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Sagot :
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]
### 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]
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