# 8: Solution Chemistry

- Page ID
- 284426

## Measuring Concentrations of Solutions

Exercise \(\PageIndex{1a}\)

What is the molarity of an NaI solution that contains 4.5 g of NaI in 21.0 mL of solution?

- 1.4 M
- 0.030 M
- 0.0047 M
- 0.00014 M
- 0.21 M

**Answer**-
a. 1.4 M

\(n_{NaI}=4.5 g NaI (\frac{1 mol}{149.89424})=0.03 mol NaI\)

\(M = \frac{mol}{L}=\frac{0.03}{21*10^{-3}L}=1.429 M\)

Exercise \(\PageIndex{1b}\)

What volume of 0.742 M Na_{2}CO_{3} solution contains 44.9 g of Na_{2}CO_{3}?

- 0.314 L
- 6.41 × 10
^{3}L - 0.571 L
- 3.53 × 10
^{3}L - 1.75 L

**Answer**-
c. 0.571 L

Exercise \(\PageIndex{1c}\)

How many moles of sulfate ions are there in a 0.650-L solution of 0.312 M Al_{2}(SO_{4})_{3}?

- 0.203 mol
- 0.608 mol
- 6.25 mol
- 0.0676 mol
- 1.44 mol

**Answer**-
b. 0.608 mol

Exercise \(\PageIndex{1d}\)

What mass of Na_{2}CO_{3} is present in 0.700 L of a 0.396 M Na_{2}CO_{3} solution?

- 29.4 g
- 74.2 g
- 42.0 g
- 187 g
- 60.0 g

**Answer**-
a. 29.4 g

Exercise \(\PageIndex{1e}\)

In order to dilute 77.1 mL of 0.778* M* HCl to 0.100 *M*, the volume of water that must be added is

- 67.2 mL
- 9.91 mL
- 6 * 10
^{2}mL - 1.01 * 10
^{-3}mL - 5.23 * 10
^{2}mL

**Answer**-
e. 5.23 * 10

^{2}mL\(M_{1}*V_{1}=M_{2}*V_{2}\)

\(V_{2}=\frac{\left (M_{1}*V_{1} \right )}{M_{2}}=\frac{\left ( 0.778 M * \left ( 77.1 * 10^{-3}L \right )\right )}{0.100 M}= .5998 L\)

Exercise \(\PageIndex{1f}\)

A dilute solution is prepared by transferring 35.00 mL of a 0.6363 M stock solution to a 900.0 mL volumetric flask and diluting to mark. What is the molarity of this dilute solution?

- 0.02474 M
- 0.04949 M
- 0.1636 M
- 0.006186 M
- 0.3182 M

**Answer**-
a. 0.02474 M

Exercise \(\PageIndex{1g}\)

What volume of 1.27 M HCl is required to prepare 197.4 mL of 0.456 M HCl?

- 5.5 * 102 mL
- 3.41 * 102 mL
- 70.9 mL
- 0.0141 mL
- 1.14 * 102 mL

**Answer**-
c. 70.9 mL

## Acid and Base Concentrations - pH Scale

Exercise \(\PageIndex{2a}\)

What is the pH of 4.1 × 10−^{3} M HCl(aq)? Is this solution acidic or basic?

**Answer**-
\(pH = -log(H^{+})=-log(4.1*10^{-3})=2.39\)

Acidic Solution

Exercise \(\PageIndex{2b}\)

What is the pH of a solution that has a H^{+} concentration of 2.4 * 10-9 M. Is this solution acidic or basic?

**Answer**-
pH = 8.62

Basic Solution

Exercise \(\PageIndex{2c}\)

The pH of a vinegar solution is 4.15. What is the H_{3}O^{+} concentration of the solution?

**Answer**-
Note Hydronium ion, H

_{3}0^{+}, and H^{+}are all the same thing.\(H^{+}=10^{-pH}=10^{-4.15}=7.08 * 10^{-5}M\)

Exercise \(\PageIndex{2d}\)

What is the H^{+} concentration of a solution with a pH of 12.05.

**Answer**-
8.91 * 10

^{-13}M

## Stoichiometry of Aqueous Reactions

Exercise \(\PageIndex{3a}\)

The concentration of sulfate in a sample of wastewater is to be determined by using gravimetric analysis. To a 100.0-mL sample of the wastewater is added an excess of barium nitrate, forming the insoluble barium sulfate (233.4 g/mol) according to the balanced equation given below. The solid barium sulfate is dried, and its mass is measured to be 0.9951 g. What was the concentration of sulfate in the original wastewater sample?

SO_{4}^{2–}(*aq*) + Ba(NO_{3})_{2}(*aq*) → BaSO_{4}(*s*) + 2NO_{3}^{–}(*aq*)

**Answer**-
\(n_{BaSO_{4}}=\frac{0.9951g BaSO_{4}}{233.4 g/mol}=4.265 * 10^{-3} M\)

Exercise \(\PageIndex{3b}\)

Zn reacts with hydrochloric acid.

Zn(s) + 2 HCl(aq) → ZnCl_{2}(aq) + H_{2}(g)

What volume of 3.05 M HCl(aq) will react with 25.0 g Zn(s)?

**Answer**-
0.251 L

Exercise \(\PageIndex{3c}\)

What minimum mass of cobalt (II) nitrate must be added to 60.0 mL of a 0.0999* M* phosphate solution in order to completely precipitate all of the phosphate as solid cobalt (II) phosphate?

2PO_{4}^{3–}(*aq*) + 3Co(NO_{3})_{2}(*aq*) → Co_{3}(PO_{4})_{2}(*s*) + 6NO_{3}^{–}(*aq*)

**Answer**-
1.64 g