How would you find the concentration M, when given pH and Ka?

Question

Ka for a given weak acid is 9.24 x 100. What concentration of acid (mM/L) will give a pH of 5.4?

How would you solve this problem?

Thank you for any help.

oops, it is supposed to be 9.24 x 10^0 not 100.......sorry.

Answer 1

Assuming this is a monoprotic acid (i.e., an acid that can only donate one proton to the solution per molecule of acid), and writing the formula for the acid as HX, where X is the conjugate base, then use the definition of Ka to write:

(eq 1): Ka = [H+]*[X-]/[HX]

where the brackets, [] mean "concentration" (in moles/liter), and [HX] is the concentration of undissociated acid in solution at equilibrium.

We also know the dissociation constant of water :

(eq 2): 10^-14 = [H+]*[OH-]

The solution must be electrically neutral. This means that the concentration of positive ions must be balanced by an equal concentration of negative ions:

(eq 3): [H+] = [X-] + [OH-]

Finally, we know that the total concentration, C, of HX acid is equal to the amount of undissociated HX in solution plus the amount that has dissociated. Because each molecule of HX that dissociates produces one ion of X-, the concentration of X- tells us how much of the HX has dissociated. We can therefore write:

(eq 4): C = [HX] + [X-]

Finally, we are told that we want a solution that has a pH of 5.4. From the definition of pH, we have:

pH = 5.4 = -log[H+]
which can be rewritten as:

(eq 5): [H+] = 10^-5.4

This gives us 5 equations in 5 unknowns (C, [H+], [OH-], [X-], and [HX]).

To solve this system of equations, we start eliminating variables. First, rewrite equation 1 as:

[HX] = [H+][X-]/Ka

and rewrite equation 3 as:

[X-] = [H+] - [OH-]

Substitute these two equations into equation 4 to eliminate [X-] and [HX]:

(eq 5*): C = [H+]*([H+] - [OH-])/Ka + [H+] - [OH-]

Now rewrite equation 2 as:

[OH-] = 10^-14/[H+]

and substitute this into equation 5* to eliminate [OH-]:


(eq 5**): C = ([H+]^2 - 10^-14)/Ka + [H+] - 10^-14/[H+]


Finally, substitute equation 5 into the above equation to get rid of [H+]:

C = (10^-10.8 - 10^-14)/Ka + 10^-5.4 - 10^-14/(10^-5.4)
C = ((1.584*10^-11)/Ka +3.979*10^-6) mol/liter

or, in mM/liter:

C = ((1.584*10^-8)/Ka + 3.979*10^-3) mM/liter

I don't understand the value for Ka that you have written. As written, Ka = 924, which is not the dissociation constant for a weak acid (it's too big). If Ka were really this large, essentially all the acid would be dissociated, and the first term in the equation above is negligible relative to the second term. Weak acids typically have a Ka < 1

You can plug in the proper value for Ka into the above equation and calculate the concentration yourself.

Answer 2

More details needed

Answer 3

So amazed that I found this question already answered! it is like you've read my mind!