Question

What is the concentration of X2- in a 0.150 M solution of the diprotic acid H2X? For H2X, Ka1 = 4.5 x 10 - 6 and Ka2 = 1.2 x 10 - 11.

Answer 1

Answer:

Concentration of  X^2- is 1.265 x 10^-6

Explanation:

• Ka1 = 4.5 x 10 - 6 for reaction

                   H2X   →   H+   +   HX^-

• Ka2 = 1.2 x 10 - 11 for reaction

                   HX^-   →   H+   +   X^2-

• Ka for overall reaction is given by multiplicative effect of Ka,s of all steps

     Therefore for reaction  H2X   →   2H+   +   X^2- is given by :-

      Ka = Ka1  x   Ka2

            = (4.5 x 10^-6)  x  (1.2 x 10^-11)

            = 2.025 x 10^-18

• Formula used:-

        Ka = (Product of concentration of product each raise to the power to                 their stoichiometric coefficient)/(Product of concentration of reactant each raise to the power to their stoichiometric coefficient)

• At equilibrium let us consider a mol /L of H2X dissociates into 2H+ and X^2-.
• Therefore concentration H+ is ax and that of X^2- is a
• Applying formula:-

                         Ka= [(a) x (2a)^2]/(0.15)]

                          2.025 x 10^-18=(4a^3)/.15

        On Solving :-a = 1.265 x 10^-6

• Hence cocentration of X^2- is 1.265 x 10^-6

Answer 2

Answer:

Answer is 2.0 × 10^-9. RATE IT BELOW