Question

Calculate the expected pH when 0.5 mL of 0.1 M HCL is added to 30.0 mL pure water. Do the same calculation when the same amount of HCl is added to 30.0 mL of your original .05M buffer. Compare the calculated pH to the actual measured pH.

My buffer has a pH of 5.09 and I used acetic acid which has a pKa of 4.76

2.
If you have 50 mL of .05M potassium phosphate buffer pH 7.0 and you added 2.0 mL of 0.1M HCl what would the final pH be? What would the pH be if 2.0 mL of 0.1M NaOH was added to the same amount of the original buffer?

Answer 1

Question 1.

Calculate the expected pH when 0.5 mL of 0.1 M HCL is added to 30.0 mL pure water.

[HCl] = [H+] = mol of HC / total V

mol of HCl = MV = 0.5*1 = 0.05 mmol of H+

V total = 0.5 + 30 = 30.5 mL

[H+] = mol/V = 0.05 /30.5 = 0.001639 M

pH = -log(0.001639 ) = 2.7854

Now,

pH changes from 7 to 7854

Question 2.

Do the same calculation when the same amount of HCl is added to 30.0 mL of your original .05M buffer. Compare the calculated pH to the actual measured pH.

My buffer has a pH of 5.09 and I used acetic acid which has a pKa of 4.76

pH = pKa + log(A-/HA)

5.09 = 4.76 + log(A-/HA)

A-/HA = 10^(5.09-4.76) = 2.1379

A-/HA =2.1379 --> A- = 2.1397*HA

A-+ HA = 0.05*30 = 1.5 mmol

2.1397*HA + HA =1.5

HA(2.1397+1) = 1.5

HA = 1.5 / ((2.1397+1)) = 0.477 mmol

A- = 2.1397*HA = 2.1397*0.477 = 1.0206 mmol

after addition:

0.5 mL of M = 0.1 M of HCl --> mmol of HCl = MV = 0.5*0.1 = 0.05

after reactions:

HA = 0.477 +0.05 = 0.527 mmol

A- = 1.0206 -0.05 = 0.9706 mmol

Substitute in pH equation

pH = pKa + log(A-/HA)

pH = 4.76 + log(0.9706 /0.527 )

pH = 5.0252