Question

Correctly classify each of the following compound as highly soluble or insoluble in water. 1. NaCl 2. CH3Cl 3. CH3OH 4. KOH 5. C6H6 6. C6H14 7. KNO3 8. CCl4 9. NaSO4 10. CH3CH2OH 11. CH2Cl2 12. CH3COOH

Answer 1

There are solubility rules for inorganic compounds in water as shown in the picture attached. For organic compounds, we use the concept of polarity.

1. NaCl: Since NaCl is a chloride and Na is not included in the list of insoluble, therefore, NaCl is highly soluble.

2. CH₃Cl: This compound is organic. Since the electonegativity between C and Cl is great, it tends to be polar. Because water is also polar, this is highly soluble.

3. CH₃OH: This is an alcohol. Alcohols with short chains of hydrocarbon are highly soluble in water because of hydrogen bonding (-OH).

4. KOH: From the table, KOH is highly soluble. In fact, this is a strong solid which readily dissociates in water.

5.  C₆H₆ is the benzene ring. This is a nonpolar molecule, so you would expect this to be insoluble with water.

6. C₆H₁₄ is hexane which is a long chain of hydrocarbons. Because there is no point of hydrogen bonding, this is nonpolar, and therefore, insoluble in water.

7. KNO₃ is highly soluble because all nitrates are soluble.

8. CCl₄ is an organic compound. Individually, C-Cl bonds are polar. But because there are 4 of them, they cancel out. As a result, there is a balance of partial charges which makes it nonpolar. Thus, this is insoluble in water.

9. NaSO₄: from the table, this is highly soluble.

10. CH₃CH₂OH: Although it is composed of the nonpolar hydrocarbon chain, the OH group is much stronger such that it is highly soluble in water.

11. CH₂Cl₂: This is polar because there is an imbalance due to the C-H and C-Cl bonds. Thus, it is highly soluble in water.

12. CH₃COOH is a carboxylic acid which is even more soluble than alcohols. Thus, this is highly soluble in water.

Answer 2

NaCl, KOH, $\text{KNO}_3$, $\text{NA}_2\text{SO}_4$, $\text{CH}_3\text{Cl}$, $\text{CH}_2\text{Cl}_2$, $\text{CH}_3\text{OH}$, $\text{CH}_3\text{CH}_2\text{OH}$  and $\text{CH}_3\text{COOH}$ are highly soluble in water whereas $\text{C}_6\text{H}_6$, ${\text{C}_6\text{H}_14}$, and ${\text{CC}}{{\text{l}}_{\text{4}}}$ are highly insoluble in water.

Further explanation:

Solubility is the property of substance as a result of which it has a tendency to dissolve in other substances. “Like dissolves like” is a general principle that is used to predict whether the substance is soluble in the given solvent or not.

Water is a polar molecule and it has $\text{H}^+$ and $\text{OH}^-$ ions in it. So ionic and polar compounds are highly soluble in water whereas the covalent and organic compounds are highly insoluble in water.

NaCl, KOH, ${\text{KN}}{{\text{O}}_{\text{3}}}$ and ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$ are ionic compounds that are composed of their respective ions. Water is a polar compound and therefore all these compounds are highly soluble in water.

${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$ and ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{14}}}}$ are non-polar hydrocarbons so these are highly insoluble in water.

${\text{C}}{{\text{H}}_3}{\text{Cl}}$ and ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$ are polar covalent compounds and water is also polar. So these are highly soluble in water.

${\text{CC}}{{\text{l}}_{\text{4}}}$ is a covalent compound so it cannot form ions and therefore it is highly insoluble in water.

${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$ and ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}$ are alcohols that contains a highly electronegative oxygen atom that is capable to form hydrogen bonds with the hydrogen atom of water molecule. Therefore both these compounds are highly soluble in water.

${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ is polar due to presence of hydroxyl group in it and therefore it is highly soluble in water.

Therefore, NaCl, KOH, ${\text{KN}}{{\text{O}}_{\text{3}}}$, ${\text{N}}{{\text{a}}_{\text{2}}}{\text{S}}{{\text{O}}_{\text{4}}}$, ${\text{C}}{{\text{H}}_3}{\text{Cl}}$, ${\text{C}}{{\text{H}}_{\text{2}}}{\text{C}}{{\text{l}}_{\text{2}}}$, ${\text{C}}{{\text{H}}_{\text{3}}}{\text{OH}}$, ${\text{C}}{{\text{H}}_{\text{3}}}{\text{C}}{{\text{H}}_{\text{2}}}{\text{OH}}$  and ${\text{C}}{{\text{H}}_{\text{3}}}{\text{COOH}}$ are highly soluble in water whereas ${{\text{C}}_{\text{6}}}{{\text{H}}_{\text{6}}}$ , ${{\text{C}}_{\text{6}}}{{\text{H}}_{{\text{14}}}}$ , and ${\text{CC}}{{\text{l}}_{\text{4}}}$ are highly insoluble in water.

Learn more:

1. Identification of ionic bonding: brainly.com/question/1603987
2. What type of bond exists between phosphorus and chlorine? brainly.com/question/81715

Answer details:

Grade: High School

Chapter: Ionic and covalent compounds

Subject: Chemistry

Keywords: NaCl, KOH, KNO3, CH3OH, CH3CH2OH, CH3COOH, C6H6, C6H14, CCl4, CH3Cl, Na2SO4, CH2Cl2, soluble, insoluble, water.