#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!
You're welcome! I'm glad I could help. And it seems like you're on the right track with your calculations. Remember, pH values are always approximate since they can be affected by many factors. Keep up the good work and have a great night too!
Great job on getting the correct values for the pH in parts a and b!
For parts c and d, you need to consider the reactions between the added solutions (HNO3 for part c and KOH for part d) and the existing solutions from part a.
In part c, you mentioned that you calculated the moles of HNO3 and the moles of the previous solution using the 10^-6.99 value. However, the moles of the previous solution should be calculated based on the initial concentrations and volumes given in problem #3a.
To solve part c, start by calculating the moles of HNO3 added (0.1 M x 0.2 L = 0.02 mol). Then, consider the reaction between HNO3 and Na2HPO4:
HNO3 + Na2HPO4 -> NaH2PO4 + H2O
This reaction will consume 0.02 mol of Na2HPO4 and form 0.02 mol of NaH2PO4. Adjust the moles of Na2HPO4 and NaH2PO4 accordingly. Then, use the Henderson-Hasselbalch equation with the adjusted concentrations to calculate the pH.
For part d, follow a similar approach. Calculate the moles of KOH added (0.12 M x 0.35 L = 0.042 mol). Then, consider the reaction between KOH and NaH2PO4:
KOH + NaH2PO4 -> Na2HPO4 + H2O
This reaction will consume 0.042 mol of NaH2PO4 and form 0.042 mol of Na2HPO4. Adjust the moles of NaH2PO4 and Na2HPO4 accordingly. Use the adjusted concentrations in the Henderson-Hasselbalch equation to calculate the pH.
It seems like you are on the right track! Just make sure to adjust the moles of the base and acid based on the reactions and recalculate the concentrations for using the Henderson-Hasselbalch equation. Let me know if you have any further questions!