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. Given the initial moles of HI(g) = 0.200 mol

Suppose at equilibrium '2y' moles of HI(g) is converted to 'y' mole each of H2(g) and I2(g).

The balanced chemical reaction is

------------------ 2HI(g) ----- > H2(g) + I2(g)

Initial moles: 0.200 --------- 0 -------- 0

Eqm.moles: (0.200 - 2y), y, --------- y

Given the moles of HI at equilibrium =  (0.200 - 2y) = 0.124 mol

=> 2y = 0.200 - 0.124 = 0.076 mol

=> y = 0.076 mol / 2 = 0.038 mol

Hence equilibrium moles of H2 = y = 0.038 mol

Hence equilibrium concentration of H2, [H2] = moles of H2 / Vt = 0.038 mol / 5.00L = 0.0076 M (answer)

equilibrium moles of I2 = y = 0.038 mol

Hence equilibrium concentration of I2, [I2] = moles of I2 / Vt = 0.038 mol / 5.00L = 0.0076 M (answer)

equilibrium concentration of HI, [HI] = moles of HI / Vt = 0.124 mol / 5.00L = 0.0248 M (answer)

Equilibrium constant, K = [H2] x [I2] / [HI]2 = (0.0076 M x 0.0076 M) / ( 0.0248 M)2 = 0.0939 (answer)

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