An aqueous solution of nacl is 350 ppm. what is the mass of nacl in 1.00 l of the solution?

Q8.1.1

Concentrated solutions have more solute per unit of solvent or solution.

Q8.1.2

\(M=\frac{mol\;solute}{L\;soln}\) \(M=\frac{0.75\;mol}{0.250\;L}\)

\(M=3.0\;M\)

Q8.1.3

\(mass\; \%=\frac{g\;solute}{g\;soln}\times100\) \(mass\; \%=\frac{12.0\;g}{40.0\;g+12.0\;g}\times100\)

\(mass\; \%=23.1\;\%\)

Remember, the mass of the solution includes both the solute and solvent.

Q8.1.4

\(volume\; \%=\frac{L\;solute}{L\;soln}\times100\) \(volume\; \%=\frac{0.200\;L}{1.60\;L}\times100\)

\(volume\; \%=12.5\;\%\)

Volumes can also be used in mL (or any other unit) as long as both volumes are in the same unit.

Q8.1.5

Write the concentration in "expanded form" which shows the relationship to then be used in dimensional analysis.

\(5.00\; \%\;m/m=\frac{5.00\;g\;glucose}{100\;g\;solution}\)

\(250.0\;mL\;soln\left (\frac{1.00\;g\;soln}{mL\;soln}\right )\left (\frac{5.00\;g\;glucose}{100\;g\;soln}\right )=12.5\;g\;glucose\)

Q8.1.6

1. \(\%\;m\diagup m=\frac{mass\;solute}{mass\;solution}\times 100\) \(\%\;m\diagup m=\frac{25.0\;g}{25.0\;g+300\;g}\times 100\)

   \(\%\;m\diagup m=7.69\;\%\)

2. The NaCl does not contribute to the volume of the solution so only the volume of the water is used for the volume fo the solution. Given the density is 1.00 g/mL, the volume of the solution is 300.0 mL.

\(\%\;m\diagup v=\frac{mass\;solute}{volume\;solution}\times 100\) \(\%\;m\diagup v=\frac{25.0\;g}{25.0\;g+300\;g}\times 100\)

\(\%\;m\diagup v=8.33\;\%\)

Q8.1.7

The parts of this problem require both the volume and mass of solute and solvent. The volume of the solute and solvent are given so first, find the mass of the solute and solvent so all the values are present before we start calculating the concentrations.

\(15.0\;mL\;\text{methanol}\left (\frac{0.792\;g}{mL\;\text{methanol}}\right)=11.9\;g\;\text{methanol}\)

\(125.0\;mL\;\text{ethanol}\left (\frac{0.789\;g}{mL\;\text{ethanol}}\right)=98.6\;g\;\text{ethanol}\)

1. \(\%\;m\diagup m=\frac{mass\;solute}{mass\;solution}\times 100\) \(\%\;m\diagup m=\frac{11.9\;g\;\text{methanol}}{11.9\;g+98.6\;g}\times 100\)

   \(\%\;m\diagup m=10.8\;\%\)

2. \(\%\;m\diagup v=\frac{mass\;solute}{volume\;solution}\times 100\) \(\%\;m\diagup v=\frac{11.9\;g\;\text{methanol}}{15.0\;mL+125\;mL}\times 100\)

   \(\%\;m\diagup v=8.50\;\%\)

3. \(\%\;v\diagup v=\frac{volume\;solute}{volume\;solution}\times 100\) \(\%\;v\diagup v=\frac{15.0\;mL\;\text{methanol}}{15.0\;mL+125\;mL}\times 100\)

   \(\%\;v\diagup v=10.7\;%\)

Q8.1.8

Write the concentration in "expanded form" which shows the relationship to then be used in dimensional analysis.

