- #1
Bohrok
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The first time I came across a chemistry problem about the pH of a solution that's been diluted, I thought there was an argument for a change and for no change. After doing some searching online, it seems that the pH does change since [H3O+] changes, which I understand. However, using the Henderson–Hasselbalch equation equation both before and after dilution would give the same pH, right?
As an example, if I have x moles each of HA and A- in 1 L solution, then[tex]pH = pK_a + \log\left(\frac{\frac{x\text{ mol }A^-}{1 L}}{\frac{x\text{ mol }HA}{1 L}}\right) = pK_a + \log(1) = pK_a[/tex]Now if the solution were at 2 L to start with, then[tex]pH = pK_a + \log\left(\frac{\frac{x\text{ mol }A^-}{2 L}}{\frac{x\text{ mol }HA}{2 L}}\right) = pK_a + \log(1) = pK_a[/tex]
How is this setup giving the wrong answer as opposed to recalculating [H3O+] after the dilution?
As an example, if I have x moles each of HA and A- in 1 L solution, then[tex]pH = pK_a + \log\left(\frac{\frac{x\text{ mol }A^-}{1 L}}{\frac{x\text{ mol }HA}{1 L}}\right) = pK_a + \log(1) = pK_a[/tex]Now if the solution were at 2 L to start with, then[tex]pH = pK_a + \log\left(\frac{\frac{x\text{ mol }A^-}{2 L}}{\frac{x\text{ mol }HA}{2 L}}\right) = pK_a + \log(1) = pK_a[/tex]
How is this setup giving the wrong answer as opposed to recalculating [H3O+] after the dilution?