This is a long question with multiple steps, so any help at all will be appreciated! Please explain how you got the answers so I will know how to do it on my own.
Suppose you have a sample of 25.0 mL of 0.01M HCl solution. HCl is a strong acid and it completely dissociates in water. Calculate the pH of this solution (pH= -log[H3O+]).
*I calculated the pH of a 25.0mL of 0.01M HCl solution to be 2. That could be wrong.
The Ka of acetic acid is 1.8 x 10^-5 at 25 degrees C. Calculate the pH of a 25.0 mL sample of 0.01M of acetic acid solution.
*I calculated pH of 25.0mL of acetic acid solution to be 3.4. This could also be wrong.
Exploring how big the pH changes near the "equivalence point"... First consider the 25.0mL of 0.01M HCl solution. Suppose we titrate this solution with 0.01M NaOH solution. The equivalence point refers that equal moles of NaOH is added to the solution. Note when NaOH is added to HCl solution, you have a solution of NaCl which is a salt. It does not produce H3O+ or OH- ions.
pH of the solution when exactly 25.0mL of 0.01M NaOH solution is added to 25.0mL of 0.01M HCl solution?
Now suppose you have only added 24.0mL of 0.01M NaOH solution to the 25.0mL of HCl solution, what is the molarity of HCl left in the solution? What is the pH of the resultant mixture?
Now suppose you added 26.0mL 0.01M NaOH to the solution, what is the molarity of excess NaOH in the mixture? What is the pH of this solution?
Consider what is the pH of the solution when 25.0mL of 0.01M NaOH is added to 25.0mL of 0.01M acetic acid. Is it the same as in the case of HCl solution?
What will the pH of the solution be if exactly 12.50mL of 0.01M NaOH solution is added to 25.0mL of 0.01M acetic acid solution?