\(10.5\;\%\;m\diagup m=\frac{10.5\;g\;\text{NaCl}}{100\;g\;soln}\)

\(150.0\;mL\;soln\left (\frac{1.00\;g\;soln}{mL\;soln}\right )\left (\frac{10.5\;g\;\text{NaCl}}{100\;g\;soln}\right )=15.8\;g\;\text{NaCl}\)

Q8.1.9

1. \(M=\frac{0.614\;mol\;\text{Na}_2\text{SO}_4}{0.500\;L\;soln}=1.23\;M\)
2. \(M=\frac{3.63\;mol\;\text{NH}_3}{7.00\;L\;soln}=0.519\;M\)
3. \(M=\frac{1.71\;mol\;\text{EtOH}}{0.500\;L\;soln}=3.43\;M\)

Q8.1.10

\(0.250\;M=\frac{0.250\;mol\;\text{KF}}{1\;L\;soln}\)

\(180.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{0.250\;mol\;\text{KF}}{1\;L\;soln}\right)=0.0450\;mol\;\text{KF}\)

Q8.1.11

1. \(3.40\;L\left(\frac{0.780\;mol}{1\;L}\right)\left(\frac{162.2\;g}{mol}\right)=430.\;g\;\text{FeCl}_3\)
2. \(60.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{4.10\;mol}{1\;L}\right)\left(\frac{158.17\;g}{mol}\right)=38.9\;g\;\text{Ca(CH}_3\text{COO)}_2\)

Q8.1.12

\(0.500\;M=\frac{0.500\;mol\;\text{NaI}}{1\;L\;soln}\)

\(113\;g\;\text{NaI}\left(\frac{1\;mol}{149.89\;g}\right)\left(\frac{1\;L}{0.500\;mol}\right)=1.51\;L\;soln\)

Q8.1.13

1. \(C_1V_1=C_2V_2\) \(2.00\;M\cdot0.125\;L=C_2\cdot4.00\;L\)

   \(C_2=0.0625\;M\)

2. \(C_1V_1=C_2V_2\) \(6.30\;M\cdot1.85\;mL=C_2\cdot5.00\;mL\)

   \(C_2=2.33\;M\)

Q8.1.14

\(C_1V_1=C_2V_2\) \(0.300\;M\cdot6.00\;L=12\;M\cdotV_2\)

\(V_2=0.15\;L\)

Q8.1.15

\(12\;ppm\;\text{Pb}=\frac{12\;mg\;\text{Pb}}{1\;L\;soln}\)

\(50.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{12\;mg}
{1\;L}\right)=0.60\;mg\;\text{Pb}\)

Q8.1.16

\(25\;ppb\;\text{Hg}=\frac{25\;\mu g\;\text{Hg}}{1\;L\;soln}\)

\(175\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{25\;\mu g}
{1\;L}\right)=4.4\;\mu g\;\text{Hg}\)

Q8.1.17

\(ppm=\frac{mg}{L}\)

\(34\;g\;\text{Fe}\left(\frac{1\;mg}{10^{-3}\;g}\right)=3.4\times10^4\;mg\;\text{Fe}\)

\(365\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)=0.365\;L\)

\(ppm=\frac{3.4\times10^4\;mg\;\text{Fe}}{0.365\;L}=9.3\times10^4\;ppm\;\text{Fe}\)

Q8.1.18

1. \(2.0\;mol\;\text{Mg}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=4.0\;Eq\)
2. \(2.0\;mol\;\text{Al}^{3+}\left(\frac{3\;Eq}{1\;mol}\right)=6.0\;Eq\)
3. \(2.0\;mol\;\text{S}^{2-}\left(\frac{2\;Eq}{1\;mol}\right)=4.0\;Eq\)
4. \(2.0\;mol\;\text{Br}^-\left(\frac{1\;Eq}{1\;mol}\right)=2.0\;Eq\)
5. \(2.0\;mol\;\text{Cs}^{+}\left(\frac{1\;Eq}{1\;mol}\right)=2.0\;Eq\)
6. \(2.0\;mol\;\text{Ba}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=4.0\;Eq\)

Q8.1.19

1. \(2.50\;mol\;\text{Mg}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=5.00\;Eq\)
2. \(2.50\;mol\;\text{Al}^{3+}\left(\frac{3\;Eq}{1\;mol}\right)=7.50\;Eq\)
3. \(2.50\;mol\;\text{S}^{2-}\left(\frac{2\;Eq}{1\;mol}\right)=5.00\;Eq\)
4. \(2.50\;mol\;\text{Br}^-\left(\frac{1\;Eq}{1\;mol}\right)=2.50\;Eq\)
5. \(2.50\;mol\;\text{Cs}^{+}\left(\frac{1\;Eq}{1\;mol}\right)=2.50\;Eq\)
6. \(2.50\;mol\;\text{Ba}^{2+}\left(\frac{2\;Eq}{1\;mol}\right)=5.00\;Eq\)

