What is the pH of this solution?

Question

What is the pH of a 2M solution of uric acid?

HC5H3N4O3 <-> H+ + C5H3N4O3-

with Ka = 1.29 x 10^-4.

I need to know the EXACT pH. Not whether it's between 0 and 7. I already know it's between 0 and 7...

Answer 1

Ka is defined as the ratio [H+][R-]/[HR]
where HR is the un-ionized form of the acid and R is the ionized form, and [X] stands for "concentration of X"

Also, since the total solution has no net charge, [H+] = [R-] (the positive and negative ions have equal concentrations)

Finally, the total concentration of the acid is [R-] + [HR], which in this case is given as 2 moles. You can then write
[R-] = 2 -[RH].

Combining these equations with the known value for Ka you get:

1.29 x 10^-4 = [H+][H+]/(2 - [H+])

This gives the quadratic [H+]^2 + Ka[H+] - 2Ka = 0.

However, a useful shortcut is to recognize that [H+] is usually very small - remember that pH = -log[H+]. So [H+] << 2 and we can write:

1.29 x 10^-4 = [H+][H+]/(2)
or in general terms Ka = [H+]^2/Ctotal

Then we solve for [H+] = sqrt(2*1.29 x 10^-4) = 0.0161

and pH = -log10([H+]) = - log10(.0161) = 1.79 or in round numbers, pH = 1.8.

Answer 2

Ka = [H+][base anion]/[unionized acid]

First lets get a rough idea by assuming that the total amount ionized is so small that it does not effect the concentration of the uric acid, so we keep uric acid concentration at 2 M, this way we avoid using quadratic equations.

1.29 x 10^-4 = [H+][base anion]/[2 Molar], note that the concentration of the h+ and the base anion are the same so we can rewrite as follows:

1.29x10^-4 = [H+]^2/[2M]

Solve for [H+] = sqrt(2*1.29*10^-4) = 0.016 molar

Is 0.016 enough to really change the calculated value of Ka to justify the messy quadratic equation, lets check

(0.016)(0.016)/[2-0.016] = ?
? = 1.29x10^-5, which is the true value of the Ka.

So, we can go with this approach and [h+] = 0.016 molar

If the calculated value of Ka was off very much we would resort to the quadratic equation below.

1.29*10^-4 = [x]^2/(2-x) which implies

x^2 - (2-x)*1.29*10^-4 = 0, solve for x

x^2 +(1.29*10^-4)x - 2.58*10^-4 = 0
x = 0.016, now lets check this answer for errors

Ka = ? = (1.6*10^-2)^2/[2 - 1.6*10^-2] = 1.29 * 10^-4
Looks like it is close enough to the actual Ka value.

So, the [H+] is 1.6*10^-2

ph then is -log[1.6*10-2] = 1.79

please check for errors.

Answer 3

This is a little tricky, but not impossible. The Ka=[H][urate]/[uric acid], and remeber that for as much uric acid dissiocates you get just as much Hydronium and urate. So just set up an algebra equation (make the ammount made/lost "x").

1.29*10^-4 = x^2 / 2-x ; now, we can assume that 2>>x, making the equation

1.29*10^-4 = x^2/2 (multiply the 2 on both sides)
2.58*10^-4 = x^2 (find the square root of both sides)
x = 0.016062 (keep two extra digits to avoid rounding errors, 5sf)

Check your assumption (that 2 is way bigger than x) 2/0.016062=124.5, the assumption is good enough for general chem

Now, the H+ concentration = 0.016062 so take the negative log of that number to get pH.

pH=1.79

Answer 4

more than 0 but less than 7

Answer 5

Acidic so more than 7.0 pH

Answer 6

is a acid base solution