Ka and pH stuff?

Question

A 0.010 M solution of aspirin, a weak monoprotic acid, has a pH of 3.3. What is the Ka of aspirin?


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Answer 1

Hey, I was just looking on this site and I saw your question. I thought this might be helpful and I apologize if it isn't:

For aspirin, Ka=3X 10-5 at 37 degrees C. There are 325 mg of aspirin in an aspirin tablet. If two aspirin tablets are dissolved in a volume of 1L in which the pH is 2, What percentage of the aspirin is in the form of molecules?

1) You can find the initial concentration of H+ in the solution from the pH.

pH = - log [H+]

2 = - log [H+]

[H+] = 0.01 M

2) Next you need to convert g aspirin to moles. The structure of aspirin is HC9H7O4. The molecular weight (molar mass) is 180.2 g/mol.

Convert 650 mg aspirin to moles. 650 mg = 0.650 g

0.650 g * 1 mol/180.2 g = 0.00361 moles

Since the total volume of the solution is 1 L, the concentration is also 0.00361 M

3) Write down the acid dissociation reaction for this compound.


HC9H7O4 ------------> H+ + C9H7O4-
4) Make a chart containing the initial amount, change and equilibrium amount of each substance.


HC9H7O4 ------------> H+ + C9H7O4-
initial 0.00361 M 0.01 M 0
Change - x + x + x
----------------------------------------------------------------------------------
equilibrium 0.00361 - x 0.01 + x x

The initial value of H+ came from step 1. Initial value for aspirin cam from step 2.

5) Set up equilibrium and solve for x.


[H+][C9H7O4-]
Ka = -------------------------
HC9H7O4


3 X 10-5 = (0.01 + x)(x)
--------------------
0.00361 - x

x = 1.083 X 10-5

To make this arithmetic easier you may assume that x that is added and subtracted in the expressions is negligible. That is x<<0.01 and x << .00361

The expression then becomes:


3X10-5 = (0.01)(x)
--------------
0.00361

Solving for x now is simple.

Now we know the number of moles of aspirin that dissociated.

6) Convert moles dissociated to g.

1.083X10-5 * 180.2 g/mol = .00195 g = 195 mg aspirin

650 mg - 195 mg = 455 mg not dissociated or remains in the form of molecules.

To find the percentage,

455/650 * 100 = 70%

Answer 2

NH4+ + H2O ==> H3O+ + NH3 for the reason that H3O+ is a robust acid and NH3 is a susceptible base, the acid results of H3O+ wins out and the answer is slightly acidic. The pH of 0.sixty 5 M NH4Cl is approximately 4.7.

Answer 3

Asprin <-> H+ + conjugate base-
Initial: .010 <-> 0 + 0
Change: -x <-> +x + +x
_____________________________
Equilibrium: .010-x <-> x + x

since you are give the pH, you can calculate the H+ which is you x in the table above

10^-pH=H+
10^-3.3=5.01e-4

Ka=concentration of products/reactants
Ka=(5.01e-4)(5.01e-4)/(.010-5.01e-4)
Ka=2.51e-5

Answer 4

Put 10 to the power of -3.3 to get [H+].

Then square it, and divide by 0.01.

Answer 5

Let's use the general formula for monoproic acids HA

.. .. .. .. .. .. .. HA <=> H+ + A-
Initial .. .. .. 0.01
Dissociate .. x
Produce .. .. .. .. .. .. .. x .. .. x
At Equil. .. 0.01-x .. .. x .. .. x

Ka = [H+][A-] / [HA] =x^2/(0.01-x)

but x=[H+]
and [H+]= 10^-pH =10^-3.3 thus

Ka= ((10^-3.3)^2) / (0.01-10^-3.3) =2.64*10^-5

Answer 6

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