Question

How many liters of 0.829 M LiOH will be needed to raise the pH of 0.327 L of 5.11 M ascorbic acid (H₂C₆H₆O₆) to a pH of 11.36?

Answer 1

Ascorbic acid (H₂C₆H₆O₆) is a diprotic acid whose pKa values are pKa1 = 4.10 and pKa2 = 11.80.

since the required pH, 11.36 is less than the pKa2 value of ascorbic acid, a pH of 11.36 can be achieved before the 2nd half-equivalence point.

The 1st neutralization reaction(1st equivalence point):

H₂C₆H₆O₆(aq) + LiOH(aq) ----> LiHC₆H₆O₆(aq) + 2H2O(l)

Initial moles of ascorbic acid = C*V = 5.11 mol/L * 0.327 L = 1.67097 mol

Mole ratio between LiOH(aq) and H₂C₆H₆O₆(aq) is: (1 mol LiOH / 1 mol LiHC₆H₆O₆)

=> Moles of LiOH required to reach 1st equivalence point = 1.67097 mol H₂C₆H₆O₆ * (1 mol LiOH / 1 mol LiHC₆H₆O₆)

= 1.67097 mol LiOH

=> Volume of LiOH required till 1st equivalence point = 1.67097 mol / 0.829 mol/L = 2.016 L

The 2nd neutralization reaction(2nd equivalence point):

--LiHC₆H₆O₆(aq) + LiOH(aq) ----> Li₂C₆H₆O₆(aq) + 2H2O(l) ; pKa2 = 11.80

I: 1.67097 mol -------------------------- 0 mol

C: -X ---------------------- -X ----------- +X

E: (1.67097 - X) --------------------------- X

Now LiHC₆H₆O₆ and Li₂C₆H₆O₆ act as buffer solution.

Applying Henderson equation:

pH = pKa + log([Li₂C₆H₆O₆] / [LiHC₆H₆O₆])

=> 11.36 = 11.80 + log[X / (1.67097 - X)]

=> log[X / (1.67097 - X)] = 11.36 - 11.80 = - 0.44

=> X / (1.67097 - X) = 10^-0.44 = 0.36308

=> X = 0.36308 * (1.67097 - X)

=> X = 0.6067 - 0.36308X

=> 1.36308X = 0.6067

=> X = 0.6067/1.36308 = 0.44509 mol

Hence we need 0.44509 mol more to reach a pH of 11..36

=> Volume of LiOH needed after after 1st equivalence point = 0.44509 mol / 0.829 mol/L = 0.5369 L

=> Total volume of LiOH needed = 2.016 L + 0.5369 L = 2.55 L (Answer)