The preparation of an aqueous solution is described in the table below. For this solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself.

Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row.

You will find it useful to keep in mind that HCN is a weak acid.

0.3 mol of KOH is added to 1.0 L of a solution that is 0.9 M in both HCN and KCN.

| Acids | Bases | Other |
|:-----:|:-----:|:-----:|
| [tex]$\square$[/tex] | [tex]$\square$[/tex] | [tex]$\square$[/tex] |



Answer :

Sure, let's analyze the solution based on the information provided and identify the species that will be present in the solution, as well as classify them as acids, bases, or neither.

### Analysis:

1. Initial Concentrations:
- HCN concentration (0.9 M)
- KCN concentration (0.9 M)
- Amount of KOH (0.3 mol in 1.0 L solution, thus 0.3 M)

2. Reactions:
- [tex]\( \text{HCN} \)[/tex] is a weak acid:
[tex]\[ \text{HCN} \leftrightharpoons \text{H}^+ + \text{CN}^- \][/tex]

- [tex]\( \text{KOH} \)[/tex] is a strong base and will completely dissociate to:
[tex]\[ \text{KOH} \rightarrow \text{K}^+ + \text{OH}^- \][/tex]

- The [tex]\( \text{OH}^- \)[/tex] from [tex]\( \text{KOH} \)[/tex] will react with [tex]\( \text{HCN} \)[/tex], neutralizing some of it:
[tex]\[ \text{OH}^- + \text{HCN} \rightarrow \text{H}_2\text{O} + \text{CN}^- \][/tex]

Since we have 0.3 mol of [tex]\( \text{KOH} \)[/tex] and 0.9 mol of [tex]\( \text{HCN} \)[/tex], the reaction will consume all 0.3 mol of [tex]\( \text{OH}^- \)[/tex] and equivalent moles of [tex]\( \text{HCN} \)[/tex], leaving:
- [tex]\( 0.9 \text{ mol} \, \text{HCN} - 0.3 \text{ mol}\, \text{HCN} = 0.6 \text{ mol} \, \text{HCN} \)[/tex] (leftover)
- [tex]\( 0.3 \text{ mol} \, \text{OH}^{-} \rightarrow \text{0 mol}\,\text{OH}^{-} (\text{since it's completely reacted}) \)[/tex]
- [tex]\( 0.3 \text{ mol} \, \text{CN}^- \, (\text{from the reaction}) + \text{0.9 mol}\, \text{CN}^- \, (\text{from KCN}) = 1.2 \text{ mol}\, \text{CN}^- \)[/tex]

### Major Species at Equilibrium:
Let's classify the resultant species:

- Acids: [tex]\( \text{HCN} \)[/tex] (as it can donate a proton, [tex]\( \text{H}^+ \)[/tex])
- Bases: [tex]\( \text{OH}^- \)[/tex] (produced from [tex]\( \text{KOH} \)[/tex]), [tex]\( \text{CN}^- \)[/tex] (produced from both HCN reaction and KCN dissociation)
- Other: [tex]\( \text{K}^+ \)[/tex] (from both [tex]\( \text{KOH} \)[/tex] and [tex]\( \text{KCN} \)[/tex])

### Summary:

- Acids: [tex]\( \text{HCN} \)[/tex]
- Bases: [tex]\( \text{OH}^- \)[/tex], [tex]\( \text{CN}^- \)[/tex]
- Other: [tex]\( \text{K}^+ \)[/tex]

Thus, the major species present at equilibrium are categorized as follows:

Acids:
[tex]\[ \text{HCN} \][/tex]

Bases:
[tex]\[ \text{OH}^-, \text{CN}^- \][/tex]

Other:
[tex]\[ \text{K}^+ \][/tex]