Question

A. Assume that the 1.0 M (NH4)2 CO3 solution that you used was 1.00 M (NH4) 2 CO3. If Ka of H2CO3 is 5.6 x 10^-11 combine K's to obtain the equilibrium coonstant for NH4 + CO3^2 = NH3 + HCO3-

B. Show that a [Co3^2-] = 0.07 sub(9) M is consistent with the Kyou calculated in (a).

C. Now Calculate the pH of this solution.

D. Compare your calculated pH with the pH of the NH3/ NH4+ buffer( pH buffer: 11.62) . Has this bufferserved its function? Discuss.

E. Using the question above and the Ksp (=6.8x10^-6) of MgCO3 is there enough CO3^2- ion in a 1.0 M (NH4)2CO3 solution to precipitate MgCO3 from a 0.2 M Mg(H2O)6^2+ solution?

F. Is the equilibrium: Mg(H2O) 6^2+ +HCO3+ NH3= MgCO3 + NH4 + 6H2O readily attained? Y or N. State how you know.

Answer 1

a ) solution:

NH4+ <===> NH3 + H+

Ka = [H+][NH3]/[NH4+] = 5.6 x 10^-10

HCO3- <=====> H+ + CO3-2

Ka = [H+][CO3-2]/[HCO3-] = 4.8 x 10^-11

[HCO3-] = [H+][CO3-2]/4.8 x 10^-11

[HCO3-]/[H+][CO3-2] = 1.0/4.8 x 10^-11

[HCO3-]/[H+][CO3-2] = 2.08 x 10^10

(III) Reaction: NH4+ + CO3-2 ⇐=⇒ NH3 + HCO3-

[HCO3-]/[H+][CO3-2] x [H+][NH3]/[NH4+] = 5.6 x 10^-10 x 2.08 x 10^10
The [H+] drop out.

[HCO3-]/[CO3-2] x [NH3]/[NH4+] = 11.64

K = 11.64

b) 0.07 M

1.0 Molar (NH4)2CO3 would initially have 2 M NH4+ and 1 M CO3-2

If the equilibrium [CO3-2] is 0.07 , then the amount of CO3-2 that reacted is 1.0 – 0.07 = 0.93

Thus, the [NH3] and [HCO3-] at equilibrium is 0.93

The amount of [NH4+] initially is 2(1.0) = 2.0 Molar

The amount of [NH4+] at equilibrium is (2.0 – 0.93) = 1.07

Substituting these values in the equilibrium constant expression for reaction (III)

[NH3][HCO3-]/[NH4+][[CO3-2] = (0.93)(0.93)/(0.07)(1.07) = 11.5 which agrees with the

K value of 11.64.

c)

HCO3- ⇐⇒ H+ + CO3-2

Ka = [H+][CO3-2]/[HCO3-] = 4.8 x 10^-11

[H+](0.07)/0.93 = 4.8 x 10^-11

[H+] = 6.3 x 10^-10

pH = - log(6.3 x 10^-10) = -(-9.2)

= 9.2