TBR Chemistry Ch3 #56 Acid Base

[drechie]

Hello - could someone dumb down the easiest way to solve this problem, and perhaps more importantly, explain the "understanding" i should have in tackling this problem.

When I first considered the problem, i thought the solution would have a pH of 2 because i thought water wouldn't affect the pH. This wasnt an answer choice, so I decided to select choice B, pH between 2 and 3, because I thought perhaps the pH would rise slightly because water is neutral and has a pH of 7, therefore increasing the pH a little bit. I clearly wasn't prepared to answer this question - please tell me your thoughts at understanding this Q.

58. If 10 mL of an aqueous solution of a strong acid with pH = 2.0 were mixed with 100 mL of pure water, then the final pH value would be:

A. pH of Less than 2
B. pH between etween 2 and 3
C. pH = Exactly 3
D. pH Greater than 3.


Answer solution:
Choice D is correct. The volume increases from 10 mL to 110 mL when 100 mL of distilled water is added to the solution, so the concentration of the acid must decrease by a factor of 11. The log of 10 is equal to 1, so the log 11 is slightly greater than 1. This implies that the pH increases by just a little more than 1.0, because the [H30+] goes down by a factor of 11. This makes the final pH greater than 3.0. The best answer is greater than 3.0, which is choice D. This is true only of a strong acid solution. If the solution were an aqueous weak acid, the pH increase would not be as significant as with the strong acid.

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[shnarf9892]

Remember that pH = -log[H+] where [H+] is the hydrogen ion concentration (molarity). Molarity, as you recall, is moles of solute (in this case, H+) per liter of solution. So although water is considered neutral at pH=7, adding water to the overall solution will change the pH because you are in essence changing the concentration of H+ in solution.

Also remember that pH is on a logarithmic scale, so a solution with a pH of 3 has 1/10th the concentration of H+ than a solution with a pH of 2. By the same principle, a solution with a pH of 5 has 1/1000th the concentration of H+ than the solution with a pH of 2.

If you wanted to, you could have solved the problem the long and tedious way by finding the number of H+ ions in solution by first finding the initial concentration of H+ [10^(-2)=0.01 mol H+/1L] and then multiplying the concentration by the actual volume you had [0.01 mol H+/1L x 0.010L]. Everything besides moles H+ should cancel out, leaving you with the number of moles you have in your 10 mL solution (1 x 10^-4 moles H+). Then 100 mL water is added to the solution, bringing the volume up from 10mL to 110 mL (or 0.110L) . Don't forget to factor in the ionization of water, where [H+]=10^-7 moles/Liter. Convert the ionization concentration of the water you added into moles of H+ added to our original 1 x 10^-4 moles H+ in solution [10^-7 moles H+/L x 0.100L = 10^-8 moles H+] Add the two values together to get the total amount of H+ you have in solution [10^-4 moles H+ + 10^-8 moles H+ = 1.001 x 10^-4]You can solve for the concentration from there [1.001 x 10^-4 moles H+/0.110L) and you end up with 9.09 x 10^-4 moles H+/L of solution. That's 0.0009 M H+. Take the -log of that to find the pH and the answer is slightly more than 3. I realize there are no calculators on the MCAT, and this method would have taken a LOT more time than you will be able to spend on one problem, but I wanted to illustrate it to be thorough.

Knowing the concepts behind the pH scale will help you save time and get the right answer. Again, if the pH increases by one, then the [H+] has decreased by a factor of 10. A solution with a pH=0 has a H+ concentration of 1.0M. A solution with a pH=1 has a H+ concentration of 0.1M. A solution with a pH=2 has a H+ concentration of 0.01M, and so on and so forth. Based off of that, you would have been able to easily deduce that by increasing the volume more than 10 fold (aka, diluting the H+ in solution), the pH would have increased by more than one unit.

I hope this helped!