If 50% of H3O+ ions are removed from a solution whose pH was originally 3...

[Ultimeaciax]

If 50% of H3O+ ions are removed from a solution whose pH was originally , what's the new pH?

Here's my attempt:

It means I have 50% remain in the solution = 50/100 = 1/2

So, log(2) = 0.30. Thus, new pH is 3.30?

Did I do that right? If not, please explain.

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[Rabolisk]

That's right.

 

[DougTPR]

If 50% of H3O+ ions are removed from a solution whose pH was originally , what's the new pH?

Here's my attempt:

It means I have 50% remain in the solution = 50/100 = 1/2

So, log(2) = 0.30. Thus, new pH is 3.30?

Did I do that right? If not, please explain.

Here's a useful shortcut since you won't have a calculator with you to get such precise values for the logs.

1 pH unit is equivalent to a 90% change in concentration. Thus...

3 -> 4 equals a 90% (90% of 100) reduction in H+

3 -> 5 equals a 99% (90% of 100 + 90% of 10) reduction in H+

3 -> 6 equals a 99.9% (90% of 100 + 90% of 10 + 90% of 1) reduction in H+

and so on...

So 50% is not enough of a change to even change the pH by one unit.

 

[timewarp424]

Another way to accomplish the same outcome, is to convert your pH to its hydronium ion concentration. Given a pH of 3, we set our concentration equal to 10^-3 of ions. That value is .0001 We then divide that in half, getting .0005. We then put it back into pH form. -log(.0005), getting 3.3 Exactly as you had.