#3
a)What is the pH when 1.25L of a 0.200 M solution of the strong electrolyte NaH2PO4 is mixed with 1.00 L of a .150M solution of the strong electrolyte Na2HPO4?
b)What is the pH if the volume of the solution in problem #3a is doubled?
c) What is the pH if 200mL of .100 M HNO3 is added to the solution in problem #3a?
d)What is the pH if 350mL of 0.120 M KOH is added to the solution in problem #3a?
I cant seem to start the problem! Are these deprotonations of one another or are they different? How do you mix 2 weak acids to find the pH?? Please help or provid some insight; it will be greatly appreciated. Thank You.
I suggest you use the Henderson-Hasslebalch equation.
pH = pKa + log [(base)/(acid)]
Use the pK for the acid in the mixture.
part b, looking at base/acid concentration---(base) = mols/volume and (acid) = mols/volume. The volume cancels so ......
parts c and d. Look at the mols HNO3 being added 0.1 x 200 = 20 millimols versus the millimols of the salts. Same for millimols KOH in part d. 350 x 0.12 M = 42 millimols KOH versus the millimols of the salts.
Check my thinking.
For both a & b I got 6.99 being the pH which I checked my answers in my book & they are correct! However, I tried to obtain c & d and I cant get there. I found the moles of each the HNO3 and found the moles of previous solution by 10^-6.99 to get me the molarity and then I found the moles by using the volume. I subtracted the limiting reagent and placed that over the total volume and obtained .0082 M as the molarity. I find the pH but it's wrong. If you wouldnt mind pointing me in the right direction please & Thanks!
I would do this.
The HNO3 is a strong acid. It will react with the base, the HPO4^-2.
HPO4^-2+ H^+ ==> H2PO4^-
So you have 0.15 x 1.0 L = 0.15 mols HPO4^-2 and you add 0.2L x 0.1M HNO3 to it. It will subtract 0.02 mols from the HPO4^-2 (the base) you started with and add 0.02 mols to the H2PO4^- (the acid) you started with. Adjust those values and redo the H-H equation and that should do it. The same process for the KOH but react it with the acid. Let me know how you come out. The final pH for the addition of the HNO3 should be SLIGHTLY more acidic and the addition of the KOH should be SLIGHTLY more basic than the 6.99. By the way, what value are you using for k2? I have an OLD OLD book here that lists k2 as 6.34E-8. I'm just wondering how much it may have changed in the last umpteen years. Not much because I obtained 6.98 for the pH of a and b. And you can save some trouble by simply using V for the volume so concn (base) = mols/V and (acid) = mols/V. But ALWAYS put the V in there (or always put the actual volume in there) even though you don't go through the actual step of calculating the REAL concentration. WHY? I always counted off if the student put mols/mols because that isn't what the equation says; i.e., it says concn/concn and without the V or the actual value of volume, it isn't concn. The fact that the V, whatever it is, cancels is beside the point. Its the PROCESS that is important as much as the answer.
Thanks so much. I ended up getting 6.89 and 7.17! Which is about right. My K value I have listed is: 6.2E-8; close to yours. Thanks again & have a good night!
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