4 Part Question

Question

You are given 60.0 mL of a 0.500 M dihydrogen phosphate / hydrogen phosphate buffer. If the buffer is 0.196 M H₂PO₄^- and 0.304 M HPO₄^2-, answer the following:

Give your answers to 3 sigfigs.

Initial moles of H₂PO₄^- :  moles

Initial moles of HPO₄^2-:  moles 

 

To the buffer from the first question, you add 1.7 mL of a 1.00 M HCl solution. Calculate the moles of hydronium added, and the new equilibrium amounts of H₂PO₄^- and HPO₄^2- in the buffer. 

Give your answers as decimals, to the nearest ten thousandth (0.0001) mole.

moles of H3O+ added:  moles

moles of H₂PO₄^- at equilibrium:  moles

moles of HPO₄^2- at equilibrium:  moles

The pKa of H₂PO₄^- is 6.64. Based on your the moles of H₂PO₄^- and HPO₄^2-above, what is the pH of your buffer after adding 1.7 mL of the 1.00 M HCl solution? 

Give your answer to the nearest hundredth.

Repeat the above calculations with a "fresh" 60.0 mL of 0.500 M buffer solution, but instead add 3.7 mL of a 1.00 M NaOH solution. What is the pH of the buffer after adding the base?

Give your answer to the nearest hundredth.

Answer 1

60.0 mls x 1 L / 1000 mls x 0.196 mol / L H₂PO₄^- = 0.0118 moles H₂PO₄^-

60.0 mls x 1 L / 1000 mls x 0.304 mol / L HPO₄^2- = 0.0182 moles HPO₄^2-

1.7 ml x 1 L / 1000 ml x 1.00 mol/L HCl = 0.0017 mols HCl = 0.0017 mols H3O+ added

mols H₂PO₄^- = 0.0118 mol + 0.0017 mols = 0.0135 mols H₂PO₄^-

mols HPO₄^2- = 0.0182 - 0.0017 = 0.0165 mols HPO₄^2-

pH = pKa + log [HPO₄^2-]/[H₂PO₄^-] = 6.64 + log (0.0165/0.0135)

pH = 6.64 + 0.087

pH = 6.73

For NaOH experiment:

mols OH- added = 3.7 mls x 1 L / 1000 mls x 1.00 mol/L = 0.0037 mols OH-

mols H₂PO₄^- = 0.0118 mol - 0.0037 mol = 0.0081 mols H₂PO₄^-

mols HPO₄^2- = 0.0182 mol + 0.0037 mol = 0.0219 mols HPO₄^2-

pH = pKa + log [HPO₄^2-]/[H₂PO₄^-] = 6.64 + log (0.0219/0.0081)

pH = 6.64 + 0.43

pH = 7.07