Säure-Base-Titration Basic
About points...
We associate a certain number of points with each exercise.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
When you click an exercise into a collection, this number will be taken as points for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit the number of points for the exercise in the collection independently, without any effect on "points by default" as represented by the number here.
That being said... How many "default points" should you associate with an exercise upon creation?
As with difficulty, there is no straight forward and generally accepted way.
But as a guideline, we tend to give as many points by default as there are mathematical steps to do in the exercise.
Again, very vague... But the number should kind of represent the "work" required.
About difficulty...
We associate a certain difficulty with each exercise.
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
When you click an exercise into a collection, this number will be taken as difficulty for the exercise, kind of "by default".
But once the exercise is on the collection, you can edit its difficulty in the collection independently, without any effect on the "difficulty by default" here.
Why we use chess pieces? Well... we like chess, we like playing around with \(\LaTeX\)-fonts, we wanted symbols that need less space than six stars in a table-column... But in your layouts, you are of course free to indicate the difficulty of the exercise the way you want.
That being said... How "difficult" is an exercise? It depends on many factors, like what was being taught etc.
In physics exercises, we try to follow this pattern:
Level 1 - One formula (one you would find in a reference book) is enough to solve the exercise. Example exercise
Level 2 - Two formulas are needed, it's possible to compute an "in-between" solution, i.e. no algebraic equation needed. Example exercise
Level 3 - "Chain-computations" like on level 2, but 3+ calculations. Still, no equations, i.e. you are not forced to solve it in an algebraic manner. Example exercise
Level 4 - Exercise needs to be solved by algebraic equations, not possible to calculate numerical "in-between" results. Example exercise
Level 5 -
Level 6 -
Question
Solution
Short
Video
\(\LaTeX\)
No explanation / solution video to this exercise has yet been created.
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Visit our YouTube-Channel to see solutions to other exercises.
Don't forget to subscribe to our channel, like the videos and leave comments!
Exercise:
Zur Titration von VHCl Salzsäure werden VNaOH Natronlauge der Konzentration cNaOH verbraucht. Berechne die Konzentration der Salzsäure.
Solution:
Geg V_ceHCl VHCl VHClC V_ceNaOH VNaOH VNaOHC c_ceNaOH cNaOH GesKonzentrationcsimolperliter Es gilt: n_ceHCl n_ceNaOH c_ceHCl V_ceHCl c_ceNaOH V_ceNaOH Die Konzentration der Salzsäure ist damit: c_ceHCl fracc_ceNaOH V_ceNaOHV_ceHCl fraccNaOH VNaOHCVHClC cHCl Die in der Probelösung enthaltene Menge an Chlorwasserstoff erhält man mit: n_ceHCl c_ceHCl V_ceHCl cHCl VHClC nHCl Die dieser Stoffmenge entspreche Masse berechnen man mit: m M n Um diese Gleichung lösen zu können muss man zunächst noch die molare Masse von Salzsäure bestimmen: M_ceHCl M_ceH + M_ceCl MH + MCl MHCl Anschliess kann dieser Wert in die obere Gleichung eingesetzt werden: m_ceHCl M_ceHCl n_ceHCl M_ceH + M_ceCl n_ceHCl M_ceH + M_ceCl c_ceHCl V_ceHCl M_ceH + M_ceCl fracc_ceNaOH V_ceNaOHV_ceHCl V_ceHCl M_ceH + M_ceCl c_ceNaOH V_ceNaOH MHCl nHCl mHCl m_ceHCl M_ceH + M_ceCl c_ceNaOH V_ceNaOH mHCl
Zur Titration von VHCl Salzsäure werden VNaOH Natronlauge der Konzentration cNaOH verbraucht. Berechne die Konzentration der Salzsäure.
