Initially, 0.200 mol of Hl is placed in a 5.00 L flask. The following equilibriu

Question

Initially, 0.200 mol of Hl is placed in a 5.00 L flask. The following equilibrium is established: 2Hl(g) H_2(g) + I_2(g). Once equilibrium is established, the flask was found to contain 0.124 mol of HI. Calculate the equilibrium concentrations of H_2 & I_2 and the equilibrium constant, K Boric Acid (monoprotic acid of type HA with K_a = 5.0 Times 10^-5) is titrated against NaOH(aq). Calculate the pH after 20.0 mL of 0.200 M NaOH(aq) has been added to 50.00 boric acid. Calculate the pH at the equivalence point.

Explanation / Answer

3)

               2HI     ---> I2    + H2

Initial                     0.2            0          0

Chagne                  -2x            x         x

Equilibrium             0.2-2x        x         x

Given

0.2-2x = 0.124 moles

so x = 0.038

so [I2] = moles / L = [H2] = 0.038 / 5 = 0.0076 M

[HI] = 0.124 / 5L = 0.0248

Kc = [H2] [ I2] / [HI]^2 = 0.0076 X 0.0076 / (0.0248)^2 = 0.0939

4) Boric acid will react with NaOH in 1:1 stoichiometric ratio

Moles of acid added = concentration x volume = 10 millimoles

Moles of base added = 20 X 0.2 = 4 millimoles

so moles of acid left = 10-4 = 6 millimoles

Concentraion of acid = moles of acid / total volume = 6 / 50 +20 = 0.0857

Let the boric acid is HA

so it will dissoicates as

HA --> H+ + A-

Ka = 5 X 10^-5 = [H+] [A-] / [HA]

let the dissociation of acid = x

so [H+] = x = [A-]

[HA] = 0.0857 - x

5 X 10^-5 = x^2 / (0.0857-x)

we can igonre x in denominator

5 X 10^-5 = x^2 / 0.0857

0.4285 X 10^-5 = x^2

x = [H+] = 2.07 X 10^-3

pH = -log[H+] = 2.68

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