# Using exponents and logarithms to calculute pH

1. Jul 28, 2013

### interhacker

1. The problem statement, all variables and given/known data

If a solution containing a heavy concentration of hydrogen ions(i.e., a strong
acid) is diluted with an equal volume of water, by approximately how much is its
pH changed? (Express (pH)diluted in terms of (pH)original.)

2. Relevant equations

I think the question requires the equation defining pH:
$pH = \log[H+]$

3. The attempt at a solution

According to the document, the answer is supposed to be:

$pH_{dilute} = pH_{original} + \log 2$

2. Jul 28, 2013

### SteamKing

Staff Emeritus
If the concentration of H+ ions in the original (undiluted) solution is [H+] and you add approximately an equal amount of H+ ions by diluting the original solution, what is the total concentration of H+ ions in the dilute solution?

3. Jul 28, 2013

### symbolipoint

interhacker,
pH means negative of the logarithm of hydrogen ion concentration; you were missing the negative sign.
pH = -log[H+]

If you just double the volume to make the dilution, this is only a fraction of a pH unit change. The concentration change must be up or down by a factor of 10 for a change of pH of one unit.

4. Aug 1, 2013

### epenguin

He must have made another mistake that cancelled out the first one and made the answer right!