Question

# What volume of 0.100 M NaOH is needed to make 100.0 mL of a buffer solution...

What volume of 0.100 M NaOH is needed to make 100.0 mL of a buffer solution with a pH of 6.00 if one starts with 50.0 mL of 0.100 M potassium hydrogen phthalate? The K a2 for potassium hydrogen phthalate is 3.1 × 10-6

Potassium acid phthalate , KHP
HP ⇐⇒ H+ + P-
Ka = 3.1 x 10^-6 ;
pKa = - log(3.1 x 10^-6); pKa = -(-5.51) = 5.51
pH = pka + log[P-]/ [HP]
6.00 = 5.51 + log[P-]/ [HP]
0.49 = log[P-]/ [HP]
[P-] /[HP] = 3.09 ; this is also the ratio of moles
Moles of HP = 50.0 mL x 1 liter/1000 mL x 0.10 mole HP/ 1liter = 0.005 moles
Addition of OH- will decrease the HP and increase the P-
OH- + HP =⇒ H2O + P-
Let Y = moles of P-
Then 0.005 – Y= moles of HP
Y moles P-/ moles HP = 3.09
Y / 0.005 –Y = 3.09 ; Y = 3.09(0.005 – Y)
Y = 0.015 – 3.09Y
4.09Y = 0.01545
Y = 0.00378 moles
OH- + HP =⇒ H2O + P-
Moles of OH- added = moles of P- that are produced.
0.00378moles of P- x 1 liter NaOH/ 0.100 moles NaOH = 0.0378 liters
0.0378 liters x 1000 mL/ 1 liter = 37.8 mL of NaOH

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