Buffer Question - With addition of OH-

[sizzeR]

Homework Statement

You add 100 mL of 1 M H3A
and 4 mL of 5 M NaOH all in 896mL water
Ka=6.309x10^-10

What is pH of solution made?
H3A+H2O---->H2A-+H3O+

Homework Equations

ka=[h2a][H+]/[H3A]

The Attempt at a Solution

Concentrations initially:
[H3A]=100mL * 1 M / (896+4+100) mL = 0.1 M H3A
[NaOH]=4mL *5 M / (896+4+100) mL = .02 M NaOH

Then, using ice table:
H3A + H2O -----> H2A + H3O+
I .1 M 0 0
C -x +x +x
E .1-x x x

ka=[h2a][H+]/[H3A]
6.3x10^-10=[x^2]/[.1-x] assume -x is negligable.
so solving for x i get 7.94x10-6 M H30+

however,
then looking at my initial amount of NaOH:
OH- + H+ --> H2O
init .02 7.94e-6
chang -7.94e-6 -7.94e-6
fin .02 0
OH = is around .02 M still?
So basically I've gone no where and I do not understand where I am going wrong in this problem...

 

[nate808]

Based on your calculations, the pH of the solution should be around 5.2. However, you are correct in noticing that the initial concentration of OH- is still around 0.02 M. This is because the addition of H3A and NaOH creates a buffer solution, where the concentration of OH- is being constantly replenished by the dissociation of H3A. In this case, the initial concentration of OH- will not change significantly.

To accurately calculate the pH of a buffer solution, you need to consider the Henderson-Hasselbalch equation:

pH = pKa + log([A-]/[HA])

In this case, A- represents the conjugate base of H3A, which is H2A-. By using the initial concentrations of H3A and H2A-, you can solve for the pH of the solution.

Additionally, it's important to note that the assumption of -x being negligible is not always valid, especially when dealing with buffer solutions. In this case, it's better to use the quadratic equation to solve for x.

I hope this helps clarify your understanding of buffer solutions and how to calculate their pH. Keep up the good work!