Q8.1.20

\(250.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{132\;mEq}{L}\right)\left(\frac{10^{-3}\;Eq}{1\;mEq}\right)\left(\frac{1\;mol\;\text{Ca}^{2+}}{2\;Eq}\right)=0.0165\;mol\;\text{Ca}^{2+}\)

Q8.1.21

\(500.0\;mL\left(\frac{10^{-3}\;L}{1\;mL}\right)\left(\frac{98\;mEq}{L}\right)\left(\frac{10^{-3}\;Eq}{1\;mEq}\right)\left(\frac{1\;mol\;\text{K}^{+}}{1\;Eq}\right)\left(\frac{39.10\;g}{mol}\right)=1.92\;g\;\text{K}^{+}\)

Q8.1.22

\(3.93\;g\;\text{Sr}^{2+}\left(\frac{1\;mol}{87.62\;g}\right)\left(\frac{2\;Eq}{1\;mol}\right)\left(\frac{1\;mEq}{10^{-3}\;Eq}\right)\left(\frac{1\;L}{128\;mEq}\right)=0.701\;L\;soln\)

Q8.2.1

The rate of the forward reaction equals the rate of the reverse reaction.

Q8.2.2

No, the concentrations are constant but the concentrations do not have to be equal.

Q8.2.3

No.

Q8.2.4

The ratio of products and reactants when the system is at equilibrium.

Q8.2.5

More products than reactants are present at equilibrium.

Q8.2.6

More reactants than products present at equilibrium.

Q8.2.7

No. The equilibrium ratio does not depend on the initial concentrations.

Q8.3.1

Le Chatelier’s principle states that a system at equilibrium is disturbed, it will respond in a way to minimize te disturbance.

Q8.3.2

temperature, change in amount of substance, change in pressure through change in volume

Q8.3.3

1. shift right
2. shift left
3. shift right
4. shift right
5. shift left
6. shift left

Q8.3.4

1. shift left
2. shift left
3. no effect
4. no effect
5. shift right
6. shift right

Q8.3.5

1. shift right
2. shift left
3. shift right
4. shift left
5. shift right
6. shift left

Q8.3.6

1. shift right
2. shift left
3. shift left
4. shift right
5. no effect
6. no effect

Q8.4.1

A semipermeable membrane allows some substances to pass through but not others.

Q8.4.2

Two solutions are separated by a semipermeable membrane. Solution A contains 25.0 g of NaCl in 100.0 mL of water and solution B contains 35.0 g of NaCl in 100.0 mL of water.

1. Solution B
2. A \(\rightarrow\) B
3. B
4. A
5. increase
6. decrease

Q8.4.3

hyper - higher

hypo - lower

iso - same

Q8.4.4

Cells contain fluid with higher concentration than solution outside the cell.

Q8.4.5

Cells contain fluid with a lower concentration than the solution outside the cell.

Q8.4.6

Water molecules will flow from the tank water into the fish because the fish has a higher concentration of salt. If the fish absorbs too much water, it will die.

Q8.4.7

The "head" region is a phosphate group and it is hydrophillic.

Q8.4.8

The "tail" is a hydrocarbon tail and it is hydrophobic.

Q8.5.1

1. TRUE
2. FALSE - Bronsted-Lowry acid-base definitions are broader.

Q8.5.2

Classify each of the following as an acid, base, or neither.

1. base (contains metal and -OH group)
2. acid (formula starts with H and isn't water)
3. acid (contains -COOH which is carboxylic acid functional group)
4. base (contains metal and -OH group)
5. neither
6. neither (-OH group has to be with metal)

Q8.5.3

Amphoteric substances can act as an acid or base.

Q8.5.4

Identify each reactant in the following reactions as an acid or a base according to the Brønsted-Lowry theory.

