Need a brainy person to help me!! lol?

Question

can someone please work this out for me and explain how they did it? i spent a really long time on it and im not getting anywhere..thank u!! (:

1. An aqueous solution of acetic acid (HOAc) was prepared by adding water to 0.10 mol of HOAc to give a final volume of 1.00 L. What is the pH of the final solution?

Ka(HOAc) = 1.78 x 10-5

2. A buffer solution was prepared by mixing 0.010 mol NH4Cl and 0.010 mol NH3 in sufficient water to give a final volume of 0.10 L. To this solution was added 2.0 mL of 1.0 mol L-1 HCl. To what value will the final pH of the resulting solution be closest?

pKb(NH3) = 4.76

Answer 1

EDITED: (Forgot the log in the H-H eq, and added sig. figures to [H+] value in the first problem.)

The dissociation eq for HOAc is

HOAc ---> H+ + OAc-

The dissociation is not complete (you are given the Ka, dissociation constant). Therefore there will be HOAc and H+ and OAc- in the solution at the same time. The initial concentration of HOAc

[HOAc] = 0.1M

Whatever part of this dissociates to H+ and OAc- reduces the concentration of HOAc from the original value. If dissociation produces [H+] = x, then [OAc] = x also, and the remaining [HOAc] = 0.1M - x. The definition of Ka is

Ka = [H+]*[OAc] / [HOAc]

putting in terms of x:

Ka = x^2 / (0.1 - x)

multiply it out to get

0.1*Ka - x*Ka = x^2, or x^2 + x*Ka - 0.1*Ka = 0

x^2 + 1.78*10^-5*x - 1.78*10^-6 = 0

solve this quadratic equation for x; you will get x = 0.001325 (ignore the negative result). Since x = [H+], the pH is -log(x) = 2.88

I'm not sure about the second problem. You would use the basic form of the Henderson-Hasselbalch equation

pOH = pKb +log ([NH4+]/[NH4OH])

to find the pH (= 14 - pOH) of the initial buffer. Then find [NH4OH] after adding the HCl (the H+ in the HCl will react with some of the OH, but [NH4+] will not change).

Answer 2

1. For a weak acid it is easy to show that

pH = 0.5(pKa - log C)

where Ka is the ionisation constant and C is the acid concentration in mol/l.

With pKa = 4.75 and C =0.1 mol/l and hence

pH = 0.5 (4.75 - log(0.1)) = 2.88

2. For ammonia Kb =[NH4+][OH-]/[NH3]

Rearranging yields pKb =pOH - log([NH4+]/[NH3]) or

pH = pKw - pKb - log([NH4+]/[NH3]) = 9.24 - log([NH4+]/[NH3]) .... Eqn 1

The charge balance in this solution is

[H+] + [NH4+] = [OH-] + [Cl-]. As [H+] << [NH4+] and [OH-] <<[Cl-], this simplifies to [NH4+] = [Cl-]

Now [Cl-] = (100x0.1 + 2x1)/102 = 0.1176 mol/l = [NH4+]

[NH3] = 100x(0.1+0.1)/102 - 0.1176 = 0.07848 mol/l

Substituting into Eqn 1 leads to

pH = 9.24 - log(0.1176/0.07848) = 9.06