long, but i really need help! please help!!! greatly appreciate!! S uppose 0.250
ID: 965177 • Letter: L
Question
long, but i really need help! please help!!! greatly appreciate!!
Suppose 0.250 moles of a weak acid, HA, is added to enough water to make a 2.00 L solution. The Ka of the acid is 2.0 x 10-5 . Identify the appropriate weak acid equilibrium. Set up an appropriate ICE table. Write the Ka expression in terms of equilibrium concentrations from your ICE table and solve for the “x”. Now determine [HA], [A-], [H3O+], and [OH-]. As a simplification, one can often use a “small x approximation” for weak acids in solution. This is because Ka’s of weak acids are generally much smaller than initial concentration so theequilibrium concentration of weak acid is approximately equal to its initial concentration. The good thing about acid-base equilibria is the small x approximation often works out ok. When can you use the small x approximation.
Equilibrium & ICE table: (draw)
[HA]= _______________ [A-]= ______________ [H3O+] = _________________ [OH-] = ___________
One way to determine Ka of a weak acid is to make up a solution with known concentration of the weak acid in water and measure the pH of the resulting solution with a pH meter as you may do next week as a practical. The pH of a solution is defined as pH = –log10([H3O+]) so for a given concentration of weak acid solution, the lower the pH of the solution, the higher the [H3O+], the stronger the acid. For acidic solution, pH is < 7, for basic solution, pH > 7, and for neutral solution, pH = 7.
As an example, consider a 1.00 M solution of a weak acid, HA. The pH of the solution is found to be 3.85.
First, calculate the [H3O+] in the solution. This would be the equilibrium concentration of H3O+ in the solution. Now write out an ICE table as before. Here, we don’t know the numerical value of Ka but we know the [H3O+] at equilibrium which you should see from your ICE table easily relates to the value of “x” in your table and knowing “x” from [H3O+], one can easily determine the other equilibrium concentrations of all species and solve for value of Ka.
[H3O+] = __________________ ICE table:
Ka =
Now suppose one were to dissolve 1.00 mole of a salt, NaA, in enough water to make 2.00 L solution where the A- in NaA is the conjugate base of the weak acid, HA, with a Ka of HA = 2.0 x 10-5. Would the pH of the solution be neutral, pH = 7? Let’s do the calculation and find out.
What would be the initial concentration of A- ? ______________________
Determine the value of Kb in this case. _________________
Next, complete the base association equilibrium below and write in the appropriate ICE table.
A- + H2O__________________
Now set up the expression for Kb and solve for the “x” in this case. Note from your table the “x” refers to [OH-] and not [H3O+]. Determine concentrations of HA, A-, H3O+, OH-. Once again, the “small x approximation” will simplify the math.
Kb = ______________
[HA] = _______________ [A-] = _______________ [H3O+] = _______________ [OH-] = ____________
Now, calculate the pH of the solution. pH = ______________
Although it’s not part of acid-base chemistry, what would be the [Na+] in this case? ______________
One last acid-base equilibrium to consider for this worksheet. So far, we looked at weak acid in water equilibrium (Ka), weak base in water equilibrium (Kb). 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.
Ka Equilibrium ICE table:
Compare the [H3O+] in this case versus that obtained by just adding weak acid, HA in water. Which solution is more acidic – one with weak acid alone or one with weak acid and its conjugate base?_______________ Would this be consistent with le Chatelier’s Principle? __________
Explanation / Answer
Moles of weak acid (HA) = 0.250 and volume of solution = 2.00 L
Hence, Initial concentration of HA = 0.250 / 2.00 = 0.125 M
Equilibrium & ICE table: Acid is weak hence change in H2O will be negligible and hence not shown here. And [OH-] we can find from [H3O+] later on.
HA + H2O < -----> A- H3O+
Initial
0.125 M
-
0
0
Change
-X
-
+X
+X
Equilibrium
(0.125 – X)M
‘X’ M
‘X’ M
[HA]= (0.125-X) M, [A-]= ‘X’M, [H3O+] = ‘X’M, and Ka = 2.0 x 10-5.
We can write expression for Ka for weak acid HA as,
Ka = [H3O+][A]/[HA]
Put all known values,
2.0 x 10-5= (X)(X) / (0.125-X)
But as HA is weak acid. X < < 0.125 M
Hence approximation holds that, 0.125 – X ~ 0.125M.
