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?
4 answers
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?
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?
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!