1. HIO3(aq) - acid; H2O(l) - base; IO3−(aq) - base ; H3O+(aq) - acid
2. F−(aq) - base; HClO(aq) - acid; HF(aq) - acid; ClO−(aq) - base
3. H2PO4−(aq) - acid; OH−(aq) - base; HPO42−(aq) - base; H2O(l) - acid
4. CO32−(aq) - base; H2O(l) - acid; HCO3−(aq) - acid; OH−(aq) - base

Q8.5.5

1. HIO3/IO3− and H3O+/H2O
2. HF/F− and HClO/ClO−
3. H2PO4−/HPO42− and H2O/OH−
4. HCO3−/CO32− and H2O/OH−

Q8.5.6

1. NO3−
2. SO32−
3. H2AsO4−
4. HCOO− (the H that is removed comes from the carboxylic acid functional group)
5. PO43−
6. HS−
7. S2−
8. CO32−
9. HCO3−
10. H2PO4−
11. Na2SO4 or NaSO4−

Q8.5.7

Write the formula of each base’s conjugate acid.

1. HBrO3
2. NH4+
3. CH3COOH
4. H2CO3
5. HCN
6. H2PO4−
7. H2S
8. HSO4−
9. HCO3−
10. H2CO3
11. PH4+

Q8.5.8

HPO42− can act as a base and accept a proton to form H2PO4− and it can act as an acid and donate a proton to form PO43−.

Q8.6.1

H2O(l) ⇌ H+(aq) + OH−(aq)

It's neutral because there are equal amounts of H+ and OH−.

Q8.6.2

Indicate whether solutions with the following pH values are acidic, basic, or neutral.

a. basic b. neutral

c. acidic

Q8.6.3

pH + pOH = 14

Q8.6.4

1. pH = 3.64; pOH = 10.36
2. pH = 9.06; pOH = 4.94
3. pOH = 8.72; pH = 5.28
4. pOH = 0.22; pH = 13.78

Q8.6.5

1. pH = 5.00; pOH = 9.00
2. pH = 10.55; pOH = 4.94
3. pOH = 8.72; pH = 5.28
4. pOH = 0.22; pH = 13.78

Q8.6.6

1. \([\text{H}^+]=1.3\times 10^{-2}\;M, [\text{OH}^-]=7.4\times10^{-13}\;M\)
2. \([\text{H}^+]=7.1\times 10^{-12}\;M, [\text{OH}^-]=1.4\times10^{-3}\;M\)
3. \([\text{H}^+]=0.11\;M, [\text{OH}^-]=8.9\times10^{-14}\;M\)
4. \([\text{OH}^-]=6.2\times10^{-7}\;M, [\text{H}^+]=1.6\times 10^{-8}\;M\)
5. \([\text{OH}^-]=3.8\times10^{-14}\;M, [\text{H}^+]=0.26\;M\)
6. \([\text{OH}^-]=9.3\times10^{-8}\;M, [\text{H}^+]=1.1\times 10^{-7}\;M\)

Q8.6.7

Given pH = 5.00, we know \([\text{H}^+]=1.0\times 10^{-5}\;M\) which means \(M=\frac{1.0\times10^{-5}\;mol\;\text{H}^+}{1.00\;L}\).

If 0.100 L of water is added to 1.00 L, then the volume changes to 1.10 L but the moles of H+ does not change. The molarity can be calclulated with the same number of moles and the new volume.

\(M=\frac{1.0\times10^{-5}\;mol\;\text{H}^+}{1.00+0.100\;L}\)
\(M=9.1\times10^{-6}\;M\)

\(pH = 5.04\)

Q8.6.8

How much water would need to be added to the original solution in question 8 in order to bring the pH to 6.00?

To get to pH = 6.00, we need \([\text{H}^+]=1.0\times 10^{-6}\;M\).

Use the dilution formula to calculate the total volume of solution.

\(C_1V_1=C_2V_2\)
\(1.0\times10^{-5}\;M\cdot 1.00\;L=1.0\times10^{-6}\;M\cdot V_2\)\(V_2=10.\;L\)

The total volume is 10. L so 9 L needs to be added to the original 1 L solution.