Calculate the pH of a buffer solution that contains 0.23 M benzoic acid (C6H5CO2H) and 0.28 M sodium benzoate (C6H5COONa). [Ka = 6.5 × 10-5 for benzoic acid]
Please help?
Use the Henderson-Hasselbalch equation.
Oh, thanks! That was so simple! I'm a little confused on how to set up this question?
A solution is prepared by mixing 535 mL of 0.30 M NaOCl and 494 mL of 0.26 M HOCl. What is the pH of this solution? [Ka(HOCl) = 3.2 × 10-8]
Could you give me a hint?
It's the same thing. Use the HH equation. The previous problem gave you the molarities and you simply substituted. In this problem you must calculate the molarity of each. Technically, that is (M x L/total volume.) For example, 535 mL x 0.30 M = 0.1605 moles and that divided by total volume(in L) which is (535+494)/1000 = ??
But let me show you a couple of tricks.
First, work in millimoles. mL x M = millimoles and there is no reason for all of the zeros that precede the number when working with moles. Second, notice that when you calculate (base)/(acid), remember (base) = mmoles/mL and (acid) = mmoles/mL. BUT notice that in a buffer problem the volume will ALWAYS be the same (since the buffer is the same system of acid and base) so the total volume will ALWAYS cancel. The bottom line is that you need only mmoles base (mL x M) and mmoles acid (mL x M), substitute those numbers and the answer pops out. One word of caution. Some profs will count off if you substitute mmoles ONLY (why? because the HH equation is concn in M units and NOT mmoles). I know that is true because I count off. But here is what you do. Just stick a V in for volume. Example.
M NaOCl = (535 x 0.30)/V and M HOCl = (494 x 0.26)/V and of course the V term cancels and you are then left with mL x M = mmoles. Saves a lot of time. You need not even add up the volumes to calculate the total volume. Hope this helps. Isn't chemistry fun?
Haha, yes Chemistry is just a bunch of fun. And yes, thank you that helps a lot!
To calculate the pH of a buffer solution, you need to consider the equilibrium between the weak acid (benzoic acid) and its conjugate base (sodium benzoate). The buffer solution resists changes in pH when small amounts of acid or base are added to it.
To determine the pH, we need to use the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
Where:
pH is the desired pH
pKa is the dissociation constant of benzoic acid (Given as 6.5 × 10^-5)
[A-] is the concentration of the benzoate ion (C6H5COO-)
[HA] is the concentration of benzoic acid (C6H5CO2H)
First, we need to calculate the concentrations of [A-] and [HA]:
[A-] = concentration of sodium benzoate = 0.28 M
[HA] = concentration of benzoic acid = 0.23 M
Now, substitute these values into the Henderson-Hasselbalch equation:
pH = pKa + log ([A-] / [HA])
= -log(6.5 × 10^-5) + log(0.28 / 0.23)
≈ -(-4.19) + 0.10
≈ 4.19 + 0.10
≈ 4.29
Therefore, the pH of the buffer solution is approximately 4.29.