So, above eqn. takes form,
2.0 x 10-5 = X2 /0.125
X2 = 2.0 x 10-5 x 0.125
X2 = 0.25 x 10-5
X = 1.58 x 10-3
Hence, equilibrium concentrations are,
[A-]= ‘X’M = 1.58 x 10-3 M
[H3O+] = ‘X’M = 1.58 x 10-3 M.
We know that, ionic product of water is
[H3O+][OH-] = 1 x 10-14
(1.58 x 10-3 ) [OH-] = 1 x 10-14.
[OH-] = 1 x 10-14/ (1.58 x 10-3 )
[OH-] = 6.32 x 10-12 M
==========================================
2) Determination of Ka of Unknown weak acid.
For, 1.00 M solution of a weak acid HA, the pH of the solution is found to be 3.85.
By definition of pH,
pH = -log10[H+] = -log10[H3O+]
i.e. -log10[H3O+] = pH
i.e. log10[H3O+] = -pH
Taking antilog to the base 10 of both side,
[H3O+] = Antilog(-pH) ……………(Or you can use exponent form as,[H3O+] = 10-pH)
[H3O+] = Antilog (-3.85)
[H3O+] = 1.413 x 10-4
From ICE table its clear that,
[A-] =[H3O+] = 1.413 x 10-4.
Here, 1M solution of weak acid used,
So,initially [HA] = 1 M
And at equilibrium,
[HA] = 1 – (1.413 x 10-4)
[HA] = 1.00 M …………..(Recall the approximation. And I think by now you understand why approximation works . See, here X = 1.413 X 10-4 << 1.
With these values now we can calculate Ka for acid,
We have,
Ka = [A-][H3O+]/[HA]
Ka = (1.413 x 10-4)( 1.413 x 10-4 )/(1)
Ka = 2.0 x 10-8.
Ka of unknown weak acid is = 2.0 x 10-8.
================================================================================
3)Moles of salt NaA = 1.00 and volume of water or solution formed = 2.00 L
Hence, Initial concentration of NaA salt = 1.00 / 2.00 = 0.50 M
Salt is strong electrolyte and dissociates completely.
I) Initial [A-] = [NaA] =0.50 M
================================
II) Determination of the value of Kb in this case :
We have given Ka = 2.0 x 10-5.
We know that,
Ka x Kb = Kw = 1. X 10-14
So, 2.0 x 10-5 x Kb = 1. X 10-14
Kb = (1. X 10-14)/(2.0 x 10-5)
Kb = 5 x 10-10.
Kb for this case is 5 x 10-10.
==============================================
Base association equilibrium below and the appropriate ICE table : Note again change in [H2O] negligible and [H3O+] we will find afterwards.
A- + H2O < -------- > HA + OH-
Initial
0.50M
-
0
0
Change
-X
-
+X
+X
Equilibrium
(0.50 –X) M
-
‘X’M
‘X’M
Kb = 5 x 10-10.
[HA] = [OH-] = ‘X’ M and [A-] = (0.50 – X) M = 0.50 M ………(small X approximation)
Now, expression for Kb,
Kb = [OH-][HA]/[A-]
Put all the values,
5 x 10-10 = (X)(X) / 0.50
X2 = 0.50 x 5 x 10-10
X2 = 2.5 x 10-10
X = 1.58 x 10-5
i.e. [OH-] = X =1.58 x 10-5.
Then,
pOH = -log10[OH-]
pOH = - log10(1.58 x 10-5)
pOH = 4.80
pH + pOH = 14
pH + 4.80 = 14
pH = 9.20
[Na+] in this case = [salt] = 0.5 M
Extremely sorry for last part that I could not finish because of time up. Sorry once again. Please try to remove theory part and instruction. Reading and understanding takes time.
Moles of weak acid (HA) = 0.250 and volume of solution = 2.00 L
Hence, Initial concentration of HA = 0.250 / 2.00 = 0.125 M
Equilibrium & ICE table: Acid is weak hence change in H2O will be negligible and hence not shown here. And [OH-] we can find from [H3O+] later on.
HA + H2O < -----> A- H3O+
Initial
0.125 M
-
0
0
Change
-X
-
+X
+X
Equilibrium
(0.125 – X)M
‘X’ M
‘X’ M
[HA]= (0.125-X) M, [A-]= ‘X’M, [H3O+] = ‘X’M, and Ka = 2.0 x 10-5.
