At 25.00 °C, What is the pH of an aqueous 2.50 x 10-8 M HCl solution? The pH of ANY aqueous acid will NEVER be greater than 7.00, REGARDLESS of the acid's concentration! This is because of the "LEVELING EFFECT" of pure water, whose [H+] = 1.0 x 107 M. Thus, the TOTAL [H+] of the solution will be: 1.0 x 107 M +2.50 x 10-8 M = 1.25 x 107 M, thus the pH of the 2.5 x 10-8 M HCl solution will be 6.90 because the contribution of H+ from the autoionization of pure water is greater than that for the very dilute acid, and MUST be accounted for!!!. NOTE: If the acid is di- (or tri-) protic, you must also take that into account!!!

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How to calculate the ph? What would be the result if the acid was a di or tripotic? How do the calculation for them?

At 25.00 °C, What is the pH of an aqueous 2.50 x 10-8 M HCl solution?
The pH of ANY aqueous acid will NEVER be greater than 7.00, REGARDLESS of the
acid's concentration! This is because of the "LEVELING EFFECT" of pure water, whose
[H+] = 1.0 x 107 M. Thus, the TOTAL [H+] of the solution will be: 1.0 x 107 M +2.50 x
10-8 M = 1.25 x 107 M, thus the pH of the 2.5 x 10-8 M HCl solution will be 6.90
because the contribution of H+ from the autoionization of pure water is greater than
that for the very dilute acid, and MUST be accounted for!!!.
NOTE: If the acid is di- (or tri-) protic, you must also take that into account!!!
Transcribed Image Text:At 25.00 °C, What is the pH of an aqueous 2.50 x 10-8 M HCl solution? The pH of ANY aqueous acid will NEVER be greater than 7.00, REGARDLESS of the acid's concentration! This is because of the "LEVELING EFFECT" of pure water, whose [H+] = 1.0 x 107 M. Thus, the TOTAL [H+] of the solution will be: 1.0 x 107 M +2.50 x 10-8 M = 1.25 x 107 M, thus the pH of the 2.5 x 10-8 M HCl solution will be 6.90 because the contribution of H+ from the autoionization of pure water is greater than that for the very dilute acid, and MUST be accounted for!!!. NOTE: If the acid is di- (or tri-) protic, you must also take that into account!!!
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