pls help!! What if one were to make a solution by adding weak acid and some of i

Question

pls help!! What if one were to make a solution by adding weak acid and some of its conjugate base. Let’s use HA with Ka = 2.0 x 10-5 for the weak acid and NaA for its conjugate base, A-. (Again, we don’t care about the alkali metal ions in aqueous solutions. They are merely spectator ions in the arena of acid-base chemistry.) Now, let’s look at adding 1.00 mole of HA and 0.50 mole of NaA in enough water to make a 2.00 L solution. Determine initial concentrations of initial HA and A-. [HA] = ___________ [A-] = ______________ The equilibrium is HA(aq) + H2O(l)H3O+ + A- as before but now there are significant initial concentrations both HA and A-. As before set up the ICE table using these two initial concentrations and solve for the “x”. Note both initial concentrations this time are much, much greater than Ka so it is reasonable to use the small x approximation and in this case one can use this approximation twice in solving for x in the Ka expression. This is sometimes referred to as the “double small x approximation”. The initial concentrations of HA and A- approximately do not change once equilibrium is achieved and so initial concentrations can be used as the final equilibrium concentrations. Be sure to show where these approximations are made in your calculations below.

Explanation / Answer

initial :
[HA] = 1mol/2 L =0.5 M
[A-] = 0.5 mol / 2 L = 0.25 M

HA + H2O ---> H3O+ + A-
0.5 0 0 0.25 (initial)
+x +x -x -x (change)
0.5+x x -x 0.25-x (at equilibrium)

use:
Ka = [H3O+][A-]/[HA]
2*10^-5 = x*(0.25-x)/(0.5+x)

use double small x approximation,
0.25-x can be written as 0.25
0.5+x can be written as 0.5

2*10^-5 = x*(0.25-x)/(0.5+x)
2*10^-5 = x*(0.25)/(0.5)
x = 4*10^-5 M

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