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# Titration of a weak base with a strong acid (continued)

Calculating the pH for titration of weak base, ammonia, with strong acid, HCl, at the equivalence point and past the equivalence point. Created by Jay.

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• In the last part of this problem (d), there will also be formation of 0.004 moles of ammonium ion ( after the equivalence point ) like in part (c).Why didn't we consider the reaction of ammonium ion with water, that I think will afect the pH ?? •  You're right. The reason you can ignore the fact that the ammonium exists in solution is that it is relatively a much weaker acid than the HCl. The HCl will completely dissociate, compared to only a very small portion of the ammonium, so the HCl will be much more responsible for the pH of the solution.
• In part a) we only did the ICE and find pH
In part b) we did the ICE and then did the HH because it is a buffer solution
In part c) we did the ICE and then did it again for NH4 to find the pH using Ka
In part d) we only found the mol of HCl and subtracted NH3 from HCl and divided it by .1 liters to find the M of H30+ to calculate pH.
Are the procedure all different because of the difference in liters compared to NH3? If so do we use d) to solve 80ml? • The procedures are different because they are all different chemical situations.
In part a), you have a solution of a weak base, so you must use an ICE table.
In part b), you use HH because you have a buffer, a solution of a weak base and its conjugate acid.
In part c), you use ICE again because you have a solution of a weak acid.
In part d) you have a solution of a strong acid.
You must use the total volume in each case to calculate the concentrations of the components.

After adding 80.0 mL of 0.100 mol/L HCl, you have in solution (0.00800 -0.00400) mol = 0.00400 mol of H₃O⁺.
The total volume of the solution is (40.0 + 80.0) mL = 120.0 mL = 0.1200 L.
[H₃O⁺] = 0.00400 mol/0.1200 L = 0.0333 mol/L
pH = -log [H₃O⁺] = -log(0.0333) = 1.48
• at the beginning of all these problems you equation HCL to H30. how and why are we able to do this? my biggest problem with titrations is just setting up the equation. • Why do we go through so much work in the previous questions b and c and then only need to do so little in the last one? • Inception like question here...

The first part of the video shows how NH3 + HCL -> NH4+ + Cl- in aqueous solution

After this reaction, 0.002 moles of 0.1M HCl (H3O+) remains, as well as 0.004 moles of NH4+

In previous videos we saw how the product from a similar reaction, NH4+, itself reacts with H2O to also form H3O+ ions... If we follow this same logic through, when solving for this reaction by using a previously given Ka value, we solve for incremental H3O+ moles = 4.73*10^-7

So is it fair to say that the total number of H3O+ moles with which we solve for pH are in reality 0.002 mol + 4.73*10^-7 mol which in effect equals ~0.002 mol given how small the latter value is?

The pH of 0.002 mol of H3O+ ions = 1.699 (rough rounding)
The pH of ~0.002 mol of H3O+ ions = 1.698 (rough rounding, but lower pH value...)

Is it correct to say then that because the incremental H3O+ mols are so negligent, we overlook these moles as well as the (ever smaller) products from continued reactions that will lead to additional H3O+ moles?

(NH3 reacts with a strong acid that produces NH4+ which in turn reacts with H2O weak base that produces NH3, that itself will react with weak H2O acid that will produce NH4+, etc., etc. etc... approaching 0) ?

If so, does a titration curve behave as if it were approaching an asymptote as the reactions within a solution never fully cease to occur?

I really hope this makes sense! I'm going nuts over here. • At , we calculate the total moles of NH4 given off by the reaction assuming it starts with 0 moles of NH4. Although, wouldn't we start with more than 0 moles because the NH3 molecule would have been disassociating before we even added the HCl? • At But x also represents NH3 why should we not take it as the concentration of NH3 ions ? • Can something that is an acid in one solution behave as a base in another solution (and vice verse for bases)?
(1 vote) • since this is a titration of a weak base and a strong acid, I assume ammonia is a weak base; doesn't that mean that its conjugate acid, NH4+, is a strong acid, therefore having a complete dissociation in part c?
(1 vote) • Nope!

While a strong acid always has a (very) weak conjugate base, moderately weak acids have, moderately weak conjugate bases (and vice versa). This is true for NH4+ and its conjugate base NH3 and will always be true for an acid-base pair that constitutes a buffer.

Remember that a strong acid (or base) more-or-less completely dissociates in water. This implies that its conjugate base (or acid) more-or-less completely fails to go in the reverse direction. In practice the conjugate base (or acid) is not really acting as a base (or acid) at all!

In contrast, moderately weak acids and bases undergo partial dissociation in water and by necessity their conjugate base or acid must also undergo partial dissociation.
• Why is the value of X negligible?
(1 vote) • x can be negligible only in comparison with something else.
It is not negligible by itself.
For example, if you had \$100, you might consider \$5 to be a negligible amount.
If you had only \$5, \$5 would not be negligible.
A common criterion is that a quantity is negligible if it is less than 5 % of something else.

Most weak bases dissociate by only less than 5%, so the initial concentration usually decreases by less than 5 %.
x is negligible only in comparison with the initial concentration.