Determining a formula of a hydrate

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The discussion revolves around determining the formula of a barium hydroxide hydrate through titration analysis. A sample weighing 3.632 g was dissolved in water, and a 25.00 mL aliquot was titrated with 0.0987 M HCl, yielding consistent titration volumes around 23.3 mL. Participants suggest calculating the moles of HCl used and relating them to the moles of barium hydroxide to find the hydrate's formula. There is confusion regarding balancing the chemical equation for the reaction between barium hydroxide and hydrochloric acid. The conversation emphasizes understanding mole relationships and gram equivalents in titration calculations to solve the problem effectively.
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Barium hydroxide froms several hydrates. A specimen of barium hydroxide, suspected of being a hydrate, was prepared and analysed 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.

I tried writing a balanced equation for the reaction, but I can't seem to balance it.

Also, I'm not exactly sure how to approach this problem so any help would be greatly appreciated.

Thanks.
 
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~angel~ said:
Barium hydroxide froms several hydrates. A specimen of barium hydroxide, suspected of being a hydrate, was prepared and analysed 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.

I tried writing a balanced equation for the reaction, but I can't seem to balance it.

Also, I'm not exactly sure how to approach this problem so any help would be greatly appreciated.

Thanks.
A good start to most titration questions is to find the number of moles of the known acid/base that was used and then find the number of moles used for the acid. Then find how many moles there are in the solution of the acid/base that you are trying to work out.

The Bob (2004 ©)
 
I know what you mean but I can't seem to balance the equation.

I thought it was BaOH + HCl --> BaCl2 + H2O

If that is right, how do you balance is because it seems to be one of those equations where no matter what number you put in front, it is different on the other side.

Thanks
 
Barium has 2+ for its charge. Hydroxide is a -1 ion.
 
Well that's explains it. Thanks.

I haven't really encountered a problem like this before. I have found the number of moles for each substance. Ba(OH)2 has 1/2 the moles of HCl. Is this for 25.0mL?

How exactly can I determine the formula from what I've got?

Thanks.
 
I'm not sure if this will be of specific help here, but still:

In titrations, a more fundamental quantity is gram equivalents (or milliequivalents). If A and B are titrated with each other, then gram equivalents of A = gram equivalents of B. To convert from moles to gram equivalents and vice versa, you need to multiply and divide by a factor. As you can see using simple examples, moles of A and B are not necessarily equal in titrations and there's usually a factor involved.
 
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