(1) A beaker with 165 mL of an acetic acid buffer with a pH of 5.00 is sitting on a bench top. The total molarity of the acid and conjugate base in the buffer is 0.100M. A student adds 7.20mL of a 0.320M HCl solution to the beaker. How much will the pH change? The pka of acetic acid is 4.760 . What is the change in pH?

(2) Imagine that you are in chemistry lab and need to make 1.00 L of a solution with a pH of 2.80.
You have in front of you

100mL of 7.00×10−2 M ,
100mL of 5.00×10−2M , and
plenty of distilled water.

You start to add HCl to a beaker of water when someone asks you a question. When you return to your dilution, you accidentally grab the wrong cylinder and add some NaOH . Once you realize your error, you assess the situation. You have 81.0 mL of HCl and 85.0 mL of NaOH left in their original containers.

Assuming the final solution will be diluted to 1.00 L , how much more HCl (in mL) should you add to achieve the desired pH?

See my suggestions from an earlier post.

http://www.jiskha.com/display.cgi?id=1247505336

The second one follows.

http://www.jiskha.com/display.cgi?id=1247505052

To get the change in pH for the first question, we need to calculate the change in concentration of HCl after it is added to the buffer solution and then use that change in concentration to calculate the change in pH. Here are the steps to solve this problem:

Step 1: Calculate the moles of HCl added.
Moles of HCl = Volume of HCl (in L) x Molarity of HCl
Volume of HCl = 7.20 mL = 7.20/1000 L
Molarity of HCl = 0.320M
Moles of HCl = (7.20/1000) L x 0.320 mol/L

Step 2: Calculate the new moles of acetic acid and conjugate base in the buffer.
Moles of acid = Initial volume of acid (in L) x Initial Molarity of acid
Volume of acid = 165 mL = 165/1000 L
Initial Molarity of acid = 0.100 M
Initial moles of acid = (165/1000) L x 0.100 mol/L
Initial moles of conjugate base = Initial moles of acid

Step 3: Calculate the new concentrations of acid and conjugate base after the addition of HCl.
Final moles of acid = Initial moles of acid + Moles of HCl
Final volume of acid = Initial volume of acid + Volume of HCl (converted to L)
Final Molarity of acid = Final moles of acid / Final volume of acid
Final Molarity of conjugate base = Final moles of conjugate base / Final volume of acid

Step 4: Calculate the change in pH.
Change in concentration of H+ = Final Molarity of acid - Initial Molarity of acid
Change in pH = -log10(Change in concentration of H+)

Now, let's go through the steps to solve the second question:

Step 1: Calculate the total moles of acid and conjugate base in the initial solution.
Moles of acid = Volume of acid (in L) x Molarity of acid
Volume of acid = 100 mL = 100/1000 L
Molarity of acid = 7.00x10^-2 M
Moles of acid = (100/1000) L x 7.00x10^-2 mol/L
Moles of base = Volume of base (in L) x Molarity of base
Volume of base = 100 mL = 100/1000 L
Molarity of base = 5.00x10^-2 M
Moles of base = (100/1000) L x 5.00x10^-2 mol/L

Step 2: Calculate the remaining moles of acid and base after the accidental addition of NaOH.
Remaining moles of acid = Initial moles of acid - moles of NaOH added
Remaining moles of base = Initial moles of base - moles of HCl added

Step 3: Calculate the remaining volume of the solution.
Remaining volume = Total volume - Volume of NaOH added

Step 4: Calculate the new concentration of acid in the remaining solution.
New Molarity of acid = Remaining moles of acid / Remaining volume

Step 5: Calculate the moles of HCl needed to achieve the desired pH.
Desired pH = 2.80
Desired concentration of H+ ions = 10^(-desired pH)
Moles of H+ ions needed = Desired concentration of H+ ions x Remaining volume

Step 6: Calculate the volume of HCl needed to achieve the desired pH.
Volume of HCl needed = Moles of HCl needed / Molarity of HCl

Now you have the answer to the second question.