Thread: why doesn't the pH go above 7 if you keep thinning an acid?

I understand the answer to the question, but my formula's seem to tell otherwise:
If you have a (strong) acid, and you decrease the concentration of the acid, by adding water of pH = 7, then the pH ofthe acid will increase by 0.3.
But wouldn't that mean that if you kept decreasing the concentration, eventually it would surpass 7?

This is obviously not true, but the formula seems to say so:
if you have a pH = 5 HCl sollution, then you have a [H3O+] = 10^-5.
if you decrease the concentration thousandfold it would become 10^-8, and the pH would be 8.

How do you do the right calculations?

Your thinking is partially correct: If you were to dilute an acid with something having zero hydrogen ion concentration, you could change the liquid's pH to over 7 ... in fact, well over 7. (Something with a zero hydrogen ion concentration would be extremely alkaline, of course.) So, your thinking along those lines is perfectly correct.

However, water has a pH of 7, which means that it has a hydrogen ion concentration of its own, and the best we could hope for would be to dilute something to a pH that approaches 7.

Originally Posted by jrmonroe

Your thinking is partially correct: If you were to dilute an acid with something having zero hydrogen ion concentration, you could change the liquid's pH to over 7 ... in fact, well over 7. (Something with a zero hydrogen ion concentration would be extremely alkaline, of course.) So, your thinking along those lines is perfectly correct.

However, water has a pH of 7, which means that it has a hydrogen ion concentration of its own, and the best we could hope for would be to dilute something to a pH that approaches 7.

That makes sense.
How will the exact calculations go?


ps. not very important, but in your example of pH over 7, the pOH will also be over 7 in that case right? (while normally, pOH = 14 -pH)

Let’s say the two volumes contain 10^7 molecules, so an acid with pH 5 contains 100 H3O+ ions, and water with pH 7 contains 1 H3O+ ion. Together they contain 2 × 10^7 molecules with 101 H3O+ ions. Half the solution would contain 10^7 molecules with 50.5 H3O+ ions. The pH =–log(50½/10^7) = 5.30. Diluting this with water and halving would yield 10^7 molecules with 25.75 H3O+ ions, giving a pH of 5.59. Repeated dilutions yields pH’s of: 5.30, 5.59, 5.874, 6.143, 6.388, 6.594, 6.751, 6.858, 6.923, 6.960, 6.979, 6.990, 6.995, 6.9974, 6.9987, 6.9993, 6.9997, 6.9998, and 6.9999.

Let an Excel spreadsheet do these calculations for you instead of doing them by hand. Start with 100 in A1, and with =–LOG(A1/10000000) in B1. Then put =(A1+1)/2 into A2, and fill down B1 into B2 (Ctrl+D). Then fill down A2 and B2 into A3 and B3 through A20 and B20. You’ll see how the diluted pH approaches, but never reaches, a pH of 7. Then go back to A2 and change it to your original thinking by changing the function simply =A1/2 (meaning no H3O+ ions in the “water”). Fill down A2 into A3 through A20, and you’ll see the dilutions approach a pH of 11.

Originally Posted by jrmonroe

Let’s say the two volumes contain 10^7 molecules, so an acid with pH 5 contains 100 H3O+ ions, and water with pH 7 contains 1 H3O+ ion. Together they contain 2 × 10^7 molecules with 101 H3O+ ions. Half the solution would contain 10^7 molecules with 50.5 H3O+ ions. The pH =–log(50½/10^7) = 5.30. Diluting this with water and halving would yield 10^7 molecules with 25.75 H3O+ ions, giving a pH of 5.59. Repeated dilutions yields pH’s of: 5.30, 5.59, 5.874, 6.143, 6.388, 6.594, 6.751, 6.858, 6.923, 6.960, 6.979, 6.990, 6.995, 6.9974, 6.9987, 6.9993, 6.9997, 6.9998, and 6.9999.

Let an Excel spreadsheet do these calculations for you instead of doing them by hand. Start with 100 in A1, and with =–LOG(A1/10000000) in B1. Then put =(A1+1)/2 into A2, and fill down B1 into B2 (Ctrl+D). Then fill down A2 and B2 into A3 and B3 through A20 and B20. You’ll see how the diluted pH approaches, but never reaches, a pH of 7. Then go back to A2 and change it to your original thinking by changing the function simply =A1/2 (meaning no H3O+ ions in the “water”). Fill down A2 into A3 through A20, and you’ll see the dilutions approach a pH of 11.

Thanks alot. It really makes sense now.