Question

A solution is prepared by adding 0.350 moles of H2PO4- 1.500 L of water (I) calculate the ph of this solution (Ka (H2PO4-) = 6.2 x 10^-8 (II) calculate the ph of this solution after adding 0.00375 moles of HCl. ka1= 7.5 x 10^-3

Answer 1

(I) The acid dissociation reaction is as below.

H₂PO₄^- (aq) <======> HPO₄^2- (aq) + H⁺ (aq)

Molar concentration of HPO₄^2- = [HPO₄^2-] = (moles of HPO₄^2-)/(volume of solution) = (0.350 mole)/(1.500 L) = 0.2333 M.

The acid dissociation constant is given as

K_a = 6.2*10^-8 = [HPO₄^2-][H⁺]/[H₂PO₄^-] = (x).(x)/(0.2333 – x)

Use the small x approximation. Since K_a is small, we can assume x << 0.2333 M and write

6.2*10^-8 = x²/0.2333

====> x² = 1.44646*10^-8

====> x = 1.2027*10^-4

Therefore, [H⁺] = 1.2027*10^-4 M and pH = -log (1.2027*10^-4) = 3.9198 ≈ 3.92 (ans).

(II) Since K_a1 is given, we shall consider the reaction

H₃PO₄ (aq) <=====> H₂PO₄^- (aq) + H⁺ (aq)

We are adding HCl, hence will shall consider the reverse reaction, i.e,

H₂PO₄^- (aq) + H⁺ (aq) -------> H₃PO₄ (aq)

The base ionization constant is

K_b1 = K_w/K_a1 = (1.0*10^-14)/(7.5*10^-3) = 1.3333*10^-12; therefore, pK_b1 = -log (1.3333*10^-12) = 11.8751.

Moles of HCl added = moles of H₂PO₄^- neutralized = moles of H₃PO₄ formed = 0.00375 mole.

Moles of H₂PO₄^- retained at equilibrium = (0.350 – 0.00375) mole = 0.34625 mole.

Since the volume of the solution remains unchanged, we can replace molar concentrations of H₃PO₄ and H₂PO₄^- but the number of moles at equilibrium.

Use the Henderson-Hasslebach equation for base as below.

pOH = pK_b1 + log [H₃PO₄]/[H₂PO₄^-] = 11.8751 + log (0.00375 mole)/(0.34625 mole) = 11.8751 + log (0.01083) = 11.8751 + (-1.9654) = 9.9097.

Since pH + pOH = 14.00, we have pOH = 14.00 – pH = 14.00 – 9.9097 = 4.0903 ≈ 4.10 (ans).