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Homework Statement:
 Recalculate the pH after the addition of 0.1 M HCl. Compare this value with that where the same 0.05 ml of 0.1 M HClis added to the equivalent amount of pH 7 water. You may assume no significant change in volume since you are only adding 1 drop to each.
Relevant Equations:

pH = pKa + [salt]/[acid]
pKa =  log (Ka)
This is for a high school chemistry class. In part a of the question, I calculated the pH of the solution to be 3.38. Part a was the question: Calculate the pH of a solution containing 0.75 M lactic acid (Ka= 1.4 *10^4) and 0.25 M sodium lactate.
For part b I am having trouble determining how to start. I believe I will be adding 5*10^6 to the 0.75 and 0.25 in the expression: pH = pKa + log (.25/.75) from the previous question. (5*10^5 L)(.1 M)= 5*10^6
Please let me know if I have not proved all of the necessary information. Any help is appreciated, and thank you for your time.
For part b I am having trouble determining how to start. I believe I will be adding 5*10^6 to the 0.75 and 0.25 in the expression: pH = pKa + log (.25/.75) from the previous question. (5*10^5 L)(.1 M)= 5*10^6
Please let me know if I have not proved all of the necessary information. Any help is appreciated, and thank you for your time.