{4} 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 H_{2}SO_{4} solution.

**Solution:**

Assume 100. g of the solution is present. This means 98.0 g of H_{2}SO_{4}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 M

_{1}V_{1}= M_{2}V_{2}(18.3856 mol/L) (x) = (0.1 mol/L) (1 L)

x = 0.005439 L = 5.44 mL

**Problem #33a:** What is the [NO_{3}¯] in 200. mL of 0.350 M Al(NO_{3})_{3}?

**Problem #33b:** What is the [NO_{3}¯] in the solution above after adding 200.0 mL of 0.150 M Ca(NO_{3})_{2}

**Solution:**

For every one formula unit of Al(NO_{3})_{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 nitrate0.210 + 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(NO

_{3})_{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 mol

2) The mass of NaOH needed is this:

40.0 g/mol times 0.0053 mol = 0.212 g

3) 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 digits

4) 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 MgCl_{2} is 1/8 M, what will its concentration be if it is diluted by 27%?

**Solution:**

M_{1}V_{1}= M_{2}V_{2}(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 M

_{1}V_{1}= M_{2}V_{2}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 M

_{1}V_{1}= M_{2}V_{2}:(11.6 mol/L) (V

_{1}) = (0.0138 mol/L) (4.30 L)V

_{1}= 5.11 x 10^{-3}L = 5.11 mL