PH of HCl solution as a function of volume.

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The pH of a 2M HCl solution is calculated as pH = -log(2), resulting in a negative value, which is valid for strong acids. When water is added, the new concentration of HCl must be determined based on the volume of water added (V litres). The pH of the diluted solution can then be calculated using the formula pH = -log([HCl]), where [HCl] is the new concentration. It's important to consider dilution calculations without needing to factor in the autoionization of water for strong acids. Understanding these concepts is crucial for accurate pH calculations in acid-base chemistry.
Bipolarity
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Homework Statement


You have a 1L of a 2M HCl solution. Assume that the HCl completely ionizes.
a) Calculate the pH of the solution right now.
b) You now add V litres of water to the solution.
Find the pH of the new dilute solution in terms of V.


Homework Equations


pH = -log[H]


The Attempt at a Solution


a) pH = -log(2) which is interestingly a negative number so I hope this is correct.
b) I'm not sure how you would do this part. Do we just find the new concentration in terms of V and then find the negative log of that? Don't we have to consider the autoionization of water?

Thanks!

BiP
 
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Yes, negative pH is possible.

Check the link I posted in your other thread for the way pH should be calculated. And read about dilution calculation.
 

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