Question

Most surface waters have silica dissolved in it in the form of silicic acid H4SiO4 (pKa = 9.5) and its ionisation product H3SiO4 ^- .

(a) Calculate the mole fraction of H3SiO4 ^- and of H4SiO4 at pHs of 5.0, 6.0, 7.0, 8.0, 9.0, 10.0 and 11.0. The total silica concentration (H3SiO4 ^- + H4SiO4) is 0.2 mM. Assume that activity = concentration.

(b) Draw a plot with mole fraction on the y-axis and pH on the x-axis. Use the pH range 55.0 to 11.0.

(c) On the plot, at what pH do the two mole fraction lines intersect? What is the significance of this pH?

Answer 1

a. H₄SiO₄ H₃SiO₄^- + H⁺

According to Henderson-Hasselbalch equation,

pH = pKa + Log(n_H3SiO4^-/n_H4SiO4)

Here, pKa = 9.5

At pH = 5

Log(n_H3SiO4^-/n_H4SiO4) = 5 - 9.5 = -4.5

i.e. n_H3SiO4^-/n_H4SiO4 = 3.16*10^-5

Let n_H3SiO4^- = x, then n_H4SiO4 = a - x

i.e. x/(a-x) = 3.16*10^-5

i.e. (a-x)/x = 31623

i.e. a/x - 1 = 31623

i.e. a/x = 31624

i.e. x/a = 3.16*10^-5 = mole fraction of H₃SiO₄^-

Since the sum of mole fractions of the H₄SiO₄ and H₃SiO₄^- = 1, the mole fraction of H₄SiO₄ = 1 - 3.16*10^-5 = 9.9997*10^-1

At pH = 6

Log(n_H3SiO4^-/n_H4SiO4) = 6 - 9.5 = -3.5

i.e. n_H3SiO4^-/n_H4SiO4 = 3.16*10^-4

Let n_H3SiO4^- = x, then n_H4SiO4 = a - x

i.e. x/(a-x) = 3.16*10^-4

i.e. x/a = 3.16*10^-4 = mole fraction of H₃SiO₄^-

Since the sum of mole fractions of the H₄SiO₄ and H₃SiO₄^- = 1, the mole fraction of H₄SiO₄ = 1 - 3.16*10^-4 = 9.997*10^-1

At pH = 7

Log(n_H3SiO4^-/n_H4SiO4) = 7 - 9.5 = -2.5

i.e. n_H3SiO4^-/n_H4SiO4 = 3.16*10^-3

Let n_H3SiO4^- = x, then n_H4SiO4 = a - x

i.e. x/(a-x) = 3.16*10^-3

i.e. x/a = 3.15*10^-3 = mole fraction of H₃SiO₄^-

Since the sum of mole fractions of the H₄SiO₄ and H₃SiO₄^- = 1, the mole fraction of H₄SiO₄ = 1 - 3.15*10^-3 = 9.97*10^-1

At pH = 8

Log(n_H3SiO4^-/n_H4SiO4) = 8 - 9.5 = -1.5

i.e. n_H3SiO4^-/n_H4SiO4 = 3.16*10^-2

Let n_H3SiO4^- = x, then n_H4SiO4 = a - x

i.e. x/a = 3.06*10^-2 = mole fraction of H₃SiO₄^-

Since the sum of mole fractions of the H₄SiO₄ and H₃SiO₄^- = 1, the mole fraction of H₄SiO₄ = 1 - 3.06*10^-2 = 9.7*10^-1

At pH = 9

Log(n_H3SiO4^-/n_H4SiO4) = 9 - 9.5 = -0.5

i.e. n_H3SiO4^-/n_H4SiO4 = 3.16*10^-1

Let n_H3SiO4^- = x, then n_H4SiO4 = a - x

i.e. x/a = 2.4*10^-1 = mole fraction of H₃SiO₄^-

Since the sum of mole fractions of the H₄SiO₄ and H₃SiO₄^- =1, the mole fraction of H₄SiO₄ = 1 - 2.4*10^-1 = 7.6*10^-1

At pH = 10

Log(n_H3SiO4^-/n_H4SiO4) = 10 - 9.5 = 0.5

i.e. n_H3SiO4^-/n_H4SiO4 = 3.16

Let n_H3SiO4^- = x, then n_H4SiO4 = a - x

i.e. x/a = 7.6*10^-1 = mole fraction of H₃SiO₄^-

Since the sum of mole fractions of the H₄SiO₄ and H₃SiO₄^- =1, the mole fraction of H₄SiO₄ = 1 - 7.6*10^-1 = 2.4*10^-1

b. You can plot all these values using OrginPro software.