What would be the proper way to solve a titration problem?

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SUMMARY

The titration problem involves a 4.36-g sample of an unknown alkali metal hydroxide dissolved in 100.0 mL of water and titrated with 2.50 M HCl, reaching the equivalence point after adding 17.0 mL of HCl. The moles of HCl calculated to be 0.0425 mol directly correlate with the moles of hydroxide, leading to the conclusion that the molar mass of the alkali metal hydroxide is 103 g/mol. This value aligns closely with rubidium hydroxide (RbOH), confirming that the alkali metal cation is likely Rb+.

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  • Basic chemistry concepts regarding acid-base reactions
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Homework Statement


"A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added and the resulting solution is titrated with 2.50 M HCl (aq) solution. The indicator changes colour signaling that the equivalence point has been reached after 17.0 mL of the hydrochloric acid solution has been added."

a) What is the molar mass of the metal hydroxide?
b) What is the identity of the alkali metal cation: Li+, Na+, K+, Rb+, or Cs+?

Homework Equations


Molarity = moles solute / volume of solution in liters

The Attempt at a Solution


First, I found the moles for HCl:
moles solute / volume of solution in liters = molarity
moles solute = molarity (volume of solution in liters)
0.0425 mol HCl = 2.5(0.017)
Now I guessed that 0.0425 mol of HCl is supposed to react with 0.425 (?)OH to produce H2O and (?)Cl. Then I tried to find the moles of all the possible alkali metal hydroxides by adding the molar mass of hydrogen and oxygen, then adding them to the different alkali metals. Then I took my given mass of 4.36 g and divided it by each of those molar masses. If the result was not 0.0425, then I eliminated it as a solution. This is my final result:
4.36/102.47514 = 0.0425 mol RbOH, where 102.47514 = 1.00794 + 15.9994 + 85.4678. I then concluded that the alkali metal was rubidium, seeing how the numbers matched up.
However, I am completely sure this is not the correct way to do this problem. Could someone tell me how they would do this titration problem?
 
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Eclair_de_XII said:

Homework Statement


"A 4.36-g sample of an unknown alkali metal hydroxide is dissolved in 100.0 mL of water. An acid-base indicator is added and the resulting solution is titrated with 2.50 M HCl (aq) solution. The indicator changes colour signaling that the equivalence point has been reached after 17.0 mL of the hydrochloric acid solution has been added."

a) What is the molar mass of the metal hydroxide?
b) What is the identity of the alkali metal cation: Li+, Na+, K+, Rb+, or Cs+?

Homework Equations


Molarity = moles solute / volume of solution in liters

The Attempt at a Solution


First, I found the moles for HCl:
moles solute / volume of solution in liters = molarity
moles solute = molarity (volume of solution in liters)
0.0425 mol HCl = 2.5(0.017)
Now I guessed that 0.0425 mol of HCl is supposed to react with 0.425 (?)OH to produce H2O and (?)Cl. Then I tried to find the moles of all the possible alkali metal hydroxides by adding the molar mass of hydrogen and oxygen, then adding them to the different alkali metals. Then I took my given mass of 4.36 g and divided it by each of those molar masses. If the result was not 0.0425, then I eliminated it as a solution. This is my final result:
4.36/102.47514 = 0.0425 mol RbOH, where 102.47514 = 1.00794 + 15.9994 + 85.4678. I then concluded that the alkali metal was rubidium, seeing how the numbers matched up.
However, I am completely sure this is not the correct way to do this problem. Could someone tell me how they would do this titration problem?

I think your reasoning looks fine. What I would say is that you know that the moles of hydroxide equals moles of acid. (All alkali metals have +1 charge as ions). You also know grams of alkali hydroxide. Molecular weight is grams divided by moles. So the molecular weight is 4.36 g/ 0.0425 mol = 103 g/mol. Since RbOH is closest to this, the sample is most likely RbOH.
 

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