Find the Error: Calculating pH of 0.5L Diluted HA Solution

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SUMMARY

The discussion centers on calculating the pH of a diluted solution of a weak acid HA, where 0.75 moles of HA is initially dissolved in 0.5 liters of water, resulting in a concentration of 1.5 M. The equilibrium expression for the dissociation of HA is established, leading to an incorrect pH calculation of 1.1 instead of the correct value of 0.8. The error arises from the assumption that the concentration of H3O+ remains constant during dilution, which is not the case. The discussion also prompts consideration of how the problem would change if the initial volume were 2 liters instead of 0.5 liters.

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  • Study the concept of weak acid dissociation and its equilibrium expressions
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this type of question should go into homework section
You prepare 0.5 liters of a solution by adding 0.75 moles of a weak acid HA to water. For HA, Ka = 10^-1 Finally you dilute this solution to a final volume of 2.0 liters. What is the pH of the diluted solution?

My solution is this. [HA] = .75 mol /.5 L = 1.5 M. HA + H2O -> H3O+ + A-. If x is the equilibrium concentration of H3O+, x^2 /(1.5 - x) = 10^-1 so solving for x gives x = 0.34 M = [H3O+].

Moles of H3O+ must be 0.34M * 0.5 L = 0.17 mol. When the new volume is 2.0 L, [H3O+] = 0.085M so the pH = 1.1. However, the correct answer to this is 0.8. Where did I go wrong?
 
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What if the question were "you prepare 2 liters of solution using 0.75 moles of a weak acid HA"? How would you approach it? Any reason why it should be a different problem from the one you tried to solve?
 

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