Determining Formula of Barium Hydrate with Titration

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The forum discussion focuses on determining the formula of barium hydrate through titration. A sample of 3.632 g of barium hydroxide was dissolved in water, and a 25.00 mL aliquot was titrated with 0.0987 M HCl, yielding an average titration volume of 23.30 mL. The calculations revealed that 0.00115 moles of Ba(OH)2 were present, leading to the conclusion that the hydrate formula is 5Ba(OH)2·H2O. Participants emphasized the importance of precise calculations and rechecking work to avoid errors.

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Barium hydroxide forms several hydrates. A specimen of barium hydroxide, suspected of being a hydrate, was prepared and analyzed as follows to determine its formula.

3.632 g of the compound was dissolved in water to give 250.0mL of solution. 25.00 mL of this solution was titrated with 0.0987 M of HCl, using methyl-orange as indicator. Precise titrations of 23.34, 23.26 and 23.29 mL of HCl were obtained. Determine the formula of the hydrate.


So far I got...

(23.34, 23.26 and 23.29)/3 = 23.30mL

Ba(OH)2 + 2HCl --> BaCl2 + 2H2O

3.632g/0.250L * (1/171.34g) = 0.08479 mols/L Ba(OH)2

0.08479*0.025L = 0.00212 mols Ba(OH)2

0.0987 * 0.0233L = 0.00230mols HCl


Ans now I have no clue what to do...:frown:
 
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How much average sample size was titrated? This comes directly from sample preparation and aloquots.

How many moles on average of HCl was needed? This comes from the titrations volume and concentration.

You have the correct reaction for ratios. How many moles and grams of Ba(OH)2 excluding hydration was then titrated?

Can you now find the difference between the amount of compound titrated and the corresponding amount of Ba(OH)2 to which this corresponds? That difference is the water.
 
symbolipoint said:
How much average sample size was titrated? This comes directly from sample preparation and aloquots.

I took the average and got 23.30ml = 0.0233 L

symbolipoint said:
How many moles on average of HCl was needed? This comes from the titrations volume and concentration.

This is just the 0.0987 M * 0.0233L = 0.00230 mols right?

symbolipoint said:
You have the correct reaction for ratios. How many moles and grams of Ba(OH)2 excluding hydration was then titrated?

So there are half the mols of Ba(OH)2 as HCl...giving me 0.00230/2 = 0.00115 mols Ba(OH)2 and 0.19704 grams.

symbolipoint said:
Can you now find the difference between the amount of compound titrated and the corresponding amount of Ba(OH)2 to which this corresponds? That difference is the water.

So then the difference is simply 3.632 g - 0.19704 g = 3.435 g H20 which is 0.1908 mols. Which is appoximately 1/5 mols?

My formula would then be 5Ba(OH)2*H20?

ok I don't get this.:mad:
 
Some of your answers follow my set of questions very well, but you obtained then the wrong hydrate ratio (or just different than mine). Try rechecking your work. In fact, you seem to have made a serious decimal error. This is why I say to recheck your work very carefully. Why did you say, "3.632 g - 0.19704 g = 3.435 g..." ?
 

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