How Do You Calculate the Solubility of Benzoic Acid in Water?

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Discussion Overview

The discussion revolves around calculating the solubility of benzoic acid in water based on an experimental setup involving titration and sample withdrawal. Participants explore the methodology for determining solubility in moles per kilogram of water, as well as how to calculate average error from multiple solubility values.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant describes an experimental procedure involving the dissolution of benzoic acid in water and subsequent titration with NaOH to determine solubility.
  • Another participant questions the relevance of certain details in the problem, suggesting that some information may be extraneous.
  • There are inquiries about the mass of benzoic acid in the solution and how to calculate it based on titration results.
  • Participants discuss the potential impact of sample volume on the calculation of moles of solute.
  • There is a query about how to compute average error when multiple solubility values are available.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations or the relevance of certain details in the problem. Multiple viewpoints on the approach to calculating solubility and average error remain present.

Contextual Notes

Some participants express uncertainty about the essential information needed for calculations, and there are unresolved questions regarding the influence of sample volume on the results.

Who May Find This Useful

This discussion may be useful for individuals interested in experimental chemistry, particularly those focused on solubility calculations and titration methods.

freedom856
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In an experiment, 7g benzoic acid is dissolving to 350ml distillated water in 500ml conical flask at 60C for a period of time. 100ml of solution is then withdrawn and place at 40C water bath. After several minutes, 10ml sample is withdrawn and the weight of sample is 9.58g. The solution is then transfer to 100ml flask and 20ml water is added. The solution is then titrated against 0.0206M NaOH. Volume of NaOH used is 28.9ml. I wonder how to calculate the solubility of benzoic acid in moles per 1kg water

Can anyone teach me how to calculate this question?

Also, if i have 2 values of the solubility. how can i calculate the average error? Thank you
 
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What does titration tell you?
 
number of moles of acid react
 
There appears to be some redundant information in this problem. You might object it is put there to confuse you :devil: but in practice in the laboratory you might have it and be confused, you do have to see what is essential.
 
freedom856 said:
number of moles of acid react

So, what is mass of the acid in 9.58g of the solution?
 
Is (0.0206x28.9/1000)x122.12 = 0.0727g ?
 
And you still don't know what to do? What is mass of solvent?
 
o i see. i wonder whether different volume of sample may or may not affect number of moles. Thank you. Is mass of solvent = 9.58-0.0727 = 9.5073g?
 
That's the correct approach. I have not checked numbers.
 
  • #10
what about average error?
 

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