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Science Chemistry Chemistry: An Atoms-Focused Approach (Second Edition)

The titration curve that results from the addition of 0.50 M N a O H to a sample containing 0.5 M K F e S O 4 2 . Solution: The titration curve that results from the addition of 0.50 M N a O H to a sample containing 0.5 M K F e S O 4 2 is shown below. Explanation: K F e S O 4 2 will dissociate into ions in water as: K F e S O 4 2 a q → K + a q + F e 3 + a q + 2 S O 4 2 - a q The F e 3 + ions actually exist in aqueous solutions as hydrated ferric ions, F e H 2 O 6 3 + . K + is a spectator ion. The highly charged F e 3 + cation draws electron density away from the water molecules of its inner coordination sphere which makes it possible for one or more of these molecules to donate a H + ion to a water molecule outside the sphere. F e H 2 O 6 3 + a q + O H - a q ⇌ F e H 2 O 5 O H 2 + a q + H 2 O ( l ) The first addition of N a O H converts the acid ion present in the solution to H 2 O . The hydrated F e 2 + ion is left with one fewer H 2 O ligand. This reaction continues until all of F e H 2 O 6 3 + , initially in the solution, has been converted to F e H 2 O 5 O H 2 + . The titration has reached its first equivalence point. In the second step of the titration, the F e H 2 O 5 O H 2 + ions formed in the first step react with additional N a O H forming F e H 2 O 5 O H 2 + by the following reaction: F e H 2 O 5 O H 2 + a q + O H - a q ⇌ F e H 2 O 4 O H 2 + a q + H 2 O ( l ) This reaction continues until all the F e H 2 O 5 ( O H ) 2 + in the solution has been converted to F e H 2 O 4 O H 2 + . This is the second equivalence point of titration. In the third step of titration, the F e H 2 O 4 O H 2 + ions formed in the second step react with additional N a O H forming F e H 2 O 3 ( O H ) 3 by the following reaction: F e H 2 O 4 O H 2 + a q + O H - a q ⇌ F e H 2 O 3 ( O H ) 3 s + H 2 O ( l ) The titration of a weak acid with a strong base titrant resembles the curve as shown below. It starts with initial low p H values and ends with high p H values. Let’s assume 25 m L of N a O H will be required for the first equivalence point of titration, then the second equivalence point will be at 50 m L o f N a O H added and third equivalence point will be at 75 m L o f N a O H added. From this, the titration curve ( p H versus volume of 0.50 M NaOH added) for a sample of 0.5 M K F e S O 4 2 is as shown below: Conclusion: From the reaction between N a O H and hydrated metal ion, the graph of p H versus volume of 0.50 M N a O H added for 0.5 M K F e S O 4 2 is drawn.

The titration curve that results from the addition of 0.50 M N a O H to a sample containing 0.5 M K F e S O 4 2 . Solution: The titration curve that results from the addition of 0.50 M N a O H to a sample containing 0.5 M K F e S O 4 2 is shown below. Explanation: K F e S O 4 2 will dissociate into ions in water as: K F e S O 4 2 a q → K + a q + F e 3 + a q + 2 S O 4 2 - a q The F e 3 + ions actually exist in aqueous solutions as hydrated ferric ions, F e H 2 O 6 3 + . K + is a spectator ion. The highly charged F e 3 + cation draws electron density away from the water molecules of its inner coordination sphere which makes it possible for one or more of these molecules to donate a H + ion to a water molecule outside the sphere. F e H 2 O 6 3 + a q + O H - a q ⇌ F e H 2 O 5 O H 2 + a q + H 2 O ( l ) The first addition of N a O H converts the acid ion present in the solution to H 2 O . The hydrated F e 2 + ion is left with one fewer H 2 O ligand. This reaction continues until all of F e H 2 O 6 3 + , initially in the solution, has been converted to F e H 2 O 5 O H 2 + . The titration has reached its first equivalence point. In the second step of the titration, the F e H 2 O 5 O H 2 + ions formed in the first step react with additional N a O H forming F e H 2 O 5 O H 2 + by the following reaction: F e H 2 O 5 O H 2 + a q + O H - a q ⇌ F e H 2 O 4 O H 2 + a q + H 2 O ( l ) This reaction continues until all the F e H 2 O 5 ( O H ) 2 + in the solution has been converted to F e H 2 O 4 O H 2 + . This is the second equivalence point of titration. In the third step of titration, the F e H 2 O 4 O H 2 + ions formed in the second step react with additional N a O H forming F e H 2 O 3 ( O H ) 3 by the following reaction: F e H 2 O 4 O H 2 + a q + O H - a q ⇌ F e H 2 O 3 ( O H ) 3 s + H 2 O ( l ) The titration of a weak acid with a strong base titrant resembles the curve as shown below. It starts with initial low p H values and ends with high p H values. Let’s assume 25 m L of N a O H will be required for the first equivalence point of titration, then the second equivalence point will be at 50 m L o f N a O H added and third equivalence point will be at 75 m L o f N a O H added. From this, the titration curve ( p H versus volume of 0.50 M NaOH added) for a sample of 0.5 M K F e S O 4 2 is as shown below: Conclusion: From the reaction between N a O H and hydrated metal ion, the graph of p H versus volume of 0.50 M N a O H added for 0.5 M K F e S O 4 2 is drawn.

Solution Summary: The author illustrates the titration curve that results from the addition of 0.50MNaOH to a sample.

Chemistry: An Atoms-Focused Approach (Second Edition)

Chemistry: An Atoms-Focused Approach (Second Edition)

2nd Edition

ISBN: 9780393614053

Author: Thomas R. Gilbert, Rein V. Kirss, Stacey Lowery Bretz, Natalie Foster

Publisher: W. W. Norton & Company

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Chemistry: An Atoms-Focused Approach (Second Edition)

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