Consider the titration of a 40.0 mL of 0.145 M weak acid HA (Ka = 2.7 x 10⁻⁸) with 0.100 M LiOH. What would be the pH of the solution after that addition of 100.0 mL of LiOH?

Answer 1

First, it may help write out the balanced equation particularly if there isn't one-to-one mole ratios from ions to acid or base, but also because a strong acid/strong base titration is calculated differently:

HA + LiOH ⇔ LiA + H2O

*remember A stands for anion after dissociation*

Then, it may help to write the net ionic equation:

1. Write out the dissociated ions (LiA and LiOH are ionic compounds): HA + Li + OH ⇔ Li + A + H2O
2. Remove spectator ions (Li in this case) on both sides of the equation: HA + OH ⇔ A + H2O
3. Use mole-to-mole ratio in calculation--in this case you don't have to worry because it is 1-to-1.

Now, you should know that pH = pKa + log([base]/[acid]) and that pKa is -log[Ka]. So, we can find the pH if we know the Ka which is given and the [acid] and [base].

*The [ ] means molarity, so we need to get everything in molarity eventually.

1. Because the volume is changing during the titration (as the two solutions mix), we need to recalculate the molarity (moles/L or M) by finding the moles of each solution: weak acid (HA) = 40mL x 0.145mol/L and LiOH = 100mL x 0.1mol/L. **Look at the units: mL will not cancel when multiplied by mol/L so we need to convert mL to L first**. So, HA = 0.04L x 0.145mol/L = 0.0058mol. And, LiOH 0.1L x 0.1mol/L = 0.01mol.
2. Find the limiting reactant. *Remember, acids and bases neutralize each other to form water, and in this case, LiA. The LiA is limited by whichever has less moles between LiOH and HA: HA has less moles and is the limiting reactant.
3. So, we know 0.0058mol of HA (our weak acid) will react with 0.0058mol of LiOH. The leftover moles of LiOH can be calculated simply by subtracting the 0.0058mol that was neutralized from the 0.01mol of LiOH calculated earlier: 0.01 - 0.0058mol = 0.0042mol. In summary: we know we have 0.0058mol of LiOH and 0.0058mol of HA that reacted to form 0.0058mol of A (and 0.0058mol of water which we don't care about). And we know we have 0.0042mol LiOH.
4. Now the tricky part, that I just realized is: because we used up all of the acid, the equation: pH = pKa + log([base]/[acid]) will not work in this instance (like I had presumed because the question gives us the Ka). Our [acid] is 0. But we know that our leftover LiOH will dissociate into its ions, so 0.0042mol of LiOH will give us 0.0042mol of OH ions. In this case you need to know that pOH = -log[OH] and pH + pOH = 14, so once we find [OH] (which is in molarity), we can solve for pOH and then solve for pH. So 0.0042mol of OH divided by our new volume of solution (by adding the two volumes given in the problem) will give us our mol/L or [OH]: 0.0042mol/(40mL + 100mL) = 0.00003mol/mL. 0.00003mol/mL x 1000mL/L = 0.030mol/L. (I adjusted for sig figs in this answer)
5. Finally, we plug this in: pOH = -log[0.030] = 1.523 (sig figs change: see sig fig rules for logs)
6. pH + pOH = 14: pH = 14 - pOH: pH = 14 - 1.523 = 12.477

The answer is a pH of 12.477. You don't even need Ka to solve this problem because the acid gets used up so its dissociation in water doesn't apply. This is either one part of a sequence of questions, or it's a trick question by giving you the Ka.