Question

1. use the solubility rules to identify the precipitate in the reaction. aqueous solutions of potassium hydroxide and cobalt(iii) nitrate are mixed.

here are the solubility rules.

1. alkali metal (li+, na+, k+, etc) and ammonium (nh4+) salts are generally soluble. exceptions include li2co3 and li3po4.
2. nitrate (no3-) salts are generally soluble.
3. silver (ag+) and lead (pb2+) salts are generally insoluble.
4. halide (cl-, br-, and i-) salts are generally soluble. exceptions include pb2+, ag+, hg22+.
5. carbonate (co32-), phosphate (po43-), and hydroxide (oh-) salts are generally insoluble.
6. sulfate (so42-) salts are generally soluble. exceptions include caso4, srso4, baso4, pbso4, and ag2so4.

Explanation:

Step1: Write the reaction equation

The reaction between potassium hydroxide ($\ce{KOH}$) and cobalt(III) nitrate ($\ce{Co(NO_{3})_{3}}$) is a double - displacement reaction. The chemical equation is:
$$\ce{3KOH(aq) + Co(NO_{3})_{3}(aq)\xlongequal{}Co(OH)_{3}(s) + 3KNO_{3}(aq)}$$

Step2: Apply solubility rules to each product

• For $\ce{KNO_{3}}$: According to rule 2 (Nitrate ($\ce{NO_{3}^{-}}$) salts are generally soluble) and rule 1 (Alkali metal ($\ce{K^{+}}$) salts are generally soluble), $\ce{KNO_{3}}$ is soluble in water.
• For $\ce{Co(OH)_{3}}$: According to rule 5 (Hydroxide ($\ce{OH^{-}}$) salts are generally insoluble). Cobalt(III) ion ($\ce{Co^{3+}}$) is not an exception - making ion for hydroxide solubility. So $\ce{Co(OH)_{3}}$ is insoluble and will form a precipitate.

Answer:

The precipitate formed is $\ce{Co(OH)_{3}}$, so the correct option is $\ce{Co(OH)_{3}}$.