Ice table ? I haven't heard of that term in awhile.
Well since acetic acid is a weak acid and NaOH is a strong base...
120ml 0.15M acetic acid
30ml 0.2M NaOH
-You need Ka for the weak acid.
(for acetic acid Ka= 1.8x10^-5)
steps:
1. Determine if moles H+ or moles OH- is in excess
2. Determine ammount of excess.
3. Determine Molarity of excess ammount.
4. Solve for pH directly or by pOH id OH- is in excess
What is the pH of a solution of 120 ml 0.15M acetic acid to which we add 30mL 0.2M NaOH?
I have a bit of a hard time with this question. I tried to get mols of each one and then use it in an ice table but didn't work.
Can someone help? Thanks
8 answers
Write the equation.
Calculate mols acetic acid.
Calculate mols NaOH.
Subtract them and see which is in excess. One of them will be left over. Probably,(my guess) acetic acid is left over and all the NaOH is added. Then you know how much sodium acetate is formed and this becomes a buffered solution. Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log[(base)/(acid)]
Post your work if you get stuck and I can help you through it.
Calculate mols acetic acid.
Calculate mols NaOH.
Subtract them and see which is in excess. One of them will be left over. Probably,(my guess) acetic acid is left over and all the NaOH is added. Then you know how much sodium acetate is formed and this becomes a buffered solution. Use the Henderson-Hasselbalch equation to solve for pH.
pH = pKa + log[(base)/(acid)]
Post your work if you get stuck and I can help you through it.
Was what I posted incorrect Dr.Bob?
Step 1 was ok.
Step 2 was ok.
Step 3 is not correct because it forms a buffered solution and that can't be solved for pH directly; i.e., pH = -log(H^+) or pOH = -log(OH^-). The HH equation will do it OR the original Ka equation will solve for H^+ which can then be converted to pH.
Step 2 was ok.
Step 3 is not correct because it forms a buffered solution and that can't be solved for pH directly; i.e., pH = -log(H^+) or pOH = -log(OH^-). The HH equation will do it OR the original Ka equation will solve for H^+ which can then be converted to pH.
Answered above. The first two steps are ok. The third is not.
Um actually it says this in my gen chem handout...not using the HH equation.
after determining the M of excess amount:
H+ + OH- => H2O
shifts right b/c product is removed.
using the numbers given in mine ..
-0.02500mol of HAc are consumed
-0.02500mol of Ac- is produced
-0.05000 - 0.02500= 0.02500moles HAc remain
then it says:
since this is a buffer (excess HAc[in this problem] and the NaOH caused it's conjugate base Ac- to be produced) we can use the initial moles (reduced aproximately) as the equillibrium moles
[H+]= Ka* moles HAc/moles Ac-= 1.8x10^-5
I just posted the general steps without elaborating but it was under determining the moles of the excess ammount but I assuem this is the original Ka equation?
after determining the M of excess amount:
H+ + OH- => H2O
shifts right b/c product is removed.
using the numbers given in mine ..
-0.02500mol of HAc are consumed
-0.02500mol of Ac- is produced
-0.05000 - 0.02500= 0.02500moles HAc remain
then it says:
since this is a buffer (excess HAc[in this problem] and the NaOH caused it's conjugate base Ac- to be produced) we can use the initial moles (reduced aproximately) as the equillibrium moles
[H+]= Ka* moles HAc/moles Ac-= 1.8x10^-5
I just posted the general steps without elaborating but it was under determining the moles of the excess ammount but I assuem this is the original Ka equation?
1. Determine if moles H+ or moles OH- is in excess
2. Determine ammount of excess.
3. Determine Molarity of excess ammount.
4. Solve for pH directly or by pOH id OH- is in excess
I copied your original post above. I won't argue with you Christina for only you know what you meant; however,that isn't what you wrote. The implication in step 3 is that you have molarity of the acid or the molarity of the base and pH or pOH = -log(M) where M is either acid or base. It is also true that solving for (H^+) from the Ka expression gives (H^+) = Ka(acetic acid)/(acetate ion) and converting to pH will give the same answer. But that is where the H-H equation comes from and it gets the answer a little faster. Either equation is good.
2. Determine ammount of excess.
3. Determine Molarity of excess ammount.
4. Solve for pH directly or by pOH id OH- is in excess
I copied your original post above. I won't argue with you Christina for only you know what you meant; however,that isn't what you wrote. The implication in step 3 is that you have molarity of the acid or the molarity of the base and pH or pOH = -log(M) where M is either acid or base. It is also true that solving for (H^+) from the Ka expression gives (H^+) = Ka(acetic acid)/(acetate ion) and converting to pH will give the same answer. But that is where the H-H equation comes from and it gets the answer a little faster. Either equation is good.
Alright alright..I should be more specific next time or just let you answer it since I don't seem to be conveying it correctly. (I just copied what I had but I feel that the HH eqzn is easier to use than what I was first taught (this))
Thus the HH eqzn IS easier.
Thus the HH eqzn IS easier.
4 answers