Help with pH problem from destroyer

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datdentist

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What would be the approximate pH of a 1 x 10^-12M HCl solution?

This is probably a dumb/easy question but can someone explain why we need to include water's pH into this problem? When are we supposed to assume a solution is being diluted?
 
Well if you take the PH without water it would be 12 obviously a strong acid won't have a Ph that high.

Usually water isn't taken into account like this but in this case you can tell it is because the concentration of HCl is so little.
 
[H] for water is 1x10^-7 .... 100,000 times more concentrated than the ACID being added.

Basically you're adding nothing to water.
 
Almost all the acid/base stuff we do in chemistry is in water. Water itself (normally, under standard conditions) has a pH of around 7 because it dissociates into [-OH] and [H+] to an extent that each has a concentration of 1 x 10^-7, or .0000001. As you can see, a very small amount.

This question is probably throwing you off because 99% of the time you've seen acid solutions, we haven't accounted for the acidity for water. The concentration of given acid solutions tends to be so high that we can effectively disregard the amount of H+ from water because it is insignificant in comparison.

For example, the pH of a .01M HCl solution is 2 because the concentration of H+ is .01 or 1x10^-2. What about the H+ from water? As you recall from earlier, the concentration of H+ already there from water is .0000001. So .01 + .0000001 ...we're basically adding a number so small that the inclusion of H+ from water doesn't make a difference.

But when the solution is 1x10^-12 HCl, that's effectively .000000000001 H+ from acid with .0000001 H+ from water. In this case it is clear the amount from HCl is insignificant.

Long story short: for an H+ concentration of 1x10^-X, where X is a number greater than 7, account for the existing H+ from water!
 
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