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# Titration of a weak base with a strong acid

In a weak base-strong acid titration, the acid and base will react to form an acidic solution. A conjugate acid will be produced during the titration, which then reacts with water to form hydronium ions. This results in a solution with a pH lower than 7. An example of this is the titration of hydrochloric acid (strong acid) into ammonia (weak base), which forms the conjugate acid ammonium and produces an acidic solution. Created by Jay.

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• How do we know in the beginning when doing the ICE procedure whether to use the molarity or the moles??
• It is always okay to use either molarity or moles in an ICE table. The only catch is that if you use molarity, you must use concentrations based on the entire mixed volume, not necessarily just the concentration of the solution that was added. For example, if you mixed 100 mL (=0.100 L) of 2.0 M CH3COOH and 100 mL (=0.100 L) of 1.0 M NaOH, then you could:

(1) calculate moles and use them in your ICE table, so you'd have 0.20 mol CH3COOH and 0.10 mol NaOH as your initial values OR

(2) use molarity in your ICE table, but you'd first need to calculate the concentration of the mixture. So, the [CH3COOH]=(2.0 M)(0.100 L)/(0.100 L + 0.100 L)=1.0 M and [NaOH]=(1.0 M)(0.100 L)/(0.100 L + 0.100 L)=0.05 M
• at will the log[A+]/[HA] always be 0 for half equivalence?
• Strong acid - strong base reactions do not have half-equivalence points nor do they have buffer regions.
• At , it's said that HCl is the same thing as H30+, how so?
• HCl is a strong acid, and therefore dissociates completely in an aqueous environment. In other words, all the H+ and Cl- break apart. H+ is like H3O+, because it is pretty much that hydrogen ion attached to a water molecule.
• hi, i was just wondering the initial question: what is the pH of the solution prior to adding the base, we didn't need to find the m of NH3. We just substituted the 0.100M into the ICE equation. However in the second part where acid is added we needed to find the m of HCL. I am not really clear as to why we did it for one but not the other....? Thanks
• At the beginning, all you had was NH₃, so you didn't have to consider the HCl.
After you added some HCl, it reacted with some of the NH₃. You had to calculate how much NH₃ was removed in order to get the pH of the remaining NH₃.
• In question (b) where we are concerned with NH4+ and NH3 doesn't finding the pOH seem more logical as opposed to finding the pH? We have a cation and the Henderson Hasselbach equation used in the video is that for anionic hydrolysis right?
• You can use either, both should give you the correct answer.
• At when Jay took the -log of Ka why did he round to 3 sig fig? I got 9.3 would this make my answer wrong?
• The digits of pH that are before the decimal point are orders of magnitude (constant), so there are only two sig figs present after the decimal point. Only the units after the decimal point count for sig figs. ex: pH=9.25 9=order of magnitude .25=2 sig figs.
• At Sal mentions how the reaction goes to completion. Since ammonia is a weak base, shouldn't this reaction be an equilibrium reaction that has the double arrow?
(1 vote)
• Take a look at both the reactants and the products.
Ammonia is a stronger base than water as it can accept a proton a bit more easier. Therefore ammonium ion is a weaker conjugate acid than hydronium ion. Strong acids/bases always react with weak base/acids till completion.
The reaction favors the side that produces the weaker conjugate acid/base as it is both thermodynamically and kinetically stable
Therefore the HCl pushes the reaction to completion.
This implies that an equilibrium arrow cannot be used as the reaction is pushed to completion.
• Can the sum of pH and pOH exceed 14?
Thanks :D
(1 vote)
• The generally taught/simple answer is no, the sum of pH and pOH is always 14. This is likely all that you will be tested on at anything below an upper level university chemistry class. However, the REAL answer is that temperature does alter the sum of (pH+pOH). At 25 C the sum of pH and pOH is roughly equal to 14.00, however as the temperature deviates (either up or down) the relationship changes. Think of it as a reaction equilibrium, as that is exactly what it is. If you increase the temperature, you can drive a reaction towards whichever side of the reaction repents a decrease in thermal energy. Here is a table for the pKw of water for reference: https://en.wikipedia.org/wiki/Self-ionization_of_water.
• If x represents molarity rather than number of moles, why did we have to divide the number of moles by the volume to get the molarity in part b?
(1 vote)
• Part a of the problem used an ICE table to solve for the hydroxide concentration and get the pH. ICE tables use molarity as their entries hence 'x' represents a molarity here. Here an ICE table is needed because we only have ammonia in solution which is a weak base.

Part b now includes the addition of a strong acid to neutralize the base. Here we don't use an ICE and instead need to know how many moles of base were neutralized by the additional acid. But once you have done that you realize you have an amount of the original base plus its conjugate acid which is a buffer solution. A buffer solution's pH can be solved using the Henderson–Hasselbalch equation which requires the molarity of both the base and conjugate acid hence the need to turn moles of each into molarity by dividing by the total volume. Hope that helps.