Use Le' Chatelier's Principle.
a solution at equilibrium:
BaCrO4(s)<> Ba^2+(aq)+ Cr)4^-(aq)recall: 2CrO4^2-(aq)+ 2H^+(aq)<> Cr2O7^2-(aq) +H20
2CrO4^2- yellow
Cr2O7^2- orange
Make predictions about colour changes or changes in the number of states
a) To a solution of 5 drops of 0.3 M K2CrO4 and 2 drops of 0.3 M NaOH add 0.1 M Ba(NO3)2 until a change is noted.
MY PREDICTION: a yellow precipitate forms
b) use test tube from step (a) add 3 M HCl until a change is noted
MY PREDICTION: colour become yelloer b/v more H+ ion r being used and more water is being produced
c) To a solution of 5 drops of 0.3 M K2Cr2O7 and 2 drops of 0.3 M HCl add a drop at a time of 0.1 M Ba(NO3)2
d) To the test tube from step (c) add 1M NaOH until a change is noted
e) Suggest a way to reverse the changes and reactions that you observed in step (b) and (d)
f) Place 5 drops of ).3 M K2CrO7 in one test tube and 5 drops of 0.3 M K2Cr)4 in another test tube. Add a few drops of 0.1 M Ba(NO3)2 to each and note results
MY PREDICTION: K2CrO7 is more soluble K2CrO4 forms a precepitate
It's so difficult to make these predictions i am very confused and i've spent almost the entire day on this
Some comments below. Don't forget to predict what happens to the colour of the supernatant (aqueous phase) and if you can the pH.
a) To a solution of 5 drops of 0.3 M K2CrO4 and 2 drops of 0.3 M NaOH add 0.1 M Ba(NO3)2 until a change is noted.
MY PREDICTION: a yellow precipitate forms.
Agreed and the colour changes of the aqueous phase is from orange to yellow to (almost) clear.
b) use test tube from step (a) add 3 M HCl until a change is noted
MY PREDICTION: colour become yelloer b/v more H+ ion r being used and more water is being produced
Adding H+ forces the chromate equilibrium to the right, turning any chromate ion to dichromate ion. At the same time some of the barium chromate ppt will dissolve (may slight change). The supernatant will turn orange.
c) To a solution of 5 drops of 0.3 M K2Cr2O7 and 2 drops of 0.3 M HCl add a drop at a time of 0.1 M Ba(NO3)2
the pH may low enough to prevent a ppt because the low pH surpresses chromate ion forming. So no change.
d) To the test tube from step (c) add 1M NaOH until a change is noted
The OH- ion will remove H+ allowing chromate ion to form. This will result in a yellow ppt of barium chromate.
e) Suggest a way to reverse the changes and reactions that you observed in step (b) and (d)
Any ideas?
f) Place 5 drops of ).3 M K2CrO7 in one test tube and 5 drops of 0.3 M K2Cr)4 in another test tube. Add a few drops of 0.1 M Ba(NO3)2 to each and note results
There is no difference.
MY PREDICTION: K2CrO7 is more soluble K2CrO4 forms a precepitate NO they are both soluble otherwise you can't have the solutions.
An interesting thing to do is to start with dichromate solution measure the pH. Add barium nitrate. Measure the pH of the supernatant. Why is there a pH change and why is it so low?
to a solution of analyte NH3nh4and h2s were add ared ppt was formed
I understand that Le Chatelier's principle can be challenging to apply in some cases. Let's go through each part of the question step-by-step and make predictions based on the given information.
a) To a solution of 5 drops of 0.3 M K2CrO4 and 2 drops of 0.3 M NaOH add 0.1 M Ba(NO3)2 until a change is noted.
Prediction: A yellow precipitate forms.
Explanation: When Ba(NO3)2 is added to the solution, it will react with K2CrO4 to form a yellow precipitate of BaCrO4, which is insoluble in water. This is based on the given equilibrium equation and the solubility rules.
b) Use the test tube from step (a) and add 3 M HCl until a change is noted.
Prediction: The solution becomes yellower.
Explanation: The addition of HCl provides more H+ ions, which will shift the equilibrium according to the reaction 2CrO4^2-(aq) + 2H^+(aq) <> Cr2O7^2-(aq) + H2O. Since Cr2O7^2- is orange and CrO4^2- is yellow, the increase in H+ ions will shift the equilibrium to the left, resulting in more CrO4^2- ions and a yellower solution.
c) To a solution of 5 drops of 0.3 M K2Cr2O7 and 2 drops of 0.3 M HCl, add a drop at a time of 0.1 M Ba(NO3)2.
Prediction: A yellow precipitate forms.
Explanation: Similar to part (a), the addition of Ba(NO3)2 will react with K2Cr2O7 to form a yellow precipitate of BaCrO4.
d) To the test tube from step (c), add 1 M NaOH until a change is noted.
Prediction: The solution becomes more orange.
Explanation: The addition of NaOH provides OH- ions, which will shift the equilibrium according to the reaction 2CrO4^2-(aq) + 2H^+(aq) <> Cr2O7^2-(aq) + H2O. Since Cr2O7^2- is orange, the increase in OH- ions will shift the equilibrium to the right, resulting in more Cr2O7^2- ions and a more orange solution.
e) Suggest a way to reverse the changes and reactions observed in step (b) and (d).
Explanation: To reverse the changes observed in step (b), you can add a basic solution, such as NaOH, to neutralize the excess H+ ions and shift the equilibrium back towards the formation of Cr2O7^2-. To reverse the changes observed in step (d), you can add an acidic solution, such as HCl, to neutralize the excess OH- ions and shift the equilibrium back towards the formation of CrO4^2-.
f) Place 5 drops of 0.3 M K2CrO7 in one test tube and 5 drops of 0.3 M K2CrO4 in another test tube. Add a few drops of 0.1 M Ba(NO3)2 to each and note the results.
Prediction: K2CrO7 is more soluble, K2CrO4 forms a precipitate.
Explanation: K2CrO7 is more soluble in water compared to K2CrO4. When Ba(NO3)2 is added to the solution containing K2CrO7, no precipitate will form as BaCrO4 is not formed. However, when Ba(NO3)2 is added to the solution containing K2CrO4, a yellow precipitate of BaCrO4 will form due to the reaction between Ba2+ and CrO4^2- ions.
I hope these step-by-step explanations help clarify your understanding of Le Chatelier's principle and its application in this specific scenario.
Making predictions using Le Chatelier's Principle can indeed be challenging, especially when dealing with complex chemical reactions. However, with a systematic approach and understanding of the principle, you can navigate through these predictions.
Le Chatelier's Principle states that when a system at equilibrium is subjected to a change, it will respond in a way that minimizes the effect of that change. Based on this principle, we can make predictions about color changes and changes in the number of states for the given reactions.
a) Predicting color change:
- The original solution contains BaCrO4(s), which is yellow.
- Adding Ba^2+(aq) and CrO4^2-(aq) ions can disrupt the equilibrium.
- According to Le Chatelier's Principle, an increase in the concentration of the products will shift the equilibrium to the left to minimize the change.
- Therefore, a yellow precipitate is expected to form, indicating the reverse reaction is favored.
b) Predicting color change and changes in the number of states:
- Adding HCl increases the concentration of H+ ions.
- According to the given reaction: 2CrO4^2-(aq) + 2H^+(aq) <> Cr2O7^2-(aq) + H2O, increasing the H+ concentration will shift the equilibrium to the right.
- This will result in an increase in the concentration of Cr2O7^2-(aq), which is orange in color.
- At the same time, the number of states increases as water is being produced.
c) and d) The predictions for these steps involve similar principles as in (a) and (b). You can apply the Le Chatelier's Principle based on the changes in reactant and product concentrations and make predictions accordingly.
e) To reverse the changes observed in step (b) and (d):
- In step (b), where the solution became more yellow and produced more water, you can add a base (such as NaOH) to neutralize the excess acid (HCl) and restore the initial equilibrium.
- In step (d), where a change was noted after adding NaOH, you can add a strong acid (such as HCl) to neutralize the excess base (NaOH) and restore the initial equilibrium.
f) Predicting the outcome of adding Ba(NO3)2 to K2CrO7 and K2CrO4:
- K2CrO7 is more soluble than K2CrO4, so it remains in the solution.
- On the other hand, when Ba(NO3)2 is added to K2CrO4, a yellow precipitate of BaCrO4 is formed because Ba^2+ ions can react with CrO4^2- ions to form the insoluble compound BaCrO4.
Remember that these predictions are based on the understanding of Le Chatelier's Principle and the given reactions. Actual observations may vary due to experimental factors.