Solution:
Geg V_ceHCl VHCl VHClC V_ceNaOH VNaOH VNaOHC c_ceNaOH cNaOH GesKonzentrationcsimolperliter Es gilt: n_ceHCl n_ceNaOH c_ceHCl V_ceHCl c_ceNaOH V_ceNaOH Die Konzentration der Salzsäure ist damit: c_ceHCl fracc_ceNaOH V_ceNaOHV_ceHCl fraccNaOH VNaOHCVHClC cHCl Die in der Probelösung enthaltene Menge an Chlorwasserstoff erhält man mit: n_ceHCl c_ceHCl V_ceHCl cHCl VHClC nHCl Die dieser Stoffmenge entspreche Masse berechnen man mit: m M n Um diese Gleichung lösen zu können muss man zunächst noch die molare Masse von Salzsäure bestimmen: M_ceHCl M_ceH + M_ceCl MH + MCl MHCl Anschliess kann dieser Wert in die obere Gleichung eingesetzt werden: m_ceHCl M_ceHCl n_ceHCl M_ceH + M_ceCl n_ceHCl M_ceH + M_ceCl c_ceHCl V_ceHCl M_ceH + M_ceCl fracc_ceNaOH V_ceNaOHV_ceHCl V_ceHCl M_ceH + M_ceCl c_ceNaOH V_ceNaOH MHCl nHCl mHCl m_ceHCl M_ceH + M_ceCl c_ceNaOH V_ceNaOH mHCl
Meta Information
Exercise:
Zur Titration von VHCl Salzsäure werden VNaOH Natronlauge der Konzentration cNaOH verbraucht. Berechne die Konzentration der Salzsäure.
Solution:
Geg V_ceHCl VHCl VHClC V_ceNaOH VNaOH VNaOHC c_ceNaOH cNaOH GesKonzentrationcsimolperliter Es gilt: n_ceHCl n_ceNaOH c_ceHCl V_ceHCl c_ceNaOH V_ceNaOH Die Konzentration der Salzsäure ist damit: c_ceHCl fracc_ceNaOH V_ceNaOHV_ceHCl fraccNaOH VNaOHCVHClC cHCl Die in der Probelösung enthaltene Menge an Chlorwasserstoff erhält man mit: n_ceHCl c_ceHCl V_ceHCl cHCl VHClC nHCl Die dieser Stoffmenge entspreche Masse berechnen man mit: m M n Um diese Gleichung lösen zu können muss man zunächst noch die molare Masse von Salzsäure bestimmen: M_ceHCl M_ceH + M_ceCl MH + MCl MHCl Anschliess kann dieser Wert in die obere Gleichung eingesetzt werden: m_ceHCl M_ceHCl n_ceHCl M_ceH + M_ceCl n_ceHCl M_ceH + M_ceCl c_ceHCl V_ceHCl M_ceH + M_ceCl fracc_ceNaOH V_ceNaOHV_ceHCl V_ceHCl M_ceH + M_ceCl c_ceNaOH V_ceNaOH MHCl nHCl mHCl m_ceHCl M_ceH + M_ceCl c_ceNaOH V_ceNaOH mHCl
Zur Titration von VHCl Salzsäure werden VNaOH Natronlauge der Konzentration cNaOH verbraucht. Berechne die Konzentration der Salzsäure.
Solution:
Geg V_ceHCl VHCl VHClC V_ceNaOH VNaOH VNaOHC c_ceNaOH cNaOH GesKonzentrationcsimolperliter Es gilt: n_ceHCl n_ceNaOH c_ceHCl V_ceHCl c_ceNaOH V_ceNaOH Die Konzentration der Salzsäure ist damit: c_ceHCl fracc_ceNaOH V_ceNaOHV_ceHCl fraccNaOH VNaOHCVHClC cHCl Die in der Probelösung enthaltene Menge an Chlorwasserstoff erhält man mit: n_ceHCl c_ceHCl V_ceHCl cHCl VHClC nHCl Die dieser Stoffmenge entspreche Masse berechnen man mit: m M n Um diese Gleichung lösen zu können muss man zunächst noch die molare Masse von Salzsäure bestimmen: M_ceHCl M_ceH + M_ceCl MH + MCl MHCl Anschliess kann dieser Wert in die obere Gleichung eingesetzt werden: m_ceHCl M_ceHCl n_ceHCl M_ceH + M_ceCl n_ceHCl M_ceH + M_ceCl c_ceHCl V_ceHCl M_ceH + M_ceCl fracc_ceNaOH V_ceNaOHV_ceHCl V_ceHCl M_ceH + M_ceCl c_ceNaOH V_ceNaOH MHCl nHCl mHCl m_ceHCl M_ceH + M_ceCl c_ceNaOH V_ceNaOH mHCl
Contained in these collections:
Asked Quantity:
Masse \(m\)
in
Kilogramm \(\rm kg\)
Physical Quantity
Eigenschaft der Materie
Unit
Base?
SI?
Metric?
Coherent?
Imperial?