Question

1. To a 100.0 mL volumetric flask were added 25.0 mL of 0.74 M KH₂PO₄ and 50.0 mL of 0.35 M Li₂HPO₄. After the flask was filled to the calibration line, a 3.75 mL sample of 2.5 M sodium hydroxide was then added. Ka1= 6.4x10^-8 and Ka2= 4.8x10^-13

1. What is the pH of the initial solution before the sodium hydroxide was added?

2. What is the pH of the final solution after the addition of the sodium hydroxide?

3. Is the solution still a buffer?

Answer 1

Sol .

(a) As Millimoles of KH2PO4 = Millimoles of H2PO4- = Conc. of KH2PO4 × Volume of KH2PO4  

= 0.74 × 25.0

= 18.5 mmol  

Also, Millimoles of HPO42- = Millimoles of Li2HPO4  

= Conc. of Li2HPO4 × Volume of Li2HPO4  

= 0.35 × 50.0

= 17.5 mmol

Reaction :

H2PO4- <----> HPO42- + H+  

As Ka1 = 6.4 × 10^-8

So , pKa = - log(Ka1) = - log ( 6.4 × 10^-8 ) = 7.19

Now , Using Henderson - Hasselbalch equation ,

Initial pH = pKa + log ( Millimoles of HPO42- / Millimoles of H2PO4- )  

= 7.19 + log ( 17.5 / 18.5 )

=  7.16

(b) Now , Millimoles of NaOH added  

= Conc. of NaOH × Volume of NaOH  

= 2.5 × 3.75

= 9.375 mmol

So , New Millimoles of HPO42-

= 17.5 + 9.375 = 26.875 mmol

and , New Millimoles of H2PO4-

= 18.5 - 9.375 = 9.125 mmol

Therefore , Final  pH = pKa + log ( New Millimoles of HPO42- / New Millimoles of H2PO4- )

= 7.19 + log ( 26.875 / 9.125 )

= 7.66  

(c) As there will be very small change in pH upon addition of NaOH to the buffer solution . So ,  the solution still remains a buffer solution .