IDNLearn.com provides a user-friendly platform for finding answers to your questions. Get comprehensive answers to all your questions from our network of experienced experts.
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]
Thank you for being part of this discussion. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Find the answers you need at IDNLearn.com. Thanks for stopping by, and come back soon for more valuable insights.