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Let's say I take $\pu{80 g}$ of a weak base, dilute it with $\pu{50 ml}$ of water and titrate it with a strong acid. I get a titration curve. Now I take again $\pu{80 g}$ of the same weak base but this time I dilute it with $\pu{200 ml}$ of water and titrate it with the strong acid. What will be the new titration curve compared to the first one?

And here is what confuses me:

On one hand, the diluted base should have lower $\mathrm{pH}$. But on the other hand, at half-equivalence point, $K_\mathrm{b}=\mathrm{pOH}$, and since it's the same base, $K_\mathrm{b}$ is the same, and thus $\mathrm{pOH}$ and $\mathrm{pH}$ at half equivalence point should be the same. I am also not sure about the endpoint.

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  • $\begingroup$ Welcome to Chemistry.SE. We’d like you to take the Tour to familiarize with the site. $\endgroup$ – Mathew Mahindaratne Apr 15 '18 at 19:36
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The equivalence is, by definition, the moment when $n_{base}=n_{acid}$. As you titrate the same mass of base (consequently the same quantity), your equivalence will be reached at the same point. The only difference, to me, between the two curves will be the fact that you will less see the equivalence when the solution is diluted. First curve is 80g in 50mL, second one is 80g in 200mL.

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  • $\begingroup$ Thanks a lot! But what about the pH at the begining? is it the same? On one hand, the diluted base should have lower pH. But on the other hand, at half equivalence point, Kb=pOH, and since it's the same base, Kb is the same, and thus pOH and pH at half equivalence point should be the same. $\endgroup$ – Shany Apr 17 '18 at 19:25
  • $\begingroup$ your initial pH will change, as it will get closer of the one of water. But keep in mind that the equivalence depends only of $n_{acid}$. Hence, $n_{acid}=n_{base} \Leftrightarrow C_{acid}V_0=C_{base}V_{eq}$. So between your two titrations, pOH at equivalence and half-eq will remain the same, but volumes will change (as the quantity of base to reach the required pOH will change) $\endgroup$ – Thomas Prévost Apr 17 '18 at 19:44
  • $\begingroup$ Thank you, so I understand that except the lower initial ph of the diluted base, other points should be pretty much the same (half equivalence point, volume untill equivalence point, ph at equivalence point). Did I get it straight? $\endgroup$ – Shany Apr 25 '18 at 5:52
  • $\begingroup$ That’s pretty much it, but your volumes won’t be the same, as you start from a different pH, and concentration. The quantities will be the same, but not volumes (they will change, as you don’t have the same concentrations of base). But all other values will remain unchanged. $\endgroup$ – Thomas Prévost Apr 25 '18 at 5:59
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At the beginning of the titration, the solution is a weak base. At the equivalence point, the solution is a weak acid (with some more spectator ions). The pH of these solutions changes with dilution, both in the direction of neutral pH.

In contrast, at the half-equivalence point (or midpoint) of the titration, the solution is a 1:1 buffer. As long as the buffer concentration is large compared to the concentration of hydronium and hydroxide ions, the pH will be close to the pKa (and the pOH close to the pKb), irrespective of dilution.

So if you dilute the weak base with water before the titration, nothing will change along the x-axis (i.e. half-equivalence point and equivalence point will occur at the same volume of added strong acid). Along the y-axis (where you plot the pH), the curve will "shrink" toward the pH of the buffer (i.e. the starting pH will be lower - less basic - and the pH at the equivalence point will be higher - less acidic). Past the equivalence point, the pH is controlled by the excess of strong acid. Because the strong acid is diluted into a greater volume if you add water to the weak base before starting the titration, this part of the titration curve will also show higher - less acidic - pH values.

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I am first going to do the ‘flip side’ of the OP’s scenario and then basically flip it back. So we start with a 0.1 M aqueous solution of benzoic acid and the initial solution volume is 25 mL. Benzoic acid is a monoprotic weak acid with $\mathrm{p}K_\mathrm{a} = 4.20$. The initial pH is 2.606. The strong base used for the titration is 0.1 M NaOH. At the equivalence point, the pH = 8.454 and the solution volume is 50 mL (= 25 mL benzoic acid solution + 25 mL NaOH solution). Thus, the pH is simply the pH of 0.05 M sodium benzoate solution. After all 50 mL of NaOH has been added, the pH = 12.523 and the final solution volume is 75 mL. At the half-equivalence point, 12.5 mL of NaOH has been added and pH = 4.185.

Now assume the initial benzoic acid solution is only 0.025 M, but the initial volume is 100 mL. The total number of moles of benzoic acid is the same as before. The initial pH is 2.912: the benzoic acid solution is only 25% as concentrated. At the equivalence point, the pH = 8.254 and the solution volume is 125 mL (= 100 mL benzoic acid solution + 25 mL NaOH solution). The equivalence point pH is closer to pH = 7 because the solution is simply a 0.02 M sodium benzoate solution, i.e., the number of moles of sodium benzoate is now in a volume that is 2.5 times larger, so the concentration is 40% of 0.05 M. After all 50 mL of NaOH has been added, the pH = 12.222 and the final solution volume is 150 mL. At the half-equivalence point, 12.5 mL of MaOH has been added and pH = 4.188, moving slightly up toward pH = 7. See the titration curves below:

Weak acid titrations

Now flipping back: the OP’s scenario involving a weak base titrated with strong acid. For reasons that will become clear, assume the hypothetical weak base has $\mathrm{p}K_\mathrm{b} = 9.80$, and that the strong acid is 0.1 M HCl solution. Note that the $\mathrm{p}K_\mathrm{b}$ is simply 14 (treated as exact throughout) - 4.20.

As before, start by assuming 25 mL of 0.1 M weak base. The initial pH = 14 – 2.606 = 11.394. At the equivalence point, the pH = 5.546 and the solution volume is 50 mL (= 25 mL weak base solution + 25 mL HCl solution). Thus, the pH is simply the pH of a 0.05 M weak base salt solution. After all 50 mL of HCl has been added, the pH = 1.477 and the final solution volume is 75 mL. At the half-equivalence point, 12.5 mL of HCl has been added and pH = 9.815.

Now assume the initial weak base solution is only 0.025 M, but the initial volume is 100 mL. The total number of moles of weak base is the same as before. The initial pH is 11.088: the weak base solution is only 25% as concentrated. At the equivalence point, the pH = 5.746 and the solution volume is 125 mL (= 100 mL weak base solution + 25 mL HCl solution). The equivalence point pH is closer to pH = 7 because the solution is simply 0.02 M weak base salt solution, i.e., the number of moles of weak base salt is now in a volume that is 2.5 times larger, so the concentration is 40% of 0.05 M. After all 50 mL of HCl has been added, the pH = 1.778 and the final solution volume is 150 mL. At the half-equivalence point, 12.5 mL of HCl has been added and pH = 9.812, moving slightly down toward pH = 7. For the titration curves, simply imagine vertically flipping the titration curves, in the figure above, about a horizontal line at pH = 7. Or just look at the plot:

Weak base titrations

One last thing: The pH values at half-equivalences were close to the respective $\mathrm{p}K$ values, as expected, but any buffer can be broken by dumping it into swimming pool full of water. So dilution may matter a great deal, or relatively little, depending on the specifics.

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