Use the henderson-hasselbalch equation to predict the change in pH when 0.50ml of 1M HCl is added to 40mL of .10M buffer at pH 6 and pH 7.5.
My two solutions are Na2HPO4 and NaH2PO4.pKa=7.2
Now I don't even know where to begin and how to use the equation.
This is a little involved. I'll pick 7.5 and go from there. First, you have two equations to solve simultaneously.
Here is how you get to equation 1.
pH = pKa + log b/a
7.5 = 7.2 + log b/a
0.3 = log b/a
b/a = 2 and b = 2a. Equation 1.
Technically, one should work with M; however, it is simpler, in my opinion, to work in millimoles and since the volume is the same in the solution for both base and acid the volume always cancels and the answer is the same no matter which way you go. I will do mmols. You have 40 mL x 0.1M = 4 mmols.
Then a + b = 4. This is equation 2.
Solve for a and b in mmols. You need to go through it but the answers are approx 2.7 for base(b) and 1.3 for acid(a). Then we add HCl to the buffer.
Then base + H^+ ==> acid
......HPO4^2- + H^+ ==> H2PO4^-
I.....2.7.......0........1.3
add.............0.5..........
C....-0.5......-0.5.......+0.5
E.....2.2........0........1.8
Then plug the E line into the HH equation and solve for the pH. Subtract from 7.5 to find the delta pH.
But doing this for 6 won't work out because you'll get a non real answer for the log....
Do you just take the absolute value for pH 6? @DrBob222
If you get a negative number it means there is more HCl added than the base can handle so that is excess HCl which is a strong acod. Then M HCl is millimols/mL and pH from there.
@Dr.Bob222 I guess I'm still not following what you're saying. Following the above method I got a=3.76 and b=.237. If I were to subtract the .5 I would get -.263 for HPO4 and 4.26 for H2PO4 (added obviously). I understand a negative number is showing excess HCl but I don't understand what you're talking about with M HCL mmols/mL and what to do from there.
Not to mention I also have to do this for 40mL of a .01M and all those will be negative following the same steps. I understand that 1M(mole/L) is the same as 1mmol/mL but I don't see how that helps as well. @DrBob222
Almost. The -0.263 (which I rounded to +0.24) is the excess HCl present (not HPO4^-). You have used all of the HPO4^- and the 0.24 mmols represents what is left of the 0.5 mmols HCl. So you need the pH of that M HCl.
Then M = mmols/mL = 0.24/40 = ? and convert to pH. Obviously this is not a buffer at these conditions.
(Note: 0.5 HCl - 0.24 H2PO4^- = 0.26 HCl left over).
So .24/40=.006 7.2+log[.006/4.26]=4.35 so that's a huge change in pH (1.47). Which makes sense because there's excess HCl.
expected pH of a substance. So
@DrBob222, would I follow this same format if I wanted to find predicted pH with the pH correction factors. for example, I would expect a solution I made with ideal pH 7.5 to have a pH of pH=7.2+log(2.2/1.8)+log(.45/.74)=7.07. Correction factors were given to us.
But now looking at pH 6, that doesn't make sense because the expected value of that would be 4.13 which is not even close to what we measured. Did I do something wrong again?