We can write expression for Ka for weak acid HA as,
Ka = [H3O+][A]/[HA]
Put all known values,
2.0 x 10-5= (X)(X) / (0.125-X)
But as HA is weak acid. X < < 0.125 M
Hence approximation holds that, 0.125 – X ~ 0.125M.
So, above eqn. takes form,
2.0 x 10-5 = X2 /0.125
X2 = 2.0 x 10-5 x 0.125
X2 = 0.25 x 10-5
X = 1.58 x 10-3
Hence, equilibrium concentrations are,
[A-]= ‘X’M = 1.58 x 10-3 M
[H3O+] = ‘X’M = 1.58 x 10-3 M.
We know that, ionic product of water is
[H3O+][OH-] = 1 x 10-14
(1.58 x 10-3 ) [OH-] = 1 x 10-14.
[OH-] = 1 x 10-14/ (1.58 x 10-3 )
[OH-] = 6.32 x 10-12 M
==========================================
2) Determination of Ka of Unknown weak acid.
For, 1.00 M solution of a weak acid HA, the pH of the solution is found to be 3.85.
By definition of pH,
pH = -log10[H+] = -log10[H3O+]
i.e. -log10[H3O+] = pH
i.e. log10[H3O+] = -pH
Taking antilog to the base 10 of both side,
[H3O+] = Antilog(-pH) ……………(Or you can use exponent form as,[H3O+] = 10-pH)
[H3O+] = Antilog (-3.85)
[H3O+] = 1.413 x 10-4
From ICE table its clear that,
[A-] =[H3O+] = 1.413 x 10-4.
Here, 1M solution of weak acid used,
So,initially [HA] = 1 M
And at equilibrium,
[HA] = 1 – (1.413 x 10-4)
[HA] = 1.00 M …………..(Recall the approximation. And I think by now you understand why approximation works . See, here X = 1.413 X 10-4 << 1.
With these values now we can calculate Ka for acid,
We have,
Ka = [A-][H3O+]/[HA]
Ka = (1.413 x 10-4)( 1.413 x 10-4 )/(1)
Ka = 2.0 x 10-8.
Ka of unknown weak acid is = 2.0 x 10-8.
================================================================================
3)Moles of salt NaA = 1.00 and volume of water or solution formed = 2.00 L
Hence, Initial concentration of NaA salt = 1.00 / 2.00 = 0.50 M
Salt is strong electrolyte and dissociates completely.
I) Initial [A-] = [NaA] =0.50 M
================================
II) Determination of the value of Kb in this case :
We have given Ka = 2.0 x 10-5.
We know that,
Ka x Kb = Kw = 1. X 10-14
So, 2.0 x 10-5 x Kb = 1. X 10-14
Kb = (1. X 10-14)/(2.0 x 10-5)
Kb = 5 x 10-10.
Kb for this case is 5 x 10-10.
==============================================
Base association equilibrium below and the appropriate ICE table : Note again change in [H2O] negligible and [H3O+] we will find afterwards.
A- + H2O < -------- > HA + OH-
Initial
0.50M
-
0
0
Change
-X
-
+X
+X
Equilibrium
(0.50 –X) M
-
‘X’M
‘X’M
Kb = 5 x 10-10.
[HA] = [OH-] = ‘X’ M and [A-] = (0.50 – X) M = 0.50 M ………(small X approximation)
Now, expression for Kb,
Kb = [OH-][HA]/[A-]
Put all the values,
5 x 10-10 = (X)(X) / 0.50
X2 = 0.50 x 5 x 10-10
X2 = 2.5 x 10-10
X = 1.58 x 10-5
i.e. [OH-] = X =1.58 x 10-5.
Then,
pOH = -log10[OH-]
pOH = - log10(1.58 x 10-5)
pOH = 4.80
pH + pOH = 14
pH + 4.80 = 14
pH = 9.20
[Na+] in this case = [salt] = 0.5 M
Extremely sorry for last part that I could not finish because of time up. Sorry once again. Please try to remove theory part and instruction. Reading and understanding takes time.
Initial
0.125 M
-
0
0
Change
-X
-
+X
+X
Equilibrium
(0.125 – X)M
‘X’ M
‘X’ M
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