Experiment Volumetric analysis - Acid base

In summary: Once you have that...For the rest of the problem, begin by understanding the neutralization equation. Try to write the neutralization equation using M^+OH^-(monobasic) and HXO_4(monoprotic). Once you have that...In summary, the lab results showed that the concentration of mineral acid, HXO4, was 1.70 g dm-3 in solution KA2. The relative molecular mass of HXO4 was determined to be 0.4 mol dm-3. The concentration of the element X was not determined.
  • #1
TheDanny
8
0

Homework Statement


Do not understand the question.


2. My Lab Results

Topic : Volumetric analysis - Acid base
purpose : To determine the exact concentration of a mineral acid, HXO4, and to determine the relative atomic mass of the element X.
Materials : KA1 is a mineral acid, HXO4
KA2 is a solution containing 1.70g of OH- ions per dm3.
Phenolphthalein as indicator.

Procedure : Pipette 25.0cm3 of KA2 into the titration flask. Add two or three drops pf phenolphthalein indicator and titrate this solution with KA1. Record your readings in the table.
Repeat the titration as many times as you think necessary to achieve accurate results.

http://img248.imageshack.us/img248/2031/rejm9.jpg [Broken]
My result up.

My problems is here the question.
a. Calculate the concentration, in mol dm-3, of solution KA2
b. Write a balanced ionic equation for the reaction between solution KA1 and the solution KA2.
c. Calculate the concentration, in mol dm-3, of mineral acid HX04 in solution KA1.
d. If the concentration of mineral acid HXO4 in solution KA2 is 20.1g dm-3, calculate the relative molecular mass of HXO4.
e. Using the answer to (d), determine the relative atomic mass of the element X.
f. Suggest and identity for element X.

Please guide me. Thx

In question (a) i tried with
mol=mass/mm
=1.7/17 = 0.1mol

mol=mv/1000
0.1=m(250)/1000
m=0.4
concentration of KA2 is 0.4moldm-3?
 
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  • #2
TheDanny said:
In question (a) i tried with
mol=mass/mm
=1.7/17 = 0.1mol

This is correct. Remember that the concentration was given as 1.70 g OH- per dm3 which is the same thing as 1.7 g OH- per liter or 0.1 moles/liter OH-, so this...
mol=mv/1000
0.1=m(250)/1000
m=0.4
concentration of KA2 is 0.4moldm-3?
... calculation isn't necessary.

For the rest of the problem, begin by understanding the neutralization equation.
Try to write the neutralization equation using [tex]M^+OH^-[/tex](monobasic) and [tex]HXO_4[/tex](monoprotic).
 
  • #3


I would like to commend you on your detailed experiment and results. Your procedure and data are well-organized and clear.

To answer your question regarding the concentration of KA2, your calculation seems to be correct. However, it would be more accurate to use the molar mass of OH- (17 g/mol) instead of the molecular mass of KA2 (17 g/mol). This would give you a concentration of 0.1 mol/dm3.

In question (b), the balanced ionic equation for the reaction between KA1 and KA2 would be:
HXO4 + OH- → H2O + X

For question (c), the concentration of HXO4 in KA1 can be calculated using the equation you mentioned, mol = mv/1000. However, in this case, the volume should be in dm3, so it would be 25/1000 = 0.025 dm3. The molar mass of HXO4 can be calculated using the relative molecular mass you found in question (d). This would give you a concentration of 0.4 mol/dm3 for HXO4 in KA1.

In question (d), the concentration of HXO4 in KA2 is given as 20.1 g/dm3. To calculate the relative molecular mass, we can use the equation molar mass = mass/concentration. Plugging in the values, we get a relative molecular mass of 100.5 g/mol for HXO4.

In question (e), to find the relative atomic mass of element X, we need to subtract the atomic mass of oxygen (16 g/mol) from the relative molecular mass of HXO4 found in question (d). This would give us a relative atomic mass of 84.5 g/mol for element X.

Finally, in question (f), based on the calculated relative atomic mass of 84.5 g/mol, we can suggest that element X is most likely sulfur (atomic mass of 32.1 g/mol) or selenium (atomic mass of 78.9 g/mol). Further experiments or analysis would be needed to confirm the identity of element X.

I hope this helps guide you in solving the questions and understanding the concepts of volumetric analysis and acid-base reactions. Keep up the good work in your experiments!
 

1. What is the purpose of Experiment Volumetric analysis - Acid base?

The purpose of this experiment is to determine the concentration of an acid or base solution by using a technique called titration, where a known volume of one solution is reacted with a measured volume of another solution of known concentration.

2. What equipment is needed for this experiment?

The equipment needed for this experiment includes a burette, pipette, conical flask, indicator, and a standardized solution of an acid or base.

3. How is the experiment carried out?

The experiment is carried out by first measuring a known volume of the acid or base solution using a pipette and placing it in a conical flask. Then, the standardized solution is slowly added to the conical flask using a burette until the indicator changes color, indicating that the reaction is complete. The volume of the standardized solution used is then recorded.

4. What is the purpose of using an indicator in this experiment?

The purpose of using an indicator is to determine when the reaction between the acid and base is complete. The indicator changes color at the endpoint of the reaction, allowing for the precise volume of standardized solution used to be recorded.

5. What are the sources of error in this experiment?

The sources of error in this experiment can include imprecise measurement of solutions, incorrect readings of the burette or pipette, and human error in determining the color change of the indicator. It is important to repeat the experiment multiple times and take the average of the results to minimize